Insights Puzzle

How Many Half-Lives Do You Have?

Gaining an intuition about half-life requires some unintuitive thinking.

Illustration: bucket half-full of amoebas.

Olena Shmahalo/Quanta Magazine

Do protons live forever or do they decay with a half-life of around 16 billion trillion trillion years? That’s an eternity considering the universe is thought to be less than 14 billion years old. Yet, as Natalie Wolchover recently described, the fate of physicists’ beloved grand unification theories — the idea that the forces of nature were unified at the beginning of time — rests on finding that protons are in fact mortal after a humongous half-life. It’s a question scientists have yet to settle.

I find the half-life concept fascinating. It forces us to think in unintuitive ways. Let’s start with two simple questions that get to the essence of what a half-life is. You have to answer both questions in just one minute (consider half a minute to be each question’s half-life).

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.

OK, start your half-life timer. Question 1:

In a bucket filled with nutrients is a bunch of amoebas. Every day, their population doubles. If it takes 48 days for the bucket to be filled with amoebas, how long did it take for the bucket to become half full?

Question 2 is based on the title of this puzzle:

How many half-lives does a pound of radioactive material have?

If you’ve never seen these types of questions and you stuck to the time limit, you may have jumped to the wrong conclusion.

The first question is very similar to one of three problems that constitute the Cognitive Reflection Test, originated by Shane Frederick, a decision theorist now at the Yale School of Management. According to Frederick, there are two general types of cognitive activity, the first of which is carried out automatically, without reflection, while the other requires conscious thought and effort.  Each of his three problems has an obvious or impulsive response, which originates from the automatic process, but is incorrect. In order to activate the second process, you must recognize that your first answer is incorrect, which requires you to reflect on your cognition. I remember encountering a question like the first one as a 7- or 8-year-old proud of my arithmetical skills. I fell for the bait hook, line and sinker, utterly confident of my answer, which turned out to be the obvious, incorrect one. I’ve been trying to reflect ever since.

In order to intuitively understand and make sense of half-life, like the Cognitive Reflection Test, you have to reject the first impulsive answer and try to consciously reach the right way of thinking, eventually making it part of your intuition or “mental heuristics” toolbox.  It’s well worth the effort. If you need to peek, click on “Answer 1” and “Answer 2” below.

Answer 1: 47

Answer 2: Theoretically, infinite. In practice, a very, very large number (we’ll discuss more precise answers in the solution column).

To think naturally about half-life, first, remember that it arises wherever there is exponential decay, or, as with the amoebas, exponential growth. We are intuitively more accustomed to linear change. Second, there is nothing magical or sacrosanct about decreasing by half — we use this fraction only for convenience. The concept and the mathematics work just as well for any other fraction, such as a third or a fifth. A more natural fraction, used in many mathematical and physical applications (such as the time constant in resistor-capacitor circuits), is 1/e, where e is the base of natural logarithms, 2.71828…. (1/e is about 0.368, or 36.8 percent). A final important concept is that the behavior that gives rise to half-life in a group is actually completely dependent on the properties of the individual. When a bunch of radioactive atoms decrease to half their number after one half-life is over, each atom knows nothing about any other atom, or the size of the group it belongs to. The group’s magical reduction to 50 percent every time one half-life passes emerges simply out of each atom’s exact same probability of decaying within the given time, independent of all the others.

The fact that particle decay is probabilistic means that, as in the case of proton decay, you just have to collect a large bunch of them and hope that some of them disintegrate within the time you have at your disposal. If the particles don’t oblige, there’s nothing you can do. That’s why one scientist compared the lack of results in the proton decay to waiting for your spouse to come home: “If they’re 10 minutes late, there’s simple explanations for that. An hour late, maybe those explanations become a little less plausible. If they’re eight hours late … you begin to worry that maybe your husband or wife is dead. So the point is, at what point do you say your theory is dead?”

While we have no grand unified theory to grind, here are a couple of questions to get the flavor of half-life calculations for those who need real numbers to sink their teeth into.

