Box of tablets, whole or broken: solution required

Question

This is a puzzle that I thought up whilst taking a course of meds. I currently haven’t solved it, and would be curious to know if anyone has a solution for it. Here goes:

Scenario: John has a box of 15 tablets. He needs to take half a tablet per day for 30 days. Each day he takes the box and removes a tablet randomly (i.e. the chance of him removing a whole tablet is equal to him removing a half tablet when the number of whole- and half-tablets are equal. E.g. if there were 5 whole and 1 half tablet in the box, then p(half) = 1/6). If the tablet he removes is a whole one, he breaks it in half, eats one half, and returns the other half back to the box. If he removes a half tablet, he just consumes it. This continues for 30 days.

Question: On which day (on average, assuming infinite re-runs of the process) is he first going to be equally/more likely to remove a half tablet than a whole one?

I already have given this some thought, but even counