# The Slippery Eel of Probability

How do you solve probability problems that appear to have more than one correct answer?

Olena Shmahalo/Quanta Magazine

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In school, we are trained to think that math problems always have one correct answer. But this is not necessarily true for problems dealing with probability, if the method used to reach the described situation is not fully specified. Surprisingly, the same problem can then have many different answers, all apparently equally valid. Take, for example, our new puzzle:

An art collector who loves pictures of sea creatures, especially eels (there’s no accounting for taste!), commissions an art project. In his living room, he arranges a set of six blank canvases in an L-shaped configuration, with four in a vertical column and three in a horizontal row (the corner canvas is part of both). He hires an artist, with these instructions: “You can paint a single sea creature of your choice on each of these canvases, with one condition. I would like to see a picture of an eel when I move my gaze vertically or horizontally. So there must be at least one eel in the vertical column and at least one in the horizontal row.” The artist does this, following her own preferences when they do not conflict with the instructions.

What is the probability that the corner canvas has a picture of an eel?

To make this concrete, if your friend offered to pay you $2 if the corner canvas did not have a picture of an eel, and you had to pay her just$1 if it did, would you take the bet? Assume that your friend knows exactly what you do about the situation. (Update: The solution is now available here.)

The first possible naïve answer that may come to mind is that the probability is one in four because there are four vertical pictures, or perhaps one in three because there are three horizontal ones. Is either answer correct? Are both?

A monthly puzzle celebrating the sudden insights and unexpected twists of scientific problem solving. Your guide is Pradeep Mutalik, a medical research scientist at the Yale Center for Medical Informatics and a lifelong puzzle enthusiast.

The last question is not facetious. There is a famous problem in probability known as Bertrand’s paradox whose statement is simple: What is the probability that a random chord of a circle is larger than the side of the equilateral triangle inscribed in the circle? To do this you have to determine the density of such chords by “counting” how many are larger than the side of the triangle and dividing that by the total number of chords. The answer can be ¼, ⅓ or ½, depending on how the chords are counted. All of these answers are correct in different circumstances.

“Wait a minute,” you might say, “that’s because there are an infinite number of chords in a circle, and counting infinities is always problematic. It can’t happen when the numbers are finite.” But it can when, as above, the method of accomplishing a task is not clearly specified to a T (or L!). Probability is a fluid beast, slippery as an eel, one that can appear in many different places at the same time if all the avenues of ambiguity are not completely nailed down.

Why should this be? One reason is that probability is actually a hybrid beast, with two faces that can sometimes be at odds with each other. The first of these two aspects is the objective or constant part — how frequently an event actually occurs over the course of multiple trials (this is known as frequentist probability). The second is the subjective or changeable part: the degree of our knowledge or ignorance about the possibility of something happening (this is known as Bayesian probability).

Another possible reason for multiple answers is that, when conditions in a problem are unspecified or unknown, mathematicians make use of something called the “principle of indifference” or “principle of insufficient reason.” If there are several different, mutually exclusive scenarios that could take place in a situation, mathematicians assign equal probability to each of them, at least as a first guess. In problems that involve picking numbers from known ranges, this amounts to choosing uniform distributions. Sometimes this is just not possible, as we saw in the previous puzzle. And sometimes there are several equally valid ways of applying the principle of indifference that may yield different probabilities.

For today’s puzzle, I want readers to shake off their school-induced “fear of being wrong” and generate as many valid answers as they can. There are many different ways the artist could have completed the task, and each of these may yield a different answer. Furthermore, all you have to do to solve this month’s puzzle is count. It may be hard to believe, but finding probabilities is just glorified counting! Of course, once you make your assumptions, you have to do the counting right. The ambiguity in such problems lies in the problems not being fully specified, not in the laws of probability, which admit only one result when all the conditions are nailed down.

So let’s try to identify all the various places where the slippery eel of probability could turn up. One or two of the answers could be termed “mathematically favored,” but the others are valid too. In fact, in real life, as with the solution to the previous puzzle, the mathematically favored answers may not always be the best ones. With your answer, also specify your assumptions — the personal preferences, if any, that the artist may have brought to her task. As you do this, you may realize that we actually apply probabilities all the time in real life, weighing psychology and context, whether consciously or unconsciously, emotionally or rationally, intuitively or analytically. Bayesian probability is the basis of all our beliefs, and to a large extent our success in life is related to how nimble we are in keeping up with the slippery beast of probability.

More mathematically adventurous readers can try and invent their own clearly stated probability problem that could have several different answers because the principle of indifference can be applied in more than one valid way. Happy puzzling, and may the insight be with you!

Editor’s note: The reader who submits the most interesting, creative or insightful solution (as judged by the columnist) in the comments section will receive a Quanta Magazine T-shirt. And if you’d like to suggest a favorite puzzle for a future Insights column, submit it as a comment below, clearly marked “NEW PUZZLE SUGGESTION” (it will not appear online, so solutions to the puzzle above should be submitted separately). Update: The solution has been published here.

An informal poll

In order to estimate real-world probabilities, we need data. The principle of indifference is just a place-holder (known in Bayesian probability as a "non-informative prior") that needs to be updated with real frequencies.

So let's collect some data. Please indicate in your comment how you would satisfy the art collector's request if you were the artist. How many eel pictures would you have?

We already have one data point. Our artist, Olena, had 2. As for me, I would probably do 2 or 3 and perhaps 4, with a much lower likelihood.

P.S. If you have never encountered Bertrand’s paradox before, take a few minutes to follow the link given or to explore other articles explaining it. It may change your view of probability forever!

• Jason says:

Thanks, Pradeep. Yet another interesting problem. Probability problems are near and dear to my heart.

I can see any number of avenues to attack this problem – all of which are based on differing assumptions (or differing priors, if you will). For instance, let us start by assuming that the artist has no preference for eel vs. non-eel (that is, she will paint either an eel, or something else, with equal probability so long as it satisfies the collectors conditions). Thus, each painting would be represented by a coin flip, and the odds of corner eel vs. no corner eel is just a matter of counting coin flips (so long as the conditions are maintained). For corner eel, there are no restrictions on the other five paintings, giving 2^5 = 32 allowable states. For no corner eel, there are again 32 possible states, but not all of these are allowable: one disallowed state with no eels whatsoever, three disallowed states with no horizontal eels (2^2-1, with the minus on accounting for the double-counting of the no eels at all state), and seven disallowed states with no vertical eels (2^3-1). This gives a total of 2^5-11 = 21 states. Since all such states are equally probable, we have the probability of corner eel P[corner eel] = 32/(32+21) = 32/53 ~= 60.4%, and a corresponding expected loss of ~$0.8 – so don't take the bet! Now, let's change the assumptions a bit, and say that the artist chooses to paint all sea creatures with equal probability (again, subject to the constraints) – let N be the number of different sea creatures of which the artist has sufficient knowledge to paint. Thus, the probability of any individual painting containing an eel is now P[eel] = 1/N. Call this fraction p. Again, we have the 53 allowable states (32 with corner eel, 21 without), but not all such states are equally weighted. Rather, each such state must be weighted by a factor of p^k*(1-p)^(1-k), where k is the number of eels in the state (a number from 0 to 6). Adding in this weighting factor makes the summation slightly more difficult – although one could reasonably easily write a program to do the summation for any value of N. Fortunately, when N is large, we don't need to do the summation, since it's easy to check that the sum will be dominated by the term containing the fewest eels. For the corner case, this is the state containing only a single eel; for the non-corner case this is the six states (2*3) containing two eels each. Thus, the probability of corner eel is roughly P[corner eel] ~= p*(1-p)^5/[p*(1-p)^5 + 6p^2*(1-p)^4] = 1/(1+6p/(1-p)) ~= 1/(1+6p). Since p is small (because n is large), this value is very close to 1. Don't take the bet! Ah, but what if the artist takes a different tactic, determining what to paint in each successive painting based on what she has already painted? Now we enter the world of conditional probability – a much messier, but more realistic, setting. Here, things would depend on the order in which she chose to paint (a non-informative prior would tell us to assume all orders are equally probably). If we consider the two cases above (eel vs. non-eel with equal probability, or eel with probability p = 1/N), gives different answers. It's too difficult to work out the complete answer here in this post, but some hand-wave type analysis can give us a clue. For case 1 (p=0.5), the probability of corner eel is >50% (don't take the bet!), since depending on order the artist will always be allowed to paint a corner eel, but sometimes will not be allowed to not paint one. For case two, where p = 1/N is small (question for the peanut gallery – what values of N are required for p to be considered small?), we can assume that the artist will always paint a non-eel until she is forced to paint an eel by the restrictions imposed by her employer (this is equivalent to the N = infinity limit). Thus, out of all 7! possible orderings, in how many of those orderings will she paint the corner after finishing either the horizontal or vertical legs of the L? This is equal to 1/3 (chance she paints the corner after finishing the horizontal leg) + 1/4 (vertical leg) – 1/7 (correction to account for the fact that we double-counted the case where the corner painting is the very last one she paints). Thus, P[corner eel] = 1/3 + 1/4 – 1/7 = 37/84 ~= 44%. Here, the expected loss is approximately$2*0.44 – $1*0.56 =$0.32. Don't take the bet!

In fact, under most sets of "non-informative" assumptions, you should not take the bet. The only way you can really shift the odds in your favor are by assuming that the artist follows certain preferred orderings (aka, starting at one end of the L and working through it in order), or that she takes into account the final state of the L when planning her pictures – and also assuming that she hates eels. Then again, considering that these are eels we're taking about, the odds that the odds are bad are probably pretty good, so perhaps you should take the bet after all.

As for myself, if I were the artist, I would paint only a single eel – across all seven paintings. They don't call it artistic license for nothing!

• o. nate says:

I'd think about it like this: let's say each canvas can either have an eel or not have an eel. So that gives 2^6 total possible arrangements. There are 2^2 possible arrangements with no eel in the vertical row, 2^3 arrangements with no eel in the horizontal row, and 1 arrangement with no eel in either row. So that leaves 2^6 – 2^2 – 2^3 + 1 = 53 possible arrangements that meet the criterion. Of those 53, there are 2^5 = 32 arrangements with an eel in the corner. So the probability of an eel in the corner is 32/53 = 60.4%. So you should take the bet if it pays 2 to 1. Of course this totally neglects human psychology, so in real life, I wouldn't bet based on this.

• Mat says:

Chance of no eel in the corner: 21/55.
Chance of eel in the corner: 32/55.

• Matthew says:

Just like Bertrand's paradox, the difficulty is what we assume the probability density is. Let's try the easy mathematical assumption that all configurations of canvases (whether they're eel-less or not) have an equal probability.

We have 6 canvases that can either have an eel or not, so that's 2^6=64 total possibilities. To widdle that down to the number of ways with at least one eel vertically and horizontally, let's count off the ways that don't work. If all 4 vertical canvases are eel-less, then we have 2 choices left for the 2 end horizontal ones, giving 2^2=4 cases that do not work. If all 3 horizontal canvases are eel-less, then we have 3 choices left for the top 3 vertical canvases, giving 2^3=8 more cases that don't work. But, we counted the case of all canvases being eel-less twice, so really we have 4+8-1=11 cases that do not follow the collector's request, leaving 64-11=53 cases to consider.

Now for the number of cases that have an eel on the corner canvas. If we fix that corner canvas to have an eel, then we have 5 choices for the remaining canvases, for 2^5=32 cases where the corner canvas has an eel. Note that all of these cases follow the collector's request, because just having an eel at the corner is all you need. 32 out of 53 cases involve an eel at the corner canvas, so we can say that the probability is 32/53.

