Probability of picking pills from a jar 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?

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 $$P(11)\doteq0.0628157\ .$$