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Consider a person having a jar of pills. Every day he picks a pill from the jar, split it in half, eats one half of the pill, and puts the other half back in the jar. Starting from day 1 with 100 pills; which one of the following days ahead has the greatest probability of that person picking the first pill that is already split in half?

Edit: Expected answer: Repeating the sceanrio described above unlimited amount of times: Each time write down on a piece of paper which day (number from start) the person took the first pill that was already split in half. Which number of day would occur most of times on that paper?

henrik
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    Welcome to Math.SE! Can you tell us what you think is the correct answer or how you would go about solving this problem? – Hrodelbert May 14 '15 at 09:49
  • @Hrodelbert - Thanks! I'm thinking like this: Repeating the scenario described in the question unlimited amount of times, and for each time remembering what number of day the person took the first pill which was already split in half. Which number of day would occur most of the time when looking at the statistics of all the scenarios? (Sorry for my bad english) – henrik May 14 '15 at 10:12

2 Answers2

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First day : 100 entire pills

Second day : 99 entire pills, one half pill

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Nth day : 100-n+1 entire pills, n-1 half pills.

Hence the probability of not drawing a half pill in the (n-1) first days and drawing one on the nth day is $$P=\left(\prod_{k=1}^{n-1}\frac{100-k+1}{100}\right)\cdot\frac{n-1}{100}=\frac{n-1}{100^n}\prod_{k=1}^{n-1}101-k$$

Hippalectryon
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  • Thanks for the formula, so which day (number) would occur most of times, if repeating the scenario described in the question unlimited amount of times and for each time writing down the day that person took the first pill that was already split in half ? – henrik May 14 '15 at 10:15
  • @henrik Have a look at http://math.stackexchange.com/questions/392444/probability-of-getting-split-pill-from-bottle?rq=1 – Hippalectryon May 14 '15 at 10:40
  • wow! Spot on! Thanks! – henrik May 14 '15 at 11:08
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According to @Hippalectryon 's argument the probability that the first half pill is picked on day $n$ amounts to $$P(n)={n-1\over 100^{\>n}}\prod_{k=1}^{n-1}(101-k)\ .$$ It follows that $$f(n):={P(n+1)\over P(n)}={n(101-n)\over 100(n-1)}\ .$$ One easily checks that $f$ is monotonically decreasing in the relevant domain. Furthermore $$f(10)={91\over 90}>1,\qquad f(11)={99\over100}<1\ .$$ This shows that the $P(n)$ are increasing from $n=1$ up to $n=11$, and then are decreasing forever. The maximal occurring $p(n)$ is therefore $$P(11)\doteq0.0628157\ .$$

Christian Blatter
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