Question 3:

Let us say you manage to obtain 5 atoms of the radioactive isotope of unobtainium, of Avatar fame. After exactly one year, 2 atoms have decayed. You want to figure out the half-life of the substance, and like a true scientist you seek a range that has a 95 percent probability of containing the true value. What is your range for the plausible half-life of unobtainium?

Question 4:

In the above scenario, for your second year, you obtain 30 atoms of the substance. You need to divide it into three portions — A, B and C — according to the following rules. When you inspect the portions after exactly one more year, you would like A to have exactly 6 undecayed atoms, B to have 7 or 8, and C to have 4 or 5. Remember, you don’t know the half-life exactly. How many atoms must be in A, B and C initially in order to maximize your chances of getting the precise results you want?

That’s all for now. Happy New Year, and happy puzzling! The solution column will arrive with a half-life of 15 days.

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. (Update: The solution is now available here.) 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).

Note that we may hold comments for the first day or two to allow for independent contributions by readers.

View Reader Comments (25)

Leave a Comment

Reader CommentsLeave a Comment

  • Q3. The way the question is phrased ("seek a range that has a 95% probability of containing the true value) means that what is required is a Bayesian posterior (or credible) interval rather than simply a confidence interval (meaning that if the experiment was repeated multiple times the confidence interval calculated would contain the true value 95% of the times). This makes the maths tricky (Poisson and gamma distributions come into it) and as I'm on holiday, I don't feel like digging into that. So, let's think a bit & try and work out something that could make sense. 0 to infinity will mean better than 95% probability. 0 itself is really silly, so we really mean >0. However, close to 0 is possible, but if we are talking only a few seconds or minutes it is unlikely that the 3 atoms that didn't decay would have hung around long enough, so probably a few months would be a good starting point.
    If the half-life was say, 3 months, then the probability of an atom decaying within 1 year would be 1-(0.5^(12/3))=0.9375 or ~94%. So with 5 atoms chosen at random 4.7 on average would decay in a year. ie mainly 5 or 4, with occasionally 3, 2,1, or even 0 very seldomly. We can play with the length of the half-life a bit and find when 97.5% of the atoms are expected to decay and come up with 68 days. This is our lower limit. We can do something similar for the upper limit and come up with about 27 years. So, even if the maths isn't exact, I think that it is plausible that there is a 95% probability the half life is between 68 days and 27 years.

  • Before going into exponential decay, some more basic math first. You asked your artist to draw a bucket half filled with amobea, but she drew a bucket filled to height 1/2. Why is this wrong, and at which height is a bucket filled to 50%? Could be a new puzzle.
    My second remark is, that exponential decay is a statistical law. Doing only one measurement and working with so few atoms I do not consider very scientific.
    Applying exponential decay gives an exponential decay constant lambda of ln(5/3) and a half life T(1/2)=ln(2)/ln(5/3)=1.36 years. As a range I suggest taking lambda from ln(5/2.5) to ln(5/3.5) resulting in half life T(1/2) = 1 … 1.94 years. Taking lambda from ln(5/2) to ln ln(5/4) yields a larger span of T(1/2) = 0.76 … 3.11years.
    I take partition 7, 10, 13, as a quick calculation yields 7.5, 10, 12.5.

    This is just on a gut level, perhaps I dig deeper into it on the weekend. But now back to work. Nice lunch break, though.

  • Hello everybody!
    I would like to submit my answers for questions 3 and 4.
    I haven't studied physics or chemistry or anything related so it's pure logic thinking. But i guess that's close enough.
    So, the halflife of the Unobtainium atoms is in the range of 1-2 years, but i prefer to stick with 1.5 years. This is because 2 out of 5 (40%) decayed in the first year, so it's plausible to think the same can happen next year, doubling the decay (have no clue if the reasoning is right!!), thus it's plausible that in 1.5 years half the atoms will have decayed. Since half of 5 is 2.5 and it's not exactly possible (yet?) the range is 1-2 years.
    Following this reasoning, question 4. I would split the 30 atoms as follows: 10 in group A, 12 in group B and 8 in group C. Following what we said in question 3, if after one year 40% decayed, we can say that there's a good chance that respectively 4,5 and 3 will decay in the groups, thus giving approximately the desired results.
    Ta daaaaaaaa
    Thanks for the questions, it was a good half hour ( no pun intended) diversion from study.