As for whether this method hits close to how a human being would decide, it seems as too large. If the artist chooses to make 2 or 3 canvases have an eel, then the chances of hitting that corner canvas are less than half. To improve this estimate we should probably weigh the chances of the artist picking only a few canvases to put eels on much higher.

• Brandon Norick says:

In an incredible stroke of luck, it just so happens that our art collector hired an artist who shared his passion for pictures of eels. In fact, her passion for eel pictures was so great that, as far as artistry goes, it's the only animal she learned how to paint. As a result, the probability that the corner canvas has a picture of an eel is 1.

I'd definitely take my friends bet.

• Jason says:

Gah; realized that I got the payouts of the bet mixed up in my above post; the probabilities for corner eel should be right, but my advice for taking versus not taking the bet is wrong. Since it's $2 win for no corner eel,$1 loss for corner eel, this means that so long as P[corner eel] < 2/3, you should take the bet. Thus, of the four cases I considered above, the answers are yes, no, maybe, and yes, respectively.

(The maybe is because I did insufficient analysis of the situation for a correct analysis for the case where we are looking at painting order with P[eel] = 0.5. A better analysis is as follows: based on case four, we know that ~44% of the time, the corner painting will be chosen after at least one of the legs has been finished. However, for the bottom leg, only 25% of the time will she actually be forced to paint an eel (P[corner last] * P[rest of leg has no eels]), for a total contribution of 1/12. Likewise, the top leg contributes 1/4*1/8 = 1/32. The double-counting factor is (1/4+1/8-1/32)*1/7 = 11/224. Adding it all up, we find 1/12 + 1/32 – 11/224 ~=6.5%. Add to this the 50% of cases where the artist is not forced to paint a corner eel but decides to do so anyway, and we get a total probability of ~56.5% corner eel. This is less than 2/3, so take the bet.)

• GlennF says:

The answer is simple: Assume all possible valid configurations are equally likely. Then just count how many valid painting configurations are there.
If an eel painting is in the corner then there are 2^5 = 32 configurations, since it doesn't matter if you put an eel or not in any of the 5 remaining positions.
If there is no eel in the corner, then there are (2^3-1) =7 independent possibilities for the vertical segment, since at least one must contain an eel. Similarly there are (2^2 -1) = 3 independent possibilities f or the horizontal squares. Hence there are 7*3 = 21 valid configurations without an
eel in the corner.
The probability of a valid configuration containing an eel in the corner is then
32 / (21 + 32) = 32 /53. Since this is more than half, you should take the bet.

• Eric K says:

6 canvases. Each canvas may be in one of two states, eel or no eel.
2^6=64 possible arrangements

The sample space of valid arrangements according to the collector's condition is a subset of the total arrangements of the six canvases. Furthermore, they can be divided into one of the following cases:
(1) No eel in middle section
(2) Eel in middle section

To count the valid arrangements for case (1) I begin by breaking up a valid arrangement with no eel in the middle section into two pieces, a vertical and horizontal section. Minus the corner canvas that makes up one end of the vertical section and one end of the horizontal section, there are 3 other canvases that make up the vertical section and 2 other canvases that make up the horizontal section. I will call the vertical section with the 3 canvases, A, and the horizontal section with the 2 canvases, B.

Now we observe that no eel in the middle section implies at least one eel in A and one eel in B. Both these conditions must be true for the arrangement as a whole to be valid and satisfy the collector's eel condition.

So if we break an arrangement into pieces, then for the whole arrangement to be valid (an eel in each piece), each piece must also be valid (at least 1 eel in the piece). And the total valid arrangements of the whole is equal to the product of the valid arrangements of the pieces. That is the strategy I will pursue to get a count of arrangements for case (1).

There are 2^3 arrangements of eel/no eel states for A and 2^2 arrangements of eel/no eel states in B.

Now counting the valid arrangements for case (1) is also the same as taking the total arrangements and subtracting the invalid arrangements to leave us with the valid arrangements.

We can also apply this to each piece of the arrangement. The count of arrangements with at least one eel in A is the same as the count of total arrangements of A minus the count of arrangements with no eel in A.
The same thing follows for B.

And for both A and B there is only one case for each where there is no eel, which is only one arrangement for each. This means the count of arrangements of a piece with at least one eel is simply the count of total arrangements of a piece minus one. For A this is 2^3-1 and for B it is 2^2-1.

Now we multiply to get the total valid arrangements of A and B where A and B each have at least one eel.

So (2^3-1)*(2^2-1) = 21 is the count for case (1)

Now to count case (2). Since there is an eel in the middle section, all arrangements with an eel in the middle satisfy the collector's conditions and are valid.

Therefore we only need to calculate the total arrangements of the other 5 canvases. So the count for case (2) is 2^5 = 32.

Adding case (1) and case (2) gives us all the different arrangements that satisfy the collector's conditions. So there are 21+32 = 53 different canvas arrangements.

So to calculate the probability of an eel being in the corner canvas we just need the count of valid arrangements with an eel in the corner canvas, which is case (2) that we already counted

and divide that by the sample space we also just counted, 53.

Alternatively, we could have taken the total arrangements of 6 canvases, 2^=64 and subtracted the invalid arrangements directly. An invalid arrangement is a subset of arrangements of case (1), in particular piece A or piece B must not have an eel.

000 0 00
000 0 01
000 0 10
000 0 11
100 0 00
010 0 00
001 0 00
110 0 00
011 0 00
101 0 00
111 0 00

There's 11 of them, so 64 – 11 = 53 which confirms our previous count.

So the probability is 32/53 = .6 that there is an eel in the middle section.

If your friend is giving you $2 for no eel and you have to pay$1 for an eel, then your expected value is 2 * .4 + (-1) * .6 = 0.2. Since your expected value is positive, you should take the bet.

• How interesting. Bertrand's Paradox is a bit easier to divorce from human preference, and deals with some interesting problems of infinities. You'd think this problem would be simpler for it, wouldn't you?

First off, I would not take the proposed bet based solely on instinct. If I were an Artist, and I only had this information, I'd only paint the one Eel in the corner, then do whatever I wanted with the rest of the paintings, which would give me more chances to make the work my own, and express my own non-eel-restrictive art.

I mean, what's the probability that an artist, given ONLY the instructions to paint a sea creature, would paint an eel? There are some 200,000 ocean species, about 800 of which are eels, so P(EEL|Sea Creature) is roughly 0.004 or 0.4%. Of course Artists may not know this so they may think of a more simplified list of sea creatures… I'd swag that I'd get tired after naming about ~100 variety of sea creature, so P(EEL) would be about 1% there. That's not counting mythical creatures or creatures made up on the spot, which could double my sample, so maybe P(EEL) ~0.5% is more likely. Fortunately these are all relatively close together, so yay for math! Or coincidence! Let's pick the middle value of 0.5% for funsies — we could do the following math with any of these estimates:

If a random artist (Artist A) who is still unable to break rules with a ~0.5% chance of painting an eel paints six canvases, there is, if my math holds up, about a 97% chance that they will NOT paint an eel. That is, if they randomly paint 5 canvases with a 97.5% chance of not Eeling(tm), the artist will panic, and be forced to paint an Eel and move it to the corner by virtue of the rules of the commission.

This is all, of course ridiculous. Any sane but slightly random artist (Artist B) would realize they had to paint an eel and not wait until the last minute to do so. They would paint an Eel first! There's a 1-in-6 chance that they'd paint an Eel in the corner, at which point they'd be relieved of all moral obligation and switch back to randomly painting the other canvases again I expect with some high probability (see 97.5 derivation above) of NOT painting an eel.

Of course, if they MISSED the corner, which is likely (5-in-6 chance via random canvas), they'd have to paint another Eel. Here again the sanity of our artist comes into question, and we must consider Artist B1 and Artist B2. Artist B1 randomly chooses between the remaining five canvasses, thus leading to a death spiral of chance in which the worst case is there is a column of three Eel paintings (all but the corner) and still the criteria are not met. If I were still in high school I'd remember how to calculate the probability of randomly choosing three specific objects out of a set of six in any order holds up, but I am not! But it's a bummer.

Slightly saner Artist B2 would simply choose from one of the canvasses which would relieve their responsibility and stave off the Eel painting madness. Of the previous 5/6 chance, there's a 3/5 chance they have already painted an Eel in the column, and a 2/5 chance that they've already painted one in the row. They could still paint the corner, so there's a 3/6 chance that their next choice has a 1/3 probability of being the corner and a 2/6 chance that their next choice has a 1/4 probability of being the corner… that's (3/6)*(1/3)+(2/6)*(1/4), which we can all agree is maniacal. Also, 0.25 (branched off of the original 5/6, which, adding back the 1/6 from the original gives like a 0.375 or 3/8 chance of the corner being an Eel, plus a 0.005*0.75 or so bump since even if there were two eels and an empty canvas there's still that 0.5% chance that we'll draw an Eel just because.

But this is all nonsense. Artists aren't random. They can be lots of things — accomplished, aspiring, starving, escape, dead (although that's often poets)… Let's assume the artists is doing this either for an acquaintance or for the money. The alternative is basically vandalism. If they're doing it for money, and they know their benefactor likes Eels, and there's some hope of a tip or a bonus if they like the results, then it's Eels all around. Maybe a lamprey (which isn't quite an eel), or a mermaid because those seem wildly popular in art (hmmm… maybe I should adjust that earlier 0.5% estimate), or something else thrown in, but certainly 4 or 5 Eels just to be safe. This more economic approach would clearly put P(EELintheCorner) much higher, near 75%, a number which I have recently imagined as being "maybe close enough" with no mathematical basis whatsoever.

Which leads the final possibility that this is being done for an acquaintance (a word which I have now mis-typed more than I've used it in this post). I have performed a non-scientific survey of a variety of people (with an embarrassing sample size that I refuse to publish, but which I will hint, when multiplied by any number returns zero) and it turns out that there's a high probability that people who would commission a painting of eels are somewhat terrifying individuals for whom their friends would probably not waste much time doing favors for (apologies to anyone who has actually done this, I'm sure you're a very nice outlier). In this case P(EelInthecorner) is about 1 (100%), because when confronted with mad eel obsessed people the best course of action is to draw six squiggly lines as quickly as possible and simply depart.

• David says:

I think it depends largely upon the way it is approached. If you treat the problem as a basic frequentist dilemma, and assume that the author is requesting one eel per column or row, then the solution is to assign equal probabilities to each canvas in said column or row. I like approaching the problem like a logic tree.

There's a 1/4 chance that the eel in the vertical column is on any canvas in that row and there's a 1/3 chance that the eel in the horizontal column is on any canvas in that row. This can be laid out as a 1/4 probability of an eel appearing in the corner from the vertical column, plus in the 3/4 chance of an eel from the vertical column not appearing there's a 1/3 probability of an eel from the horizontal row appearing in the corner, or:
(1/4) + (3/4)(1/3) = 0.50 or 50%

Based upon this logic, I would go for the bet since my opportunity is $2.00 * 0.50 =$1 and my cost is $1.00 * 0.50 =$0.50.

However, who says there only has to be one eel per row/column. I think a much more interesting and, in my mind realistic approach, would be to start with specifically what the art collector asked for. He wants to see an eel when he pans horizontally or vertically. To pan horizontally or vertically and see one, he must be starting from the corner, so this implies that there is not necessarily an eel in the corner. Unless the artist has a particular predilection to eels, so unless her prior Bayesian probability of painting eels is high, I wouldn't expect that she'd waste her creative options on more eels. Though this is based upon my prior assumption that she has a separate creative motif in mind – she may in fact be pre-disposed to more homogenous motifs.

