The second Insights puzzle, “The Slippery Eel of Probability,” was designed to show how some probability problems have more than one correct answer. The puzzle went as follows:

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?

As in the previous Insights puzzle, this can be analyzed both as an idealized, pure mathematical question and as a real world one. The idealized view ignores aesthetics and artistic preferences: the objects could just as easily be chickens in boxes or stones on a game board. The only thing that matters is the ratio of the configurations that satisfy a given condition divided by the total number of all configurations. This abstract approach is what gives mathematics its power and generality. But it’s also responsible for math’s reputation of other-worldliness in popular culture, as expressed in the popular meme: “Math. The only place where people buy 60 watermelons, and nobody wonders why.” Art collectors who love eels? No one would look at them twice in the mathematical universe!

But even in this abstract world, there are different methods to making these configurations. We know that there are six objects that must be arranged in an L shape with a constraint: There has to be at least one object vertically and one horizontally that have a common property (an eel in this case). What frequency do we assign to the common property? When do we address the constraint?

Let’s consider two possible methods to accomplish the task.

**The Canvas Method**

On every canvas, the artist either draws an eel or not, at random (with fifty percent probability per the principle of indifference). She arranges the pictures in the L shape, and rejects the configuration if it doesn’t satisfy the constraint. When she finds a configuration that works, she stops.

Let’s call a picture of an eel a 1 and any other picture a 0 and represent the L shape, from top left to bottom right, as a string of six binary digits. The corner canvas is at the fourth position. The arrangement in the illustration at the top of the puzzle can be coded as 010010, with the zero in fourth position representing no eel in the corner. Now there are 2^6 = 64 possible configurations, which are nothing but the binary numbers from 0 to 63 represented with six digits. Eleven of these configurations do not satisfy the constraint of one vertical and one horizontal eel: 000000, 000001, 000010, 000011, 001000, 010000, 011000, 100000, 101000, 110000 and 111000. Out of the remaining 53 configurations, 32 have a corner eel (those with their fourth bit set to 1) and the remaining 21 do not. The probability of a corner eel is thus 32/53 or 60.4 percent.

This method, mentioned in 15 comments, was by far the most popular among readers.

**The Number-of-Eels Method**

The artist decides how many eels she wants in the drawing and then considers only those arrangements that satisfy the constraints.

Using the principle of indifference, we assign an equal probability of 1/6 to every number of eels that can be used from one through six. Within each of these scenarios, we then determine whether there is an eel in the corner or not.

With one eel, there is only one way to satisfy the condition — it has to go in the corner. The probability is 1.

With two eels, there are 11 possibilities of which five have a corner eel. The probability is 5/11.

With three eels, the probability is 10/19.

With four eels, it is 10/15.

With five eels, it is 5/6.

With six eels, the probability is 1.

The combined probability of a corner eel is thus 1/6 of (1+5/11+10/19+10/15+5/6 + 1), which works out to be 74.7 percent.

Note that the total number of arrangements here is again 53, but they are weighted quite differently. This method was used by three commenters: Kyle Sinclair, Davide Bassanini and Malachi.

The original problem included the following bet:

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.

According to the canvas method you should take the bet: Your winning expectation is $0.19. By the number-of-eels method, you should shun the bet: You would expect to lose $0.24.

So what we have here are two simple mathematical models that could both be correct in different circumstances. It all depends on which method is used by the artist. One of the pitfalls of applied mathematics is trying to make the real-world situation fit the mathematical model, rather than the reverse.

It was with this in mind that I had suggested a poll to see which method was more appropriate here. Here is my analysis of the number of readers who prefer a given number of eels (I assigned fractional numbers when readers gave more than one alternative):

Eels Readers (out of 24 total)

1 7.5

2 8

3 1

4 1

5 1

6 5.5

It’s clear that the number-of-eels method, though still a long way off, is a far better approximation than the canvas method. The canvas method gives a weight of 1/53 (or 0.02) to the unique cases of one eel and six eels. The poll weights them at 0.31 and 0.23, respectively. The number-of-eels method weights all eel numbers at 0.17. It’s highly likely that the first decision an artist would make is how many eels to paint.

Still, our mathematical model requires significant tweaking to accurately reflect reality. As I said in the comments, probabilities are only a reflection of our ignorance: If you know something for certain, you don’t need probabilities. But we can get fairly accurate results with good data. Our probability model will be more accurate if we incorporate how people (specifically artists) behave, as commenter Moray mentioned. 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 assumptions to a “posterior probability” based on real-world data. Perhaps we can poll artists to explore these methods, as Moray is trying to do.

What about the bet? Is there a way of thinking about it rationally, rather than just going with a hunch? First we should construct a probabilistic model that considers all possible points of view and not just our own personal preferences. Ensemble models tend to be far more accurate than individual guesses. A famous example of such a model is the one built by Nate Silver for the 2012 U.S. presidential election, which outperformed the individual polls it used and, of course, vastly outperformed the personal pronouncements of political pundits.