  • @NB Thank you for bringing this issue to our attention. You should now be able to click (or tap) to reveal the answers. Please let us know if you run into any additional problems.

  • The correct answer to question 1 is "no time," as it is clearly stated at the beginning that the bucket is already "filled with nutrients."

  • With 5 atoms decaying to 3 atoms, statistics may be assumed. Going with Gaussian, for a sample of 5, Confidence and Reliability are very low and perhaps meaningless for determining A, B, and C as required.

  • Question 3:

    For this problem I used a Bayesian approach.

    t = half-life in years
    x = probability an atom does not decay in one year = 2^(-1/t)

    Therefore, t = -1 / log2(x)

    Given n nuclei and x, then the probability that k nuclei do not decay in one year is given by the binomial distribution:

    P(k|n,x) = (n choose k) x^k (1-x)^(n-k)

    Now P(k|n,x) is the probability mass function that gives the probability of observing k undecayed nuclei *given* x, but we actually want a probability distribution function (PDF) for x *given* k. So we need to use Bayes's Theorem:

    P(x|n,k) = P(k|n,x) P(x|n) / P(k|n)

    Without information about k we would have no reason to prefer particular x values, so we should pick the flat prior over x=0..1, which is P(x|n) = 1. Likewise with no information about x we would have no preference for a particular k, so we pick the flat prior over k in {0, 1, …, n}, which is P(k|n) = 1 / (n+1).

    So the PDF for x given a measurement k is:

    P(x|n,k) = (n+1) (n choose k) x^k (1-x)^(n-k)

    Integrating P(x|n,k) to find the mean x yields x_mean = (k+1)/(n+2). However the most *likely* value of x is where the derivative of dP/dx(x|n,k) is 0, giving x_max = k/n. The most likely answer is also the naive answer.

    To find a 95% confidence interval, we can take different approaches. The usual way is to find an x0 and x1 such that the integral of P(x|n,k) for x=0..x0 is 0.025 and the integral of P(x|n,k) for x=x1..1 is also 0.025. That is, (x0, x1) bounds the middle 95% of the PDF. However, because P(x|n,k) is not symmetric in general, P(x0|n,k) may be greater or less than P(x1|n,k), so more likely values of x might be excluded from the inner 95th percentile.

    Instead we find a probability threshold y such that P(x0|n,k) = P(x1|n,k) = y and the integral of P(x|n,k) for x=x0..x1 is 0.95. This means the 5% least likely values of x will be excluded. This required the use of a computer. In the end, I found P(x0|5,3) = P(x1|5,3) = 0.472 gives (x0, x1) = (0.239, 0.895) and an integral of 0.95. This interval for x leads to a half-life interval of (0.484, 6.26) years.


    Question 4:

    Now we have a group of a different size m. The probability after a year that j do not decay is given by the binomial distribution:

    P(j|m,x) = (m choose j) x^j (1-x)^(m-j)

    However, we do not know x. Instead we know its probability distribution based on our previous measurement:

    P(x|n,k) = (n+1) (n choose k) x^k (1-x)^(n-k)

    So the appropriate distribution for j is:

    P(j|m,n,k) = integral((m choose j) x^j (1-x)^(m-j) P(x|n,k) dx, x=0..1)
    = (n+1) (n choose k) (m choose j) integral(x^(j+k) (1-x)^(n+m) dx, x=0..1)
    = (n+1) (n choose k) (m choose j) / ((n+m+1) ((n+m) choose (j+k)))

    The probability of ending up with the given outcomes is:

    P(m_A, m_B, m_C) = P(6|m_A,5,3) (P(7|m_B,5,3) + P(8|m_B,5,3)) (P(4|m_C,5,3) + P(5|m_C,5,3))

    If we can leave out some atoms from the three groups, this probability can be maximized by the following grouping:

    m_A = 8 or 9
    m_B = 11
    m_C = 6

    If we need to include all 30 atoms in the three groups, then the probability is maximized by the following grouping:

    m_A = 10
    m_B = 13
    m_C = 7

    Thanks for the interesting puzzle!