So I would have to look at her prior work. Does she often paint eels (this may even be why this art collector requested she be the artist)? Does she tend to have a tendency towards homogenous motifs in her work. If either of these are true, I wouldn't go for the bet. However, if neither are true, I would assume that the chance of an eel in the corner canvas is relatively low and I would gladly accept the bet.

… Or, since the bet is only going to cost me a $1 at most, I might just impulsively guess based on the feeling I get. • John Riegsecker says: The probability there is an eel in the corner canvas is 32/53. A canvas has an eel, or it doesn't, so there are 2^6 = 64 ways to paint the canvases with or without eels. If the three bottom canvases do not have an eel, then there are 2^3 ways to paint the canvases that DON'T satisfy the conditions. Similarly, if the 4 canvases in the column do not have eels, then there are 2^2 ways to paint the canvases in the bottom row so that it DOESN'T satisfy the conditions. Since the canvas with no eels has been double counted, there are 64 – 12 + 1 = 53 ways to satisfy the eel condition. Suppose we have one of the 53 valid arrangements without an eel in the corner. Of the 2^3 = 8 ways to paint the canvases in the column, only the one with no eels is not allowed. Therefore, there are 7 ways to paint these canvases. Similarly, there are 3 ways to paint the canvases in the bottom row. Therefore, there are 7 x 3 ways to satisfy the condition without an eel in the corner, or 53 – 21 = 32 ways with an eel in the corner. So the probability of an eel in the corner is 32/53. Obviously, this assumes any canvas is as likely to have an eel as any other. • Ben says: I would paint an eel in every picture; since the art collector loves eels, nothing should make him happier than seeing nothing but eels! What's the probability that a random triangle is acute? The largest angle can be anywhere between 0 and 180 degrees, so by PIR we will say it is uniform. If the largest angle is <90 degrees it is acute, and if it is >90 degrees it is not, so the probability must be 1/2. But we also know that an acute triangle is exactly one which contains the center of the circle which circumscribes it. Given the circumcircle, by the PIR we can say that each of the vertices of the triangle are distributed uniformly around the circle. If we choose the location of the first point, then the diameter of the circle through the first point splits the circle into two sides. If the triangle is to contain the circle's center, then the other two points must each lie on different sides of the circle (with respect to the diameter through the first point); the probability of this is 1/2. Suppose that the other two points do indeed lie on different sides; then for every such triangle which contains the circle's center (i.e., is acute), if we flip the first point across the diameter along which it lies, we get a triangle which satisfies the same condition that the last two points are on different sides, but which does not contain the circle's center (i.e., is obtuse). By symmetry, this means that the probability that the triangle will be acute, given that the other two points lie on different sides, is 1/2. Therefore, the unconditional probability that the triangle is acute is 1/4. • Kyle Sinclair says: As Pradeep advertised while introducing this excellent puzzle, there are many ways to do this, and the conflicting results are due to the assumptions that one makes. Here are some examples that show how different strategies can lead to different answers. Strategy #1 – The Lazy Painter The painter follows the rule: minimize the amount of work needed to satisfy the requirements. In this case, she can accomplish the job by drawing one eel on the corner canvas. There is only one way in which to draw only one eel that meets the requirements, so the probability of an eel on the corner canvas is 1 if the painter employs this (rather efficient) strategy. Strategy #2 – It Takes Two to Conger The painter realizes that the collector might not be too happy with strategy #1, so she decides that she will instead draw exactly two eels. Her new strategy is: 1. Choose a canvas at random along the vertical direction 2. Choose the remaining canvas at random along the horizontal direction In this case, the probability that the corner contains an eel is 1/4 + (3/4)(1/3) = 1/2. (Note that swapping the order gives the same probability.) Strategy #3 – The Indecisive Painter 1. Randomly choose how many eels to draw (between 1 and 6) 2. Randomly choose a canvas from the remaining canvases (the last choice must simultaneously satisfy the collector's requirements as well as the new requirement established in Step 1) The probability in this case is (1/6)[1 + 1/2 + 11/20 + 2/3 + 5/6 + 1] = 91/120. Strategy #4 – The Blind Painter The painter randomly selects canvases until the minimum requirements are satisfied, then stops. This strategy yields a probability of 1/6 + 1/4 + 1/12 + 1/60 = 31/60. Strategy #5 – A Wild Loophole Appears! This one might be considered cheating since I am purposefully exploiting ambiguities in the phrasing of the problem. I will include it only because the author specifically asked for many examples. Consider the requirement that: "I would like to see a picture of an eel when I move my gaze vertically or horizontally. So there must be at least one eel in the vertical column and at least one in the horizontal row.” Suppose an eel (mod rotations) looks like this <—-) and a canvas looks like this [ ]. The painter could draw a "sea creature" (notice that does not exclude partial sea creatures since the word "whole" is omitted, another technicality and maybe a stretch) on each horizontal canvas such that when they are strung together, it looks like an eel: [<-][–][ -)]. Even though no single canvas contains an entire eel, the horizontal requirement is technically satisfied since the collector still sees an eel (a giant one at that) when moving his eyes in the horizontal direction. The painter could use a similar strategy for the vertical direction. In this case, there are infinitely many such arrangements. Let's consider just the horizontal direction for now. Suppose the painter uses the following strategy: 1. Select a value E at random (say using the uniform distribution over [0,L], where L is the standard length of an eel in centimeters) 2. Paint the first E cm of the eel on canvas 1, the next (L-E)/2 on canvas 2, and the remaining (L-E)/2 on canvas 3. Using these rules, an entire eel appears in the corner only when E=L in either the horizontal or vertical case. The probability of selecting any specific number for E is exactly 0 (yay, continuous distributions!), hence the probability is 0. —————————————————- Would I bet on it? No, since this is not an exhaustive list, and there may be many more strategies that make my position unfavorable. However, if the painter may only employ one of the strategies outlined above and she must select one randomly (call this Strategy #6 if you must), I would bet because my expected value would be$67/200 > 0. Which strategy would I use if I were the painter? In order to find a nice balance between effort exerted and creative expression, I would use Strategy #4 — randomly select canvases until the requirements are satisfied. Most of the time, I would not use much energy, but occasionally, I would be able to express my artistic ability by drawing four beautiful eels. • Davide Bassanini says: To fulfill the request our artist can pick any number of eels from 1 to 6. We cannot control art, so we have to say that the choice of eels will be random. At this point we need to count the number of possible ways the artist can fulfill the request with each number of eels.. Wih one eel we have only one configuration and the eel will be at the corner, so the probability is 1 With 2 eels we have 15 configurations, but 4 would not fulfill the request of having one eel in vertical and one in orizontal. In this case the probability of having a eel at the corner is 5/11. A similar reasoning can be done for the other cases: P(3) = 30/57 P(4) = 10/15 P(5) = 5/6 P(6) = 1 So, we can calculate the chance of having an heel at the corner by multiplying each probability for 1/6 and then summing up. I get something close to 0.75 Thanks Bye • Robert Garner says: While an art collector may have no accounting for taste, if an artist on the other hand does, (s)he will mount but a single eel canvas in the corner, so the probability of finding an eel there will be 1. However, in an anarchistic world where each artist feels (s)he must be different from every other potential artist, the idealistic frequentist will feel compelled to count all possible valid canvas arrangements and assume they're all equally likely. Case 1: Eel in corner, there are 2^5 possible arrangements of remaining eel or no eel canvases. Case 2: No eel in corner, there are 2^3 – 1 possible vertical canvas arrangements with at least one eel, and similarily 2^2 – 1 possible horizontal arrangements. The total possible sample space of all canvas arrangements is 2^6, so, if my math is right, Pr{eel in corner} = (32+7+3)/64 = 21/32 ~= 0.66 for a perfect universe of anarchist artists. • Ben says: Oops, I made a mistake in my previous comment. The smallest that the largest angle of a triangle could be is 60 degrees, not 0. And so actually assuming a uniform distribution between 60 and 180 degrees also gives a probability of 1/4, just as in the case using the circumcircle. This paper [1] details the probability that a triangle is acute with many different ways of sampling a triangle. Many of them are 1/4, but many also differ from 1/4. [1] http://www.math.northwestern.edu/~diana/math/acute.pdf • Michael DeLyser says: There is a very long list of sea creatures that I could paint. I feel like I could find six of them I would rather paint instead of an eel. So, if I was the artist, in order to paint more of what I wanted and still meet the collector's request, the eel goes in the corner, and that would be the only eel. • I believe the probability of the corner canvas has an eel in it is 32/53. Let me explain: If we assume that the artist picks with uniform distribution among the valid orientations that satisfy the collector's needs, we simply have to divide the number of valid orientations with an eel in the corner by the total amount of valid orientations. Let's count the number of valid orientations with an eel in the corner (X marks no eel and O marks an eel): X X X O X X O X X O X X X O X O X X . . . X X X O X O As you can see, the amount of valid orientations is simply a counting of combinations which breaks down to: (5 choose 0) + (5 choose 1) + (5 choose 2) + … + (5 choose 5) = 2^5 = 32 Now lets count the number of valid orientations with no eel in the corner: O X X X O X O X X X X O O X X X O O One again, this breaks down into a counting of combinations which has the equation: ((3 choose 1) + (3 choose 2) + (3 choose 3))*((2 choose 1) + (2 choose 2)) = 7*3 = 21 The reason we start at 1 and not 0 is because we must have at least 1 along the vertical and horizontal since we are counting only those that do not have an eel in the corner. The reason why we multiply is because the orientations on the vertical are independent of the choices on the horizontal. So the total amount of valid orientations is 21+32 = 53 and 32 of those have an eel in the corner so: 32/53 ~= 0.604 is the probability. As for the bet, the expected outcome is 2*0.396-1*0.604 = 0.188 > 0, therefore I would take the bet. • Fred says: Small ugly Matlab program enumerating all possible binary vector of length 6. We count how many of these vectors have at least one 1 in the first 4 elements, and at least one 1 in the last 3 elements; we call these vectors "valid". And we count among the valid ones, how many have a 1 in the 4th position: 32/53 = 0.6038. b = dec2bin(1:2^6-1,6); C = b-'0'; nb_valid = 0; nb_valid_corner_is_1 = 0; for k = 1:size(C,1) if sum(C(k,1:4)) > 0.5 && sum(C(k,end-2:end)) > 0.5 nb_valid = nb_valid + 1; if C(k,4) > 0.5 nb_valid_corner_is_1 = nb_valid_corner_is_1 + 1; end end end p = nb_valid_corner_is_1 / nb_valid • Nathan Ritchey says: I would take the bet, here's why. Taking into consideration that no one likes to paint eels and that more than one eel can be painted, please see my mathematical proof. http://imgur.com/S4oLw0k There are 32 possible combinations with an eel in the bottom corner, and 21 possible combinations with out it. However, I don't think the probability is evenly distributed among each case. The verbage of the request will effect how the painter will paint. I find this to be me more of a psychological problem than a mathematical one. Specific words such as "single", "one", and "an" will guide the person towards picking the corner. If only just to avoid painting more than one eel. • eJ says: The probability of an eel in the corner could be 0, 1, or anything in-between: you are entirely at the mercy of your artist's tastes. I will assume that you have already hired the artist and you are asking for the objective probability of the corner being an eel. That is, it's the probability conditional upon the fixed-but-unknown tastes of the artist, not anything to do with your personal uncertainty about the type of painter you hired. Thus, for any p in [0,1] (the range 0 to 1 inclusive), you might happen to have commissioned an artist that behaves as follows: pick a random number R in [0,1]; draw an eel in the corner if R<p, otherwise draw a shark; now draw an eel on every other canvas (just to be safe). The probability of an eel in the corner is p … any p. In practice, I imagine that an artist would have a strong preference for a particular pattern of eels and would fill in the remaining canvases more-or-less at random. So, in practice, the answer should be close to one of: * 0% : the generous artist might interpret your commission to mean that you wanted a "horizontal eel" and a "vertical eel" and that you'd feel cheated if they turned out to be one and the same. * 100% : the literal, or perhaps slightly mean, artist interprets your commission to mean that you want "a" (1) picture of an eel, necessarily on the corner canvas. Personally, I think I'd be the 0% guy, not least because eels are easier to draw than just about anything else. • francois says: Some though on Bertrand Paradox : It is clear that the Principle of Indifference is not enough as "Taking a random chord" can be achieved in many way. We say that extremity are uniformly distributed over the circle. We can say that middle point of chords are distributed uniformly over the radius of the circle ect. Also sometime the Indifference does not lead to a uniform distribution : for example if the random variable is the output of some recombinatory process, we should use a normal (gaussian) distribution instead. If this random variable is the return of some other variable, we should use a log-normal distribution, and so on. • Francois says: I think we should note that the artist is not picking its eels paint at random. It appears random to us because we dont have the information on its decision logic, but if we could repeat the exact same experience over and over, the artist will always choose the same configuration. This is why Pradeep introduced us to Bayesian probability. We cant really apply principle of indifference to the way the artist is choosing its paint in mathematically rigorous manner. So the way I will take to solve this problem will simply to say : what should I do if I were the artist ? Of course, like Pradeep suggested in its opening comment, if you can have access to a bigger sample, then you could apply the principle of indeference to all choice, and then determine the correct answer ! So let everybody gives its answer as what he or she will do as the artist, and then let us determine quanta's aggregated solution to this problem, which will be by construction better than each solution taken individually ! As the artist I will simply paint in the eels in the corner first, then I will go on for dolphin for on other paint (latest part is obviously not relevant 😉 ) • Colin P says: We know the preference of the art lover but what about the preference of the artist? The artist might think, Yay! My dream customer! I love painting eels, I'll do 6! Then the probability of having an eel in the corner is 1. Or the artist might think, Yuk! Just my luck! I hate painting eels, I'll do the minimum