Second, if we are forced to make a choice on the basis of limited information, we should weight our ensemble model with elements that reflect aesthetics, artistic choices and so on, as some readers have done. Some people will be able to do this better than others, and it will be subjective, but it’s the best we can do. Beyond choosing the number of eels, using the principle of indifference for the choice of arrangement is probably not quite correct either. This may well be a critical factor. I think that artists who choose three eels will prefer a corner eel, while those with two or four will not. Putting all these considerations together, I estimate the probability of a corner eel to be somewhere between our two estimates above. The bet? A tossup. In the absence of hard data, you have to stop taking the numbers too seriously, no matter how much you love them. Using numbers to a precision beyond what their reliability warrants is another common pitfall in applied mathematics.

The reference to chickens in boxes above was not random. A similar problem to this one, in fact called “Chickens in Boxes,” was posed by a reader to the famed columnist Marilyn vos Savant. The problem asked for the probability of having a chicken in the corner box of an L-shaped arrangement of four boxes, with three in the vertical column and two in the horizontal row, given that both the row and column contained one chicken. I came across it in Paul J. Nahin’s excellent book on probability, “Will You Be Alive 10 Years From Now?” According to Nahin, vos Savant never answered the problem. Nahin’s answer was the conventional one, represented by the canvas method above. Changing the “one object in both directions” condition to “at least one in both directions” in the current problem,highlights the question of how many total objects are used, which draws attention to the other variable that we can apply the principle of indifference to.

The *Quanta* T-shirt for this puzzle goes to Malachi, who clearly contrasted the two main methods of approaching the problem that we detailed above and gave the nod to the second method. Congratulations!

Sleeping Beauty problem very interesting. Personally I think it shows that the results from a detection apparatus can be misinterpreted pending more knowledge from other experiments. As someone somewhere said (not me) there are two different questions: What is the chance of a Head, and What is the chance of being woken by a Head?

Thanks for proposing the puzzle, Pradeep. I was able to sample 7 of the 80 pro artists in our area. Only 3 of the 7 respondents would have painted an eel on the corner canvas. So that's 43%, significantly lower than either of the proposed solutions. (We have a saying on the reef where I live: "Reality bites.") Anyway, a larger sample is needed to 1) get a more accurate estimate of the probability and 2) put a reasonable confidence interval on that figure. Suppose the sample size could be increased to 21. And suppose 9 of those sampled would paint an eel in the corner canvas. That's again the 43% figure. Now we can get an approximate 90% confidence interval on the estimated probability that one of the randomly selected 80 painters would paint an eel on the corner canvas. That interval is 43% +/- 18%. So the number of eels model almost certainly over-estimates the probability for my local reef. And the canvas model looks highly likely to also provide an over-estimate. As you've said, there's just no substitute for getting data whenever possible.

@Moray, 3 out of 7 is not significant evidence against the canvas method: you would get 3 or lower (out of 7) 28% of the time if the true probability was 32/53.

running this matlab script I chalked up led to neither of the results…perhaps there is something wrong with how I decided to solve it?

L = zeros(1,6)

cnt = 0;

iter = 100000

for i = 1:iter

h_or_v = randperm(2)(1);

%choose 1 1 to be horizontal

if(h_or_v == 1)

cnt = randperm(4)(1);

L(cnt) = L(cnt) + 1;

else

cnt = randi([4 6]);

L(cnt) = L(cnt) + 1;

endif

if(cnt != 4 & h_or_v == 1)

cnt = randi([4 6]);

L(cnt) = L(cnt) + 1;

endif

if(cnt != 4 & h_or_v == 1)

cnt = randperm(4)(1);

L(cnt) = L(cnt) + 1;

endif

endfor

L= L/iter

Update to the original comment there was an error, this code seems to converge to

P(vertical) + P(horizontal) – P(vertical and horizontal) ~ .50

L = zeros(1,6)

cnt = 0;

iter = 100000

for i = 1:iter

%horizontal or verticle pick

h_or_v = randperm(2)(1);

%choose 1 to be vertical

if(h_or_v == 1)

cnt = randperm(4)(1);

L(cnt) = L(cnt) + 1;

else

cnt = randi([4 6]);

L(cnt) = L(cnt) + 1;

endif

%if the index wasnt 4 ( the common canvas ) then we need to pick one from

% the horizontal 3 if the original picked was vertical

% or pick one from the vertical if the horizontal was originally picked

if(cnt != 4 & h_or_v == 1)

cnt = randi([4 6]);

L(cnt) = L(cnt) + 1;

endif

if(cnt != 4 & h_or_v == 2)

cnt = randperm(4)(1);

L(cnt) = L(cnt) + 1;

endif

endfor

% calculate the probabilities of L canvas 1-4 being vertical 4-6 being horizontal

L= L/iter