  • Q3:
    Applying exponential decay gives an exponential decay constant lambda of ln(5/3) and a half life T(1/2)=ln(2)/ln(5/3)=1.36 years. As a range I suggest taking lambda from ln(5/2.5) to ln(5/3.5) resulting in half life T(1/2) = 1 … 1.94 years. Taking lambda from ln(5/2) to ln ln(5/4) yields a larger span of T(1/2) = 0.76 … 3.11years.
    I take partition 7, 10, 13, as a quick calculation yields 7.5, 10, 12.5.

  • Q3: For the two atoms that have already decayed by exactly one year, their half life is less than or equal to half a year. For the three that have not, their half life is greater than half a year. The atoms are identical so the half life is probably around half a year. The sample size is small (5) so it's hard to say what the 95% confidence interval should land. If I have to guess I'd put down (0.3,0.8) years. It feels like there is not enough information to make that call.

    Q4: For A, I would put 10 atoms since 3 out of 5 atoms decayed within a year in Q3. That's the only piece of information I have so I have to "maximize" the likelihood by assuming it's going to happen again. For B&C, we have 20 atoms to allocate and need to end up with 11 – 13 atoms. If I allocate 12 atoms to B then B could end up with 3/5*12 = 7.2 undecayed. The remaining 8 atoms to C so I could end up with 8*3/5=4.8 undecayed in C.

  • I see you used my junior high school science project as a beginning reference point to discuss something that just is not pertinent to my mind. Well, I have my reasons… here is my half life equation dilemma:
    If the sun in 4.5 be years old and by all other agreements in the research has a life span of 8 billion years. What do we know about power plants and half-life cooling periods with exponential scenarios where degrees and seconds means utter disaster.
    Am I approaching half life conversations properly?

  • Question 3
    2 atom =1 year at 95% probability
    1atom = 6 month at 95% probability
    0,5 atom= 3 month at 95% probability
    0,1666 atom = 1 month at 95% probability

    1 year and 3 months is the solution

    Question 4
    5-2= 3

    3 : 0,1666 = 6: a
    a= 0,3332
    a*12= 4
    Group A= 10 atoms

    3 : 0,1666 = 8: b
    b= 0,4442
    b*12= 5
    Group B= 13 atoms

    3 : 0,1666 = 4: c
    c= 0,4442
    c*12= 3
    Group C= 7 atoms

  • Question 3:
    To answer this question, I think we need a prior idea on what the distribution of half-lifes are in nature. I have no idea what this is, so I'll use a remarkably uninformative prior: I'll assume that if I'm given an atom of a new substance, it has an x chance of decaying in a year, where x is uniformly distributed between 0 and 1. (Yes, I'm putting my prior on decay rate rather than half-life, as I find decay rate easier to think about. A prior on decay rate is equivalent to a prior on half-life.) In other words, some substances have a half-life that means they almost instantaneous decay; some practically never decay; the "median substance"'s (whatever the heck that is!) half-life is one year.

    Now I observe 2 out of 5 atoms of a new substance decay in a year. I need to incorporate this data into my distribution of x. Previously, the probability of an atom of this substance having a decay rate of x was the same for all Xs. Now, we need to change the probability; the new probability of the substance having a decay rate of X is proportional to x^2 * (1-x)^3 for the two atoms that decayed and the three that didn't. (Actually, there are combinatorics in there, like 3 choose 5, but we can ignore them because they're the same for all Xs. That's why I said the probability is "proportional to". We can ignore those multiplicative constants as long as we normalize the distribution so it sums to 1.)

    To normalize the distribution, we take the integral from 0 to 1 of x^2 * (1-x)^3. This yields 1/60. Thus, the actual distribution of Xs is x^2 * (1-x)^3 / 60.

    Now we just need to find the lower and upper bounds of the 95 percent confidence interval. I'll find a symmetric confidence interval, with equal probability of erring on either side. This means I need to find a bound (a, b) such that (0, a) has 2.5% of the probability distribution of X and such that (b, 1) has 2.5% of the probability distribution of X.