possible to satisfy my new customer. Then the artist will work out he can get away with just 1 if he puts it in the corner. So in this case the probability of having an eel in the corner is… 1 also! Or the artist might think, OK… I don't mind doing eels, but for my own sanity I'd rather do a mix with other sea creatures! I know, I'll do an eel on every other canvas starting at one of two ends. How shall I decide which end to start at? I'll toss a coin and if it lands heads up I'll start at the top of the vertical and work down and along. But if it lands tails up I'll start at the end of the horizontal and work back and up. Then the probability of having an eel in the corner is 50%. (The canvas in the corner will have an eel on it if and only if the coin lands tails up.) Or the artist might think, Aha! A chance to use my lottery ball random number generator! I know, there are 6 canvasses all of which can have eels on or not, I'll think of this as eel canvasses and non-eel canvasses, or 1's and 0's. Since there are 2^^6 = 64 possible arrangements of 6 1' and 0's, I'll list all the numbers in the range decimal 0 – 63 inclusive, write them as 6-digit binary numbers of 1's and 0's with leading zeros where relevant, I'll map those numbers to the canvasses starting from the top of the vertical, and by inspection I'll cross out all the numbers that do not meet my customer's requirement (for example any that start with 4 zeros). After this I'll write each remaining number on a lottery ball and add just those balls into my random number generator. Then I'll use it to generate a random number which will be one that with the mapping described meets the customer's requirement and is somewhere in the range 0-63. In this case the probability of having an eel in the corner is 32/53. (Of the 64 numbers in the range 0 – 63, 11 do not conform to the customer's requirement, meaning there are 53 balls in the random number generator, and of those 32 result in a ball in the corner.) Additional puzzle: In the last option the artist rejected in advance all the numbers that would not meet the customer's requirement. He could employ other strategies. Will the probability of having an eel in the corner be different from 32/53 for either of these two different strategies: a) He puts all 64 balls in the random number generator, generates a number, and if it does not conform to the customer's requirement he puts the number back in the random number generator and makes it generate another number. He keeps going until he gets a number that conforms and he uses that number. b) He puts all 64 balls in the random number generator, generates a number, and if it does not conform to the customer's requirement he does not put the number back in the random number generator and makes it generate another another number. He keeps going until he gets a number that conforms and he uses that number. • eJ says: Did you hire a Chinese painter or a Westerner? It could make a difference. Let us suppose that the painter knows how to draw 2 types of eel and n-2>=4 other sea creatures. She (say) works from one end of the L to the other, drawing a new random sea creature unless forced, by the requirements you set down, to paint an eel. For example, she might work down the L, drawing jellyfish, starfish, octopus, then (as she's not drawn one yet) an eel in the corner, then whale, then (unforced, at random) the other eel. The Westerner works top to bottom, left to right. She'll put an eel in the corner with probability (n-3)(n-4)/(n(n-1)) [forced] + 6/(n(n-1)) [unforced] = 1 – 6(n-3)/(n(n-1)). The Chinese painter works right to left, bottom to top. She'll put an eel in the corner with probability (n-2)(n-3)/(n(n-1)) [forced] + 4/(n(n-1)) [unforced] = 1 – (4n-10)/(n(n-1)). The latter is larger. Assuming the same "vocabulary" of 2 types of eel and n-2 other sea creatures, the Chinese painter is more likely to put an eel in the corner! • Deb says: The probability is 1. • Micah says: Let us assume that our painter is a careful planner, with a strong test-driven mentality. He wishes first to consider every way in which he can satisfy his benefactor's request with regard to eels- he will first arrange before him mockups of all 64 possible arrangements assigning whether or not each of the 6 squares will contain an eel. He will find that of these 64, 53 will meet his clients needs with regard to eels. The artist proceeds to discard of the other (non-satisfactory) arrangements, and considers only those that fit the eel criterion. Being now an artistic sort of person, and subject to whimsy, he makes what will appear to any person not sharing his unique aesthetic sense a choice completely without reason- such that to any unsavvy observer, no preference could be given to any one of these 53 arrangements. Since 16 of the remaining 53 arrangements possess an eel in their corner spot, we note that the odds of the painter placing an eel in the corner is 16/53, or about 30.2%. • John Hebbe says: Massaging the numbers in the ‘L’ pattern of canvasses seemed straight forward. Labeling the column A – D and the row D – F revealed (to me) 15 column and 7 row combinations which include ‘D’ totaling 105 overall. With ‘D’ removed, the number of combinations reduces to 7 and 3 respectively resulting in 21 possibilities. There are 84 row and column combinations remaining that contain ‘D’. The probability that the ‘D’ cell may contain an eel image is therefore 84/105 or 80 %. Ignore that the ‘D’ box appears in both column and row data sets. This is the simple mathematical answer and you will see that it reduces to 1.2 %, all things considered. There are other rubbery considerations to apply here. In reading Bertrand’s paradox , you see how that more than one nicely calculated answer appears. Was it 1/2, 1/3, or 1/4? While those approaches utilized fixed values, this approach does not. You are invited (by the moderator) to look beyond numerical input. Big pallet here so you have to make assumptions about several items. The collector and the artist are foremost. An entire plethora of other factors have input but these may be assumed to be incidental. . . .like Aj when you read through ”Summary: learning, reasoning and action” as presented in http://users.ics.aalto.fi/harri/thesis/valpola_thesis/node18.html . In this, Aj is assumed to have no effect on the state Si. I’ll limit my assumptions to the collector and artist. These assessments will drive the final configuration of paintings on the wall and generate the 1.2% final solution. First, the collector. Collectors most often have comfortable means which often include developed taste in artistic quality. His/her interest in sea creatures suggests comparative familiarity with other fields among which he makes this choice. He didn’t merely ‘hire’ an artist, he ‘commissioned’ an artist. Not the same. He has other pictures of sea creatures including eels. The ‘L’ shaped collage –already formatted with this arrangement of blank, framed pieces-already occupies space in his living room. Central room in most homes. He doesn’t ‘look up, down and across’, he ‘gazes vertically and horizontally’. We can visualize this person. Pierce Brosnan. Second, the artist. Already chosen and available. A struggling artist is available at the drop of a hat and for any price but is unlikely to have the talent to please this collector. A recognized artist would not likely be available on short notice or willing to accommodate the collector’s demands. But we have a wealthy collector searching for a talented artist and willing to pay the price. The mediocre group is overlooked. A talented artist accepts the position because he is versatile and accomplished. . . fully capable of rising to the occasion. The fee is grand. And so on. His name will appear on every piece. Seen by important guests. Now then. What is necessary and what will be undertaken. Minimally, the central ‘D’ position. If this is done, the collector’s contracted demands seem satisfied even though he may feel otherwise. He fully expects the other five canvases to be rendered. And in tasteful fashion. There will be no blank canvases. All in all, there will be six quality paintings. How many will have eels as the primary focus? Not all of them because neither the collector nor the artist wish to bore the on looking guest. Balance becomes an issue. The row will contain two images of eels. The column will contain three. The vertex plus two to differentiate it from the row. There will definitely be an eel at the ‘D’ position since any other arrangement would not be tolerable if you have viewed these arrangements before. The blank position here is like a void. So we envision a set of either three or four. With ‘D’ filled, you must leave ‘E’ as ‘not an eel’. Otherwise, the row eels would be unbalanced in the horizontal sense. All that remains is the column. The ‘C’ position could hold an eel but this would make the composition bottom heavy in eels. Down to ‘A’ and ‘B’ now. The Collector loves eel pictures. We actually need two to dynamically balance this setting. ‘A’ and ‘B’. Our presentation is almost complete. ‘C’ and ‘E’, the non-eels, remain to be determined. This can only be done after the first four have been finished since their appearance will drive the non-eel paintings. So, this is what the artist has determined: ‘A’, ‘B’, ‘D’ and ‘F’ will be eel pictures. Four. To answer an earlier question, the vertex position must be filled to retain ensemble balance. This reduces our choices to four fixed positions. Column = 3 and Row = 2, vertex in use. Here is where we reduce the earlier 80 % figure. Revisiting the beginning layout, we are left with only two choices now. Column paintings will appear in positions ‘A’, ‘B’ and ‘D’ while the Row paintings will appear in “D’ and ‘F’. ‘D’ exists in both legs. The possibilities of this occurring are 1 X 1 (with the vertex utilized). One. The calculated probability for this choice is 1 out of 84 (all of the ‘D’ permutations) or 1.2 %. And there is a 98.6 % probability that the rest of the collector/artist world may languish with poorly chosen gallery compositions. Let me know if you want more numbers. I’m probably off base but everyone needs amusement at times. • John Hebbe says: There is a 0.02 % probability of error in the above as you may have noticed (100 – 1.2 subtraction problem). • Robert Garner says: "This branch of mathematics [Probability] is the only one in which good writers frequently get results which are entirely erroneous." — Charles Pierce. In my rushed answer above (excuse: I was doing this puzzle to relax while accompanying someone in the Stanford Hospital ER yesterday!), the state space of all acceptable canvas arrangements with both a vertical and a horizontal eel, as various folks noted, is (2^3-1) * (2^3-1) + 2^5 = 53. As I noted, there are 2^5 corner eel arrangements acceptable to the collector, so assuming random anarchic artists, Pr{eel in corner) = 32/53 ~= 0.60 (and NOT 0.66 as I erroneously calculated above). It's interesting that while I made two errors in my calculation, I would still have won the bet! Culpable probability calculations frequently seem to get away with this, although they quickly become irrelevant as reality plays out its course (there aren't multiple universes any one art collector experiences). I still prefer Pr{eel in corner} = 1 because artists will likely either despise or adore painting eels — so in either case they've made their mark — whereas no normal world would contain an equiprobable number of non-conforming anarchic artists. • Malachi says: The frequentist approach that produces a probability of 32/53 of an eel in the corner location seems entirely justifiable. So does the following frequentist approach (also described by Davide Bassani), and neither seems to require arbitrary assumptions, such as the artist's personal preferences, other than how the principle of indifference should be applied. (Reasoning on the basis of the artist's state of mind without evidence is, in my opinion, effectively adding an additional constraint and, thus, not addressing the original problem.) A) The artist paints six pictures. The artist may have a specific arrangement in mind, but we do not know it. So, by the principle of indifference, we consider the pictures without regard to their arrangement. So long as at least one picture depicts an eel, there is a possible arrangement that satisfies the constraints. Five paintings may or may not depict an eel, so there are a total of just six possibilities. Keep in mind that pictures of eels are freely interchangeable when satisfying the arrangement rules; likewise for the other pictures. The count is therefore not 2^5 = 32, because this would count equivalent combinations more than once. B) The artist arranges the pictures. There are only the six cases to consider: * 6 eels: Probability of an eel in the corner = 1 * 5 eels: 6 positions for the picture without an eel, only one in the corner = 5 /6 * 4 eels: (4+3+2+1) / (5+4+3+2+1) = 10 / 15 * 3 eels: 10 / 19 * 2 eels: 5 / 12 * 1 eel, which must be in the corner: 1 By the principle of indifference, all cases are equally likely. So, if the counting for each case and some miscellaneous arithmetic is correct, the final probability of an eel in the corner is: (1/6) + (5/36) + (10/90) + (10/114) + (5/72) + (1/6) = ~ 0.74 The more popular approach produces 32/53 = ~ 0.60. I don't find the reasoning for that outcome quite as satisfying, because it seems to rely on a subtle, extra assumption about the unknown intentions of the artist concerning the final arrangement. • Malachi says: In practice, I would not bet on the basis of statistical reasoning. Not all arrangements are compositionally equal. Intuitively, I would expect a large percentage of artists to favor an arrangement like one of these, which avoid a "static" horizontal or vertical alignment of eels: [_] [E] [_] [_] [E] [_] [E] [_] [_] [_] [E] [_] [_] [_] [E] [_] [_] [E] I would not bet on an eel in the corner. • Colin P says: @John Hebbe I believe this puzzle shows no probability answer between 0 and 1 is "wrong", because any answer can be arrived at depending on the preferences of the collector and artist. But I believe your calculation 15 x 7 double-counts some of the eel/non-eel arrangements skewing your final probability of 80% to be higher than the probability of 32/53 when the arrangement is chosen randomly from a uniform distribution. If you intended it, fine, you're just saying you want to skew the probability by using a particular double-counting method. • John Hebbe says: Thank you, Colin. Upon review, I notice many, Many flaws in my solution. Admittedly, much of that entry was unworthy of this column. Apologies. Here is an improved approach. Does this remove the skewness you addressed? This is the best I can do. Using the layout of the column IDs, top down, as A, B, C, D and the Row IDs as D,E and F, here are the possibilities I counted. ‘x’ suggests that the canvas exists but no picture has been painted upon it. In the root case (only one eel picture), we agree that the ‘D’ canvas must be an eel picture. The remaining canvases may be sea creatures (non-eel) or any other rendering including blanks (these are shown as ‘x’). This matrix is complete: Axxx – Bxxx – Cxxx – Dxxx – – -xxxx is not allowed since at least ABxx – BCxx – CDxx – – – – – – – one column entry must be a sea creature. ACxx – BDxx ADxx – BCDx ABCx ABDx ACDx ABCD – – – – – – – – – – – – – – – There are 15 possible Columnar possibilities. 8 contain the ‘D’ position. Dxx – Exx – Fxx- – – – – – – – – -xxx is not permitted here either. DEx – EFx DFx DEF – – – – – – – – – – – – – – – – -There are 7 possible Row possibilities. 