    This means the integral from 0 to a of x^2 * (1-x)^3 / 60 has to be 0.025 and, likewise, the integral from b to 1 of x^2 * (1-x)^3 / 60 has to be 0.975. Evaluating these equations (I did the integral symbolically then used an online high-order polynomial solver) yield a = 0.11(rounded) and b = 0.77 (rounded).

    Converting these back to half-times yields bounds of ln(2) / 0.11 = 6.3 years and ln(2) / 0.77 = 0.9 years. I.e., the 95 percent confidence interval is 0.9 years to 6.3 years. Note that the best guess of a half-time is based on a probability of decay of 0.4 (2 atoms out of 5), which is a half-life of ln(2) / 0.4 = 1.7 years.

    Frankly, I don't like using such a naive prior (though I'm happy it made the math easy! 🙂 ). I wish I had something better to use but I don't have a good idea of what's more plausible. I certainly don't have enough of a physics background to try to guess the half-life of whatever new particles can hypothetically exist… 🙂 I wonder if there's a frequentist answer to this problem; my answer is clearly Bayesian.

  • The answer to Q2 is given as (ROT13) "Gurbergvpnyyl, vasvavgr. Va cenpgvpr, n irel, irel ynetr ahzore". It appears the author has once again fallen for the bait; the actual answer is 88 or less, hardly so very large. The pound contains 0.454 grams, or 0.454*6.022e23 atoms (assuming hydrogen). This is equal to about 2^88 atoms, so they will decay to one in 88 half-lives. For materials other than hydrogen, the number will be slightly smaller.

  • Mark P. I think you should use a Poisson Distribution.

    See Poisson distribution about Page 14 etc.…/Lectu…/phys2150_lect6_sp12.pdf

    Joseph F.

  • Question 4:
    Given the wide range of possible decay rates / half-lifes for unobtainium (see answers to question 3), I kind of think that the obvious division of 10, 12, 8 or 10, 13, 7 is misguided. Those answers assume that the decay rate is exactly 0.4 per year. If it's much higher or much lower than that–which is reasonable given the tiny sample size–using 10, 12, 8 or 10, 13, 7 is likely to overshoot or undershoot the mark by a wide margin. I wonder if 10, 10, 10 is actually more likely to obtain the answer. That way, if we're a little bit off, we still have a chance to get lucky. I think this would work if we don't have to decide beforehand which pile is A, B, and C and instead see if we get a set of three piles that fit the constraints after seeing how many atoms remain in each pile. It don't think the 10, 10, 10 answer is right if we have to decide which pile is which beforehand.

    I'm curious if someone can do the math to check my assertion. (max_{a, b, c} of integral over all possible decay rates, weighted by their probability given the earlier observations, of the probability of resulting a, b, and piles fitting the constraint.)

  • Just used some logic and excel for doing this quick and rough.

    Question 3: 2/5 in 1 year so in 50% of that time it can decay another 2/5. 3/5 remains after one year. After 150% years 3*2/5 can remain, which is 1,8; after 175% years 1,8*2/5 can remain, which is 1,08; after 187,5% years (1,08 * 3) / 5 = 0.648; After 193.75% years (0.648 * 3) / 5 = 0.3888. Here we passed the 50% mark of the last unobtainium.

    So an estimate of the decay rate of unobtanium is between 187,5% years and 193,75% years. Between 684 days, 19 hours 53 minutes 56 seconds and 227,2 milliseconds and 707 days, 15 hours, 45 minutes 44 seconds and 64 milliseconds. I didn't use a 95% interval, others could enhance this answer by throwing that in.

    Question 4:

    30 after a year 2/5th decays. So 18 remains. This nummer fits with A 6+ B 7,5 (7+8/2) and C 4,5 (4+5/2), because 6+7,5+8,5=18. Short form: 30*3/5 -(6+7,5+4,5)= 0. So lets use 18 as number, this means B and C can't both be the upper value OR the lower value. So either B or C is an upper value.

    (6/18)*30 =10; (7/18)*30 = 11 2/3; (8/18)*30 = 13 1/3; (4/18)*30 =6 2/3; (5/18)*30 = 8 1/3

    A 10, B 8, C 12 OR A 10, B 7, C 13. I prefer the last one, since it's easier to obtain 1extra unobtanium in the bigger group. 5/8 for B = 0.625 needed, but 8/13 for C =0.61538461538. Then we can hit 4/7 =0.57142857142 for B and 6/10=0.6 for A.

    My answer is A=10, B=7, C=13

  • Found a few typing mistakes, sorry. Because 6+7,5+8,5=18 needs to be 4,5.

    I also mixed up B and C in the final parts of my answer to question 4.
    Corrections: A 10, B 12, C 8 OR A 10, B 13, C 7. 5/8 for C = 0.625 needed, but 8/13 for B =0.61538461538. Then we can hit 4/7 =0.57142857142 for C and 6/10=0.6 for A.

    My answer should be A=10, B=13, C=7

  • Question 3: Does this mean you want the two values of half-life which have a 5% chance of having the measured decay profile?

  • Using the information that n(0) = 5 atoms with an observed decay rate of 2 per year, lambda is: ln(5/3) = 0.5108256238… or 0.51082562377 (more or less).

    This corresponds to a half-life of ln(2)/ln(5/3) = 1.3569 years.

    Using the Poisson calculator below:
    the following are the probabilities of N decays per year.

    For the original case when N = 5, the probability of from 0 to 5 decays are:

    # decays n(0) = 5 Prob. Cumulative Prob.
    0 0.6000000000 0.6000000000

    1 0.3064953743 0.9064953743

    2 0.0782828454 0.9847782196

    3 0.0133296278 0.9981078474

    4 0.0017022789 0.9998101263

    If the number of decays were 3 vs. 2, lambda would increase to ln(5/2) = 0.9162907319 and half life would be reduced to ln(2)/ln(5/2) = 0.754671 yrs.

    More later.


  • If zero atoms out of 5 decay in 1 year, its half life is not necessarily infinite.
    You have to wait a billion years to make sure!

    But a range from 0 to 3 decays per year (starting with 5 atoms) gives a 99.8% confidence level.

    If 1 atom decayed, half-life would be ln(2)/ln(5/4) = 3.106 years, for two atoms, half-life is ~1.357 years and for three atoms, half-life is 0.755 years. That's a more plausible range ( 0.755 to 1.357 yrs) than a half life of from 0.755 years to infinity.


  • "Before going into exponential decay, some more basic math first. You asked your artist to draw a bucket half filled with amobea, but she drew a bucket filled to height 1/2. Why is this wrong, and at which height is a bucket filled to 50%? Could be a new puzzle."

    Not sure I follow? How is a half full bucket and a bucket that is 50% full different?

  • As to question 4, you will not always get exactly 6 atoms when you start with 10 atoms. You might even end up with all 10 original atoms.

    The collection starts with a total of 30 atoms, Group A should start with 10 atoms, and Groups B and C should have a total of 20 atoms with several atoms of unobtainium shared between them. (Via an Entanglement machine, yet to be built.)

    My configurations look as follows (without use of Visio Software):
    Group A: 10 atoms
    Group B1: 10 atoms
    Group B2/C1: 5 atoms |
    Group C2: 5 atoms
    The 5 atoms in Group B2/C1 are shared and are "in" (shared) both groups B and C.

    Effectively, 2.5 atoms are in Group B1 and 2.5 atoms are in Group C2.
    Thus, the year starts with 12.5 atoms in B1/B2 and 7.5 atoms in Group C1/C2.

    Multiplying by the factor 0.6, Group A will end up with 6 atoms, group B1/B2 will end up with 0.6*12.5 = 7.5 atoms (OK, maybe 7 or 8.) and Group C1/C2 will end up, after a year, with 0.6*7.5 = 4.5 atoms (OK, 4 or 5).


  • The answer to question 2 is wrong. The correct answer to the question given is that it depends on the definition of the word material. If the radioactive material is made of a single isotope then it will have a single half life. If it is a mixture of different isotopes each will have its own half life.

    For the answer given in the text to be correct the question would have to be different.

Comments are closed.