4 contain the ‘D’ position. I calculated that there are, overall, 15 x 7 or 105 total sea creature possibilities. There are 8 x 4 or 32 combinations which contain ‘D’. Perhaps I assumed incorrectly that Dxxx and Dxx are separate (although equal) but I think both must be considered in the total possibility calculation. If I work with this ratio (32/105), I see that, lacking other restrictions, there is a probability of 30 % that the ’L’ arrangements will contain an eel (‘D’) at the vertex. At a maximum, no ‘x’ pictures, the probability becomes 100 % since there is a ‘D’ in every choice. The question is to determine which of the other canvases contain eels. In the previous dissertation, I concluded that the most likely final wall selection would be an arrangement which contained eel pictures in the ‘A’, ‘B’, ‘D’ and ‘F’ positions for several reasons. A new set of possibilities occurs. Now the calculations to determine the probability of the vertex changes as the number of eel pictures changes. . .but not up to the maximum of 6 since that is unlikely (discussed previously). Other choices are eliminated. The new count is as follows: The number of all possibilities remains the same. The allowable Columnar count now includes ‘A’, ‘B’ and ‘D’. . .all combinations excepting those containing ‘C’. ‘A’ and ‘B’ are both eel pictures (concluded previously). This number is 7. Check the matrix. The allowable Row count now includes ‘D’ and ‘F’. . .all excepting ‘E’. This number is 3. The maximum combinations of eel pictures is now (7 x 3) or 21. The calculation now, which is the probability that the corner canvas has an eel picture becomes 21/105 or 20 %. Final answer. This is based upon development of the ‘L’ arrangement which has all sea creature paintings and eels in positions ‘A’,’B’, ‘D’ and ‘F’. Final. Final. Final. Final. Final. • Adam says: Assuming the artist enjoys painting many different sea creatures, she would intentionally paint an eel in the corner canvas. Then she would only need to paint one. That's what I would do. If the collector is really annoyed by that, I could throw in another painting at half-price, but as an earlier commenter noted, I would rearrange the two eel painting for symmetry, taking the original eel out of the corner. • Matthew Petersen says: Combinatorics is fun! I solved this using two steps. Firstly, I used an "all minus bad" approach to find how many arrangements are good (at least one eel in column, and at least one eel in row). Then, I divided the number of "eel in the corner" arrangements by the total number of good arrangements (at least one eel in column, and at least one eel in row). All (64 ways): There are 6 paintings which have 2 possible states ("eel" or "not eel"). Thus, there are 64 unique ways to paint everything. Bad (4 + 8 = 12 ways): 1) No eels in column There are 2^2 ways because of the 2 empty row slots. 2) No eels in row There are 2^3 ways because of the 3 empty column slots. Good (64 – 12 = 52 ways) Eel in the corner ways (32): Assume an eel is in the corner. Then there are 5 empty paintings, which can be eels or "not eels." That makes 2^5 = 32 unique ways to have an eel in the corner. Probability of eel in the corner (32 / 52 = 61.5%) That's how I think the probability should be 😛 • John Hebbe says: (please ignore the earlier entry starting with 'Thank you Matthew'. Thank you. Thank you Matthew, for showing me what I overlooked. ‘D’ may be exchanged with ‘A’, ‘B’ or ‘F’. There are four eel pictures to consider for the corner position. These exchanges add a total of 3 more combinations resulting in (21 + 3)/105 or 22.9 %, the probability of finding an eel picture at the vertex of the arrangement. Last correction. When I get big, I’ll be as organized as all of you. . Never say Final again • Fred Viole says: Just have her paint the living room 10,000 times (in R): Eels <- function (x){ canvas_1 <-numeric(0L) canvas_2 <-numeric(0L) canvas_3 <-numeric(0L) canvas_4 <-numeric(0L) canvas_5 <-numeric(0L) canvas_6 <-numeric(0L) valid<- numeric(0L) vert<- numeric(0L) horiz<- numeric(0L) corner<- numeric(0L) for (i in 1:x){ if(runif(1)<.5) {canvas_1[i]<- 1} else {canvas_1[i]<-0} if(runif(1)<.5) {canvas_2[i]<- 1} else {canvas_2[i]<-0} if(runif(1)<.5) {canvas_3[i]<- 1} else {canvas_3[i]<-0} if(runif(1)<.5) {canvas_4[i]<- 1} else {canvas_4[i]<-0} if(runif(1)<.5) {canvas_5[i]<- 1} else {canvas_5[i]<-0} if(runif(1)<.5) {canvas_6[i]<- 1} else {canvas_6[i]<-0} vert[i] = sum(c(canvas_1[i],canvas_2[i],canvas_3[i],canvas_4[i])) horiz[i] = sum(c(canvas_4[i],canvas_5[i],canvas_6[i])) if (vert[i]*horiz[i] >=1) {valid[i]<-1} else {valid[i]<-0} if(canvas_4[i]*valid[i] >=1) {corner[i]<- 1} else {corner[i]<-0} } return(sum(corner)/sum(valid)) } > Eels(10000) [1] 0.6018687 • Colin P says: @John Hebbe I love the blank canvasses! (I had not thought of those.) But there are still only eel and non-eel canvasses. What I meant by double-counting is that each of the 15 options for the 4 vertical canvasses ties down the D canvas, so that leaves only 2 horizontal canvasses to be varied when computing the total number of unique arrangements. In other words I believe the first term in your (compound) calculation should 15 * 3, not 15 * 7. • AlexP says: Before any counting of probabilities, the puzzle definition needs to be understood. The collector's "I would like to see a picture of an eel when I move my gaze vertically or horizontally" is primary, with the following "So there must be at least one eel …" merely a secondary consequence. And any "move" must be taken to be in the context of the art project, since nothing wider (such as the entire wall) is mentioned. Therefore the row and column ends, and the corner, not only fail to satisfy the conditions, they are positively contrary to instructions, since they are the natural starting points of a gaze, and eels there would be detected before any "move". Indeed the corner of an L is its anchor point, and an eel there would be the central feature of the set of paintings, whereas the instructions are clearly to have eels be a secondary part of the overall effect. And I expect commissioned artists to be good at picking up on such detail in their client's wishes. The probability of a corner eel is then nearly 1 (only "nearly" since people are unpredictable, and there was nothing in the puzzle statement that removed human vagaries). So I would take the bet, expecting to collect2 with no eel in the corner, while accepting a small risk of paying $1. • Cyprus says: Probability of eel in corner is 1 of 7. If no eel in corner, then painter must use at least one eel in each side, and those combos add up to 6, so 7 options in all and corner option is 1 of 7. This is most likely incorrect, but I suspect it's no more incorrect than some of the much more sophisticated posts above. The important thing to realize, if this question was important (example, probability of forest fire in Southern California this month), is that even a lower estimate of 14% (1/7) is something that requires preparation the same as 50% or greater. To paraphrase the bumper sticker saying, "1/7 happens". • Fred Viole says: Incorporating the artist's preferences yields even more interesting results. Substituting the following lines of R code: Eels <- function (x,pref){ . . . if(runif(1)< pref) {canvas_1[i]<- 1} else {canvas_1[i]<-0} if(runif(1)< pref) {canvas_2[i]<- 1} else {canvas_2[i]<-0} if(runif(1)< pref) {canvas_3[i]<- 1} else {canvas_3[i]<-0} if(runif(1)< pref) {canvas_4[i]<- 1} else {canvas_4[i]<-0} if(runif(1)< pref) {canvas_5[i]<- 1} else {canvas_5[i]<-0} if(runif(1)< pref) {canvas_6[i]<- 1} else {canvas_6[i]<-0} . . . Increasing or decreasing the artist's "pref" from 0.5 (indifference) results in *higher* probabilities of an eel on the corner canvas when satisfying the collector's condition, "So there must be at least one eel in the vertical column and at least one in the horizontal row.” Even substituting a Normal distribution (coupled with the artist's strong negative preference for eels) for the Uniform distribution used above does not lower the probability of an eel on the corner canvas to below 0.5. Take the bet. • Fred Viole says: One (mis)assumption I made was symmetry of outcomes from the artist preferences. However, the range of preferences for a positive expected value of the bet are very biased to the artist's dislike of eels ~[0.11, 0.62] yielding corner canvas eel probabilities of ~0.66667. Symmetry from the initial 0.5 would have made the 0.5 (artist's indifference) more likely and consequently a better bet. Seeing (finally) as this is really a bet on the artist's preferences and the revealed bias, do not take the bet. • Mattia Landoni says: The probability is one hunded percent. As an economist, I must point out that this problem involves an optimizing agent, the artist. If the artist dislikes eels, she will paint only one in the corner in order to minimize the number of eels. If the artist likes eels, she will paint seven eels and there will still be one in the corner. Since I have no prior reason to believe that the artist prefers eels to other aquatic animals, the two scenarios are equally probable. But that doesn't matter. I'm just writing this to keep up the facade that this has anything to do with probability. Also, if anyone asks "what is the expected number of eels", this information is needed. The expected number of eels is four. Averages can be deceitful. A reader may ask, "why do you rule out that the painter will paint between two and six eels?" The answer once again has nothing to do with probability. Seven is a prime number. To keep balance in the composition, you should have the same number of every type of fish. Since seven is not divisible by two, the seven canvases cannot be filled with pairs. The artist would start with two eels, then add two porpoises, two humuhumunukunukuapuaas, and then what? Same applies for three, four, five and six eels. No artist would go for a number other than zero, one, or seven. Since zero is in contravention of the artist's commission, the number of eels must be one or seven. QED • Mattia Landoni says: It appears I misread – the canvases are actually six. This introduces all kinds of mathematical problems akin to the ones that have been studied above. Using whatever is left of the structure of my previous solution, I can still say that the probability is greater than fifty percent (i.e., 100% if the artist dislikes eels, and x>0% otherwise). I'm too lazy to calculate x. I still like the idea of seven eels. • eJ says: My comment about "Chinese" vs "Western" painting contained a mistake: the formulae are nastier (ratios of cubics) than I'd realised. A better demonstration of the fact that order matters is given below. Set-up. Assume that any painter can be described by (a) the order in which he paints the canvases and (b) the probability, u, that he'll paint something other than an eel when free to do so. On each canvas in turn, he is free to paint what he likes, except when he's about to start the last of an eel-less horizontal or vertical line, in which case he must paint an eel. There are 6! = 720 possible orderings, but they boil down to just four cases according to whether or not the last horizontal and vertical canvases are the corner: * "Corner first" — equivalent to 420/720 orderings. Pr(eel in the corner) = 1-u. * "Corner last" — equivalent to 120/720 orderings. Pr(eel in the corner) = 1-u+u^3+u^4-u^6. * "Chinese" — equivalent to 120/720 orderings. Pr(eel in the corner) = 1-u+u^3. * "Western" — equivalent to 60/720 orderings. Pr(eel in the corner) = 1-u+u^4. For a given u, the "corner last" ordering gives you the biggest Pr(eel in the corner), followed by "Chinese", then "Western", then "corner first". If the artist picks his order uniformly at random from the 720 possibilities, the probability of an eel in the corner works out as 1-u+u^3/3+u^4/4-u^6/6, always at least 5/12 no matter how strongly the painter would rather draw anything else. • Nathan Gantz says: First I will show why two of six canvasses must contain eels. The collector's gaze is limited in height and width. His gaze is not tall enough to see all four rows at once. His gaze is not wide enough to see all three columns at once. This is counterintuitive because human gaze tends to be wider than it is tall. But these constraints — upon the collector's gaze — are implied by the idea that there must be an eel painting in each column and each row. For if his gaze was of greater area, he could sweep his vision across either dimension and see all of the six paintings. Thus a single eel painting in one corner would not satisfy his gaze requirement. In fact, the corner eel painting is useless in the gaze sense: there must be an additional eel painting above the corner and an additional eel painting to the right of it. Thus two of six canvasses must contain eels. What sea creatures the artist chooses to paint in the other four canvasses is her arbitrary decision. In this sense there is no predictable difference between the corner and the other three non-required-eel canvasses. Maybe she wants to paint eels everywhere because she prefers the aesthetic of one type of sea creature (100% probability of corner eel). Maybe she chooses the corner canvas randomly from all species of "fish" (no mollusks or mammals), 97.5% of which are not eels (=2.5% probability of corner eel). Maybe she decides to balance the sea-creature diversity by excluding eels from all four optional canvasses (=0% corner eel). Applying the principle of indifference to these three scenarios, we get roughly 34% probability of the corner canvas having an eel. These three scenarios are far from comprehensive, but that is just what happens when you have subjective psychological scenarios. So with 34% probability of losing a dollar and 66% probability of gaining two dollars, I will take the bet. When there is such a wide conceivable range of approaches available to the artist, it would be irresponsible to assign probability to the outcome. • AlexP says: CORRECTION I of course meant to say "nearly ZERO" not "nearly 1", ie The probability of a corner eel is then nearly zero (only "nearly" since people are unpredictable, and there was nothing in the puzzle statement that removed human vagaries). — Alex • AlexP says: Before any counting of probabilities, the puzzle definition needs to be understood. The collector's "I would like to see a picture of an eel when I move my gaze vertically or horizontally" is primary, with the following "So there must be at least one eel …" merely a secondary consequence. And any "move" must be taken to be in the context of the art project, since nothing wider (such as the entire wall) is mentioned. Therefore the row and column ends, and the corner, not only fail to satisfy the conditions, they are positively contrary to instructions, since they are the natural starting points of a gaze, and eels there would be detected before any "move". Indeed the corner of an L is its anchor point, and an eel there would be the central feature of the set of paintings, whereas the instructions are clearly to have eels be a secondary part of the overall effect. And I expect commissioned artists to be good at picking up on such detail in their client's wishes. The probability of a corner eel is then nearly zero (only "nearly" since people are unpredictable, and there was nothing in the puzzle statement that removed human vagaries). So I would take the bet, expecting to collect$2 with no eel in the corner, while accepting a small risk of paying $1. • Nathan Gantz says: David has argued that sweeping gaze must begin at the corner. Similarly Alex has argued that sweeping gaze must begin at one of three extremities, including the corner. Since the three of us reach similar conclusions, I want to point out the difference. I believe the sweeping gaze can start from outside of the system of 6. Perhaps this has something to do with my visualization of the gaze as having more "area" than a single canvas and/or a different interpretation of what the original problem is asking. • Nathan Gantz says: So for example the gaze could fall from above upon the rightmost two blocks, missing the column on the left. Like Tetris. • William Metcalfe says: Quote: “You can paint a single sea creature of your choice on each of these canvases, with one condition. I would like to see a picture of an eel when I move my gaze vertically or horizontally. So there must be at least one eel in the vertical column and at least one in the horizontal row.” The collector does not say that each canvas should show a different creature from the others. He says again: "a single sea creature of your choice on each of these canvases" Or one creature per/canvas. Therefore the number of eels could be between 7 & 1. Someone other than myself can do the probabilities. • TomB says: Let’s start with a similar scenario using a different profession. I propose you and I ask a math professor chosen at random to write out a function that maps x to itself. What is the probability he will write either f(x) = x OR g(x)=x? To make this concrete, you pay me$100 if he writes f(x) = x OR g(x)=x. If he write down anything else… k(x) = x, j(x) = x, u(x) = x, f(x) = 2x/2, z(x)=x… I pay you \$200. Will you take the bet take?

Most artists like most professionals follow standard conventions most of time. This leads us to the “rules of composition”.

If you google “rules of art composition” you will find several generally agreed upon techniques to be considered while deciding on the composition of a painting, photo, etc.

Below are a brief descriptions of two of these “rules”:

1. There should only be one main element. Other objects/elements should NOT distract from the main element.

2. Line “like" elements should draw or lead the viewers attention to the main element.

While the artist could view each canvas a separate works of art and their placement on the wall as irrelevant. This would be a very non-artistic interpretation. It is much more likely he will view the 6 canvases as parts of one larger single composition. For example, he might decide the six paintings viewed together should depict a single large sea scape.

Next he considers the lines of "L" shaped canvas configuration. Both lines lead to (and create an arrow pointing at) the corner canvas. Based the rules of composition, this spot is the prime candidate for the main element of the composition.

What should be the main element of the composition he ponders? Well, the artist knows that the guy paying the bill especially likes eels thus making it a prime candidate for the focal point. Also, the artist needs to research eels to properly render the creature. After viewing multiple images and reading the wikipedia entry, he learns among other things, that eels are primarily bottom dwellers. The artist wants to mimic the natural order of world and decides the bottom row will represent the sea floor and that he will restrict all eel images to this row. After considering the original requirements, his master plan begins to take shape. It includes an eel on corner canvas.

I would NOT take the bet.

• John Hebbe says:

Act III. Another day. Another diagram. Shorter. Shorter is nice. Aesthetics rule.
Sticking with the ‘all sea creature’ arrangement containing four eel pictures (E) and two non-eels (N). And arranged so that the two pictures closest to the vertex are non-eels. All other arrangements of these six pictures, when ‘gazed’ upon as the collector is wont to do, will not provide a satisfying sense of balance. The ends and the vertex must hold eels and the second down position emphasizes the longer extension of the column. Otherwise you have either a short or a stubby ‘L’ arrangement. The basic format remains:
.
E – – – Let the eel pictures be A, B, D and F. – – – – A
E – – – – – – – – – – – – – – – – – – – – – B
N – – – Let the non-eel pictures be C and E. – – – C
E N E- – – – – – – – – – – – – – – – – – – – D E F
.
With this 4 – 2 mix, there are 24 possible combinations satisfying all of the restrictions are:
.
A – – – – A – – – – A – – – – | A – – – -A – – – – A – – – – | A – – – – A . . . .
B – – – – B – – – – B – – – – | D – – – – D – – – – D – – – – | F – – – – F . . . .
C – – – – E – – – – C – – – – | C- – – – E – – – – C – – – – | C – – – – E
D E F – – D C F- – F E D – – | B E F – – B C F – – F E B – – | D E B – – D C B. .
.
You can see the pattern. I’m not aware of a nice set of simple equations to demo this.
.
Here is the best part: Driven by the basic format of the arrangement, the non-eel pictures are always adjacent to the vertex but never in the vertex position. Earlier logic dictated that there would be no blank or non-sea creature pictures. Every combination permitted contains an eel picture at the vertex. The probability of the corner canvas having an eel picture is 1 or 100 %. This was evident the moment the basic format was designed. All of you noticed that before I did. Having read your submissions, I feel less embarrassed with my offerings.
I’m taking the rest of the month off.

• John Hebbe says:

All solutions to date make traditional assumptions about the fundamental ‘L’ shape. Bear in mind that ‘L’ arrangements can be inverted, reversed or oriented in many ways. Nothing stipulates that the canvases are identical in size. Everywhere you look, there are Wild Cards. There are tons of artsie-fartsie gimmicks the collector could have set for you and some of these could easily call for methodology not examined to date. Who knows, he could be in love with fore-shortened, oblique art to catch the wandering eye. Mathematicians and artists beware.

• Themba says:

No person in their right mind would paint more than one eel if they do not absolutely have to. So the corner piece has to contain the eel with probability 1.0…

• Colin P says:

@Themba

I am usually in my right mind (I think) but sometimes I'm not, so I reckon if I was hired to do the art the probability would be 0.8…

• Ian says:

We know nothing about the artist's preferences and personal motivations. Any assumptions we make are as likely to be wrong as right. The only FACT we know is that the final pattern will match the given criteria. In the absence of further information, we must assume that every valid pattern is equally likely.

There are 32 possible patterns with an eel in the corner. There are 21 possible patterns with no eel, but matching the given criteria. The probability that one pattern chosen at random will have an eel in the corner is therefore, in the absence of further information, 32/(32+21).

In practice, of course, we know that people en masse are not entirely random, but have preferences. People doing the lottery will tend to avoid sequences of numbers because they're somehow less "random", yet prefer low numbers because they correspond to meaningful dates. People asked to pick a 1-digit number at random are more likely to choose 7 than any other. And so on; there are many other such examples. So I have no doubt that, if we were to go out and gather statistical information on this precise problem, by asking many people to do this task, we'd find that some patterns occur far more often than others, and the actual probability is not as I've calculated. Certainly, it's quite likely that the availability of many different subjects for the "non-eel" pictures would bias the selection of patterns away from those having a large number of eels. And it's also possible that, if we were to limit our sample to people classifiable as "artists for hire", we might well find a subtly different set of preferred patterns – an "artistic set", if you will – to those chosen by random individuals. It's even possible that that "artistic set" will vary according to sociological factors such as age, gender and nationality. But none of that alters the fact that, until and unless we have more information, attempting to factor such criteria into a calculation of probability is ultimately nothing more than, basically, playing a hunch.

• Dave says:

I tend to favor simple, if tedious solutions rather than clever more complex ones. So I've just plotted out all the possibilities into a Truth Table:

Which yields a roughly 60% probability of an eel in the corner.

I believe speculation about the artist's motivations to paint or not paint eels beyond the requirements are completely subjective and are not really fit for a mathematical calculation. The problem then becomes a psychological question, and one that we really couldn't begin to answer without running batteries of tests with lots of people. Even this result isn't necessarily very valid across cultures and time periods, as some of the commenters have suggested.

To keep this problem out of the psychological realm, the only reasonable assumption is that all valid combinations are equally likely.

• John Hebbe says:

From the outset, the personalities of the collector, the artist and puzzle-solvers in general have been brought under the microscope. Explicative analyses abound. The surprise is that Pradeep seems to have spread before us a problem with no discernible mathematical solution. I’ll be interested in his pronouncement come month’s end.

• Colin P says:

Consider Bertrand's Paradox via the link in the slippery eel puzzle and with the benefit of hindsight here is how I noted the probabilities 1/4, 1/3 and 1/2.

By inspection, for a chord on a circle to be longer than the side of an equilateral triangle inscribed exactly in said circle the chord must pass through the inner part of the equilateral triangle, in fact it must pass through an inner circle inscribed exactly in the equilateral triangle. So now we have an outer circle, an equilateral triangle inscribed within it, and an inner circle inscribed within that, and by geometry one can show that the inner circle is half the radius of the outer circle.

Given this geometry consider the probability that a chord will have a length longer than the side of the equilateral triangle…

If we hold some pre-drawn lines by their midpoints and drop them individually and randomly from a great height above the outer circle then one quarter of the midpoints will fall into (and one quarter of the chords will pass through) the inner circle because the area of the inner circle is one quarter the area of the outer circle. Probability = 1/4.

If we draw all our chords starting from a single point on the circumference of the outer circle, (bearing in mind the choice of point is arbitrary) then one third of them will pass through the inner circle because viewed from the arbitrary point the inner circle subtends an angle of 60 degrees but we have 180 degrees of freedom in which to draw out each line so that it makes a chord. Probability = 1/3.

If we draw all our chords as a parallel set of lines across the outer circle (bearing in mind the choice of direction is arbitrary) then one half of them will pass through the inner circle because the diameter of the inner circle is one half the diameter of the outer circle. Probability = 1/2.

An intriguing point is that the method of deriving the size of the inner circle is arbitrary, any inner circle treated in the same three ways will result in three different probabilities.

• Albieri Garcia says:

I would say that the probability that the corner canvas has a picture of an eel is 1:

As the problem is presented, we can assume the following:

The art collector hires an artist to make just an art work (just one), which consist on painting 6 pictures in a concrete arrangement, an L-shaped configuration. Because of the conditions given by the art collector (to see a picture of an eel when he moves his gaze vertically or horizontally) the artist has mainly two options to make the art work:

1. Painting six different sea creatures, one on each canvas, so to keep the art collector condition the eel needs to be necessary in the corner.

2. Painting an eel six times, that is, to make a study on the motive of the eel. Artists use to make a study of a motive (as Cezanne with Mont Saint-Victorie, or Rouen Cathedral, Van Gogh on The Langlois Bridge at Arles, Monet on the Water Lilies and so on…). We can presuppose that for an artist is more important how to paint that what to paint: what the artist want to express and how to do it is more important that the motive he/she paints.

For both options there will be an eel on the corner, so the probability of having an eel on the corner should be 1.

• John Hebb says:

Jason, this may sound like sour grapes but it is, ultimately, appreciation for your sound logic in both this and the preceding two-number puzzle. Had I read your initial solution I might have skipped the August offering and counted the days until the next puzzle.

In your first response, you state (in the conditionals, case two), “where p=1/N is small, we can assume” etc. And the peanut gallery was challenged to determine what values of N are required for p to be considered small. My statistics classes were in the early 50s at RPI but I recall that Poisson distributions are similar to binomial distributions in that if N is large and nP is fairly small (Np<5 is the requirement usually stated), it will give the same probabilities (Citation available). Would a value of N<5/p be acceptable as ‘small’? Continuing. . .

Pradeep published the puzzle at 1:40 pm. Your solution was tendered at 3:14 pm the same day. Let’s say that you eagerly awaited the announcement of the new puzzle and instantly attacked it. Hmmm. 905 words in 94 minutes. And not Jack and Jill writing. Is this that simple for you, Jason? Most annoying is that, as I re-read your Mach Two response, I see elements of it that I buy into after only the third or fourth reading. I think this is marvelous in itself. Read it several times again today with the growing levels of appreciation and acceptance.

So there IS a mathematically sound way of approaching this problem which I can abide with. This is not to say that all of your assumptions are correct.

Easy to disregard the ‘seven’ typo in the first assumption. Your numbers correctly show three. Then you showed 2Λ2 -1 for the horizontal and 2 Λ3 -1 for the vertical. When you combined these, you showed 2 Λ5 -11. Did you mean 2 Λ5 -2 or 30? 32/(32+30) = 32/62 = 51.6 %. Chances are that I’m wrong again. Your expected loss is different but it’s still not bet worthy.

The math in the next paragraph ending with P(corner eel)~=1/(1+6p) looks good. Very good.

You moved then to approaches using conditional probabilities. In class, when we took this, the professor drew an inverted V on the board. He wrote Bayes at the top of the vertex and said, “In the world of statistics, this is where conditional probabilities exist. We can’t prove it but we know they’re there.” Something like that. Instantly, all of us revered conditionals. I enjoyed your walk-through. The next paragraph was a non-serious close. Nice. Your final statements reveal our differences. One eel for you. A million eels for me. . .each grander than the one before!

• br says:

To me, the coverage of the gaze is important. As the collector states that 'So there must be at least one eel in the vertical column and at least one in the horizontal row', it implies that his gaze only covers one canvas at a time. However, he could start from anywhere in the L shape and move his gaze either horizontally or vertically. I assign each canvas position a (row, column) number to get a (4, 3) size matrix, where the corner canvas is at (4, 1), as shown in the illustrating picture. If he starts off at (1, 1) (the top most canvas) and moves his gaze horizontally, he must see an eel. But there are no other horizontal pictures, so (1, 1)=eel. Similarly for (2, 1) and (3, 1). For the bottom row, if he starts at (4, 2) and moves his gaze vertically he must see an eel. But as there are no more pictures above (4, 2) to see then (4, 2)= eel. Similarly for (4, 3). That just leaves the corner square. As we have filled the other canvases with eels, then it actually doesn't matter whether (4, 1) is an eel or not. Hence without further information the probability is 0.5. Maybe the artist is sick of painting eels, or maybe she just photocopies another one, but as we have no prior knowledge of this then p=0.5 and you should take the bet.

• John Hebbre says:

Your development is interesting, br. At the front of the train you state: If he starts off at (1, 1) (the top most canvas) and moves his gaze horizontally, he must see an eel. But there are no other horizontal pictures, so (1, 1)=eel.
The collector states, “I would like to see a picture of an eel when I move my gaze vertically or horizontally.” If he has the option, I’m not certain 1,1 must be an eel since the collector could then look vertically. 1,1 could remain blank it seems.

• Alan Finn says:

The problem is obviously ill posed. The definition of “moving my gaze {horizontally | vertically}” does not specify the starting point.

Suppose the bottom row of pictures are denoted h_i, i=1,… and the vertical column of pictures are demoted v_i, i=1,…

Assumption: “moving my gaze” may start at any h_i or v_i.
Assumption: The artist’s preference for depicting an eel is p_0 from everything he/she might depict.

Since moving my gaze vertically from h_i, i=2,… traverses only 1 picture, that picture must be an eel. Analogously all v_i, i=2,… must depict an eel. Therefore, h_1 (= v_1) can depict anything. This will be an eel with probability p_0 by assumption.

• Neil says:

If I were the painter, I would simply paint an eel on every canvas and collect the dough.

• Colin P says:

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay." (Sherlock Holmes, The Adventure of the Copper Beeches)

• Mr. Crumbs says:

We're using math, not philosophy to solve this right? hmmm… I don't know, because nothing can determine what will actually happen.

• Ravi Kumar Meduri says:

I would not agree that the artist would draw randomly eel or any other creature with equal probability. Prior probabilities do not have a meaning as the artist is now working in constrained space. So the answer of 32/53 does not suit here.

For a moment, ignore that the corner canvas can be both in the vertical and horizontal. In the back of the mind of the artist, it is clear to him that he has to draw at least 2 eels out of 6 canvases. That works to one in a three and now we enter a path dependent estimation of probability. Suppose the artist draws the first canvas as eel (with a probability of 1/3), then he is free to draw what he wants in the remaining 2 canvases in the vertical line. Amongst the three horizontal canvasses again, the artist has to draw one eel and that is again 1 in 3. So the probability of the corner canvas being eel is:

1/3*1/3 (first canvas eel and horizontal random) + 2/3(first canvas not eel)*2/5 (second canvas eel)*1/3 + 2/3*3/5 (first two canvases not eel)*1/2*1/3+2/3*3/5*1/2*1(now if the first three canvases vertically are not eel, the fourth one must be).

This works out to 0.4666 and assumes that the artist has no special perspective or odds for the corner canvas. This is fairer than the 32/ 53 estimate as the latter underestimates the probability at which eels are drawn in the non-corner canvases.

• Moray says:

Since probability measures a set U to see what proportion of its members have a certain characteristic, I will begin by defining that set. The question posed is about what an artist will do. So in this case U is a set of artists. Practically speaking it's a subset of all artists in the world. U can consist only of those in the art patron's labor market of artists. U is not some set of frame configurations nor is U some set of sea species. It's artists. So what proportion p of U would execute the project in such a way that an eel is painted in the corner picture? Ask them. I agree with Pradeep Mutalik: if you want a real world answer you have to get data. I know that both "data" and "poll" are just four-letter words, but sometimes you have to get your hands dirty. Once you have a decent estimate of n, the number of members in U, standard sampling techniques and your desired level of confidence can be used to decide on a sample size, s. You then ask a sample of U of size s how they would execute the project. The proportion in the sample then lets you infer (plus or minus some uncertainty) the proportion p of U that would paint an eel on the corner canvas. When my wife goes to her art class tomorrow I'll try to get some data and work the numbers.

• Colin P says:

@Moray
Bravo Eel!

• eJ says:

We should probably take a moment to reflect on where the "probability" is coming from here. It is composed of at least two factors: (1) the probability of hiring a particular type of artist, and (2) the probability, once hired, of that particular artist drawing an eel in the corner. Which of those has already happened? There are three possibilities:

1. The artist hasn't been hired yet. Then, as others have said, a survey is about the only reasonable way to get data. Be careful though! If you survey too few people, you'll likely end up with a badly skewed response. So there's a subjective part here, too, namely in setting a reasonable prior distribution on the probability of an eel in the corner.

2. The artist has been hired and has produced the work. Here, we hit ambiguity in the definition of "probability". A frequentist would say that the objective probability of there being an eel in the corner is 0 or 1, we just don't know which. Alternatively, we might rather know the subjective (Bayesian) probability, i.e. a measure of the strength with which we believe an outcome to have occurred, in which case we're back to case 1 above.

3. The artist has been hired but has yet to start work. If it's subjective probability you want, see case 1. Otherwise, the frequentist answer would be that it's some probability (objective, because the event hasn't happened yet) that the hired artist will paint an eel in the corner: so there are as many different answers as there are types of artist, including any of the quite reasonable models of artist behaviour given in the responses above.

I suspect Pradeep had interpretation #2 in mind, and is a Bayesian at heart, though I preferred the frequentist interpretation of #3 for the fun it gave me in the analysis of hypothetical painting styles.

• Bob64 says:

I have yet to see a conclusive solution of the infamous "Sleeping Beauty Problem" with both "thirders" and "halfers" seeming to claim victory. It also seems to satisfy your idea that issues arise when we confuse frequentist and Bayesian probability. Any thought?

• Colin P says:

My original hypothesis under the current state of knowledge at the time (not long after this puzzle was published), and there being no better information available, was that the probability sought would be based on random placement of eels and non-eels. Under that hypothesis the probability of an eel appearing on the corner canvas I believed would be 32/53 or about 0.6. But data gathered since then suggests that hypothesis is somewhat lacking. My broad reading of the answers above suggest four hypotheses for how the artist might work: "One eel in the corner", "An eel on every canvas", "Random frequentist", "Other". But many of the answers are non-artistic and are written purely by statisticians surmising how artists might work. Amongst the artist-like actual answers above I think there is a preponderance of would-be artists who would prefer to act under either of the first two of these four hypotheses, in either case of which an eel lies in the corner.

Anyhow, looks like the probability that my hypothesis is right needs to be adjusted. I'll adjust it by a factor equal to (Probability of an eel in the corner under my hypothesis) divided by (Weighted probability of an eel in the corner under all the hypotheses). The logic being that if the lower term (weighted probability) is less than my hypothesis probability then the likelihood that my hypothesis is right is increased because the probability arising from my hypothesis is higher than the weighted average; but if the lower term is higher than my hypothesis probability then the likelihood that my hypothesis is right is decreased because the probability arising from my hypothesis is lower than the weighted average; and finally if the lower term is equal to my hypothesis probability than my hypothesis is as good as the average.

Regarding this lower term, i.e. the weighted probability of an eel in the corner under all the hypotheses, my feeling is it's about 0.8. I arrive at this figure by listing the results and by a "feel" for the weightings to attach to this data. I recognise this is subjective, a more objective approach would be to tabulate the data according to the various hypotheses, and work out the weightings for each.

Anyhow, given my figure of 0.8 I can also say that the sum (Probability of an eel in the corner under my hypothesis) divided by (Weighted probability of an eel in the corner under all the hypotheses) is (32/53) / (0.8) ≈ 0.6/0.8 = 0.75, and therefore under this reading the probability that my original hypothesis is correct is diminished by 25% or a quarter. It looks like I need a new hypothesis, one that results in a somewhat stronger probability of an eel in the corner at about 0.8.

• Fred Viole says:

Bradley Efron gives an excellent example of the influence of the prior distribution (quote and link below). Seeing as the comments here are all over the place from "I would paint 1 eel in the corner" to "I would paint 6 eels" is not doing much to move us from the Uniform prior of the artist's preferences over the range [0,100] for the probability of painting an eel on a given canvas.

If everyone shared the preferences of Pradeep, "As for me, I would probably do 2 or 3 and perhaps 4, with a much lower likelihood." that would put us in the sweet spot of artist preferences required for a positive expected value of the bet ~[0.11,0.62].

"Figure 3 graphs three different prior distributions for p: the doctor’s delta function
at p = 1/3, Laplace’s flat prior, and Jeffrey’s U-shaped prior density. Of course
different prior distributions produce different results. My answer to the physicist,
that she had 50% chance of identical twins, changes to 58.6% with Jeffreys prior,
and to a whopping 61.4% with a flat Laplace prior."

http://www.ams.org/journals/bull/2013-50-01/S0273-0979-2012-01374-5/S0273-0979-2012-01374-5.pdf

I would like to make 4 points:

1. Single data points do not a probability estimate make.
2. The probability of a given outcome may be different depending on the method by which the outcome is achieved.
3. We should not try to shoehorn real-world situations into simplistic probability models.
4. Since the accuracy of probability models improves with increasing knowledge, collecting more data is always helpful.

Single data points do not make probability estimates

Probability is the science of considering all possible outcomes, and by using the principle of indifference when necessary, coming up with a model that gives the best results across the ensemble of all eventualities.

So, if you consider only your own possible preference or guess or speculation or opinion or hunch, and declare with complete confidence that that the corner canvas definitely will or will not have an eel, you ignore the fundamental fact that different people will do things in different ways. This is amply illustrated here: just peruse the comments! The likelihood of a hunch being right will likely be much lower than a well-constructed probabilistic model that considers other points of view. This is the fundamental reason why ensemble models can be far more accurate than individual guesses. A famous example of such a model is the one built by Nate Silver for the U.S. presidential election of 2012 (http://www.theguardian.com/science/grrlscientist/2012/nov/08/nate-sliver-predict-us-election) which was far more accurate than the individual polls it used, and of course much, much more accurate than the personal pronouncement of political pundits.

On the other hand, every personal preference is a valuable data point, and the more data you examine, the better your probabilistic model can be in practice. Like the previous puzzle, this one two has two aspects: the purely mathematical and the real-world. The importance of data for the real-world models is obvious. But even if you do not use the data or cannot collect enough of it, it can give you insight about whether your mathematical approach and assumptions are justified.

The probability of a given outcome may be different depending on the method in which the outcome is achieved.

This, of course, is the lesson of Bertrand’s paradox. The principle of indifference can be applied in different ways: each one of the solutions of Bertrand’s paradox is the correct solution for a particular way of “counting” the chords – and each of these will be a correct mathematical model for an actual real-world situation.

So what we need to do here to solve the mathematical problem is try to come up with different ways in which the task could be done, pick the one that is likely to be used (or use a weighted sum of different methods), and then apply the principle of indifference, assigning equal probability to all the different ways that the method could actually be applied. Nothing else can be assumed. As Malachi put it “Reasoning on the basis of the artist's state of mind without evidence is…effectively adding an additional constraint and, thus, not addressing the original problem.”

Readers have come up with several possibilities.
1) The canvas method (the majority of commenters)
Assumption: Each canvas may or may not have an eel with equal probability.

2) The number-of-eels method (Kyle Sinclair’s “Indecisive Painter”, Davide Bassanini, Malachi);
Assumption: Each possible number of eels from 1-6 is equiprobable.

3) Kyle Sinclair’s “Blind Painter”:
Assumption: The painter paints eels on randomly selected canvases until the requirements are satisfied, then paints non-eels.

Which one is best?

Do not try to shoehorn real-world situations into simplistic probability models.

The canvas method would have been the right answer if this were a mathematics textbook and the problem were described as follows.

There are a large number of paintings, half of which have pictures of eels. Six of these are picked at random and assembled into an L shape with 4 vertical pictures and 3 horizontal pictures. Both the vertical column and the horizontal row have at least one eel. What is the probability that the corner painting is an eel?

But this is not how our problem was described. The constraint of at least 1 vertical and at least 1 horizontal eel was clearly present from the outset. Does anyone seriously believe that any artist would ignore that constraint, assemble the canvasses at random and then say – “oops that doesn’t satisfy the instructions, let me try again”?

On the other hand the number-of-eels method seems eminently reasonable. Without further information, the next mathematical step is to apply the principle of indifference, consider each case to be equiprobable and construct the probability model as Kyle, Davide and Malachi did.

Interestingly, this method flips the result for the bet question compared to the canvas method.

As for the third method, like the first, it takes a rather dim view of human intelligence ☹.

Since the accuracy of probability models improves with increasing knowledge, collecting more data is always helpful.

Of course, the mathematical models we’ve discussed probably still do not accurately reflect what will actually happen. Probabilities are only a reflection of our ignorance: if you know something for certain, you don’t need probabilities any more. But as Nate Silver has shown, we can get pretty accurate results with good real world data. Our probability model can get more accurate if we incorporate how people and specifically, artists behave. No amount of speculation about aesthetics and artistic preferences can replace actual data. This is the heart of Bayesian probability – we convert our “prior probability” based on our assumptions to a “posterior probability” based on real-world data.

This was my motivation for starting the poll, and I see that some intrepid readers such as Francois and Moray (nice name!) have thought along the same lines. It’s unlikely in practice that we can get enough reliable data, but even the small number of responses we have from readers so far supports the plausibility of the number-of-eels method as a first approximation. My subjective speculation is that the graph of the actual probability of number of eels painted by actual artists will be U-shaped with a peak at 2 or 3, decreasing through 4 and 5, with a small bump at 6 to account for creative interpretations such as Jason’s one eel across all 6 canvasses.

Anyone up to totting up the numbers here? And any result from the artist’s poll, Moray?

• eJ says:

@Bob64, let's up the ante. Upon waking, Sleeping Beauty is presented with two bags of beans, marked "H" and "T", and told she must reach into one bag, grab a bean, and eat it. She is told that the bags are filled with jellybeans, J, or lethal kill-pills, K, as follows:
* if the coin was Heads: bag "H" = 6J, bag "T" = 6K
* if the coin was Tails: bag "H" = 1J+5K, bag "T" = 5J+1K
The "halver" thinks "H" is the right choice, giving a survival probability of 7/12; the "thirder" thinks "T" is the right choice, giving a survival probability of 5/9. If, unluckily, you were the participant in this experiment, which would you pick?

@eJ and Bob64,

Thanks for bringing up the Sleeping Beauty problem. This is an amazing problem, well worth a Quanta Insights column, the more so because it doesn't seem to have produced a clear consensus.

The reason, I think, is similar to Bertrand's problem and this one: both answers are correct depending on the real-world method by which you are going to test the probability. In eJ's scenario, you would need to know how often Sleeping Beauty would be tested with the beans – every time, or randomly say, once in 10 wakings. The answer would be different in the two cases.

I am a thirder, but if tested with the beans on every waking, I would agree with the halvers: H is the right choice.

• Moray says:

No results from the poll of artists yet. One of the area's professional artists is helping with the survey. I hope to pick up some results this coming Friday. I have learned from him that there are an estimated 80 pro and 500 semi-pro artists in the labor market. The sample size will probably not come in big enough to provide a narrow confidence interval around the point estimate of p. But it's the approach and methodology I'm hoping to illustrate.

• eJ says:

@Pradeep and Bob64, I added the beans experiment to the Wikipedia "Talk" page for the SB problem here:
https://en.wikipedia.org/wiki/Talk:Sleeping_Beauty_problem#Beans_experiment
It's a better-specified version than I offered to Bob64, testing SB on every wakening, which I think gets more directly to the heart of the "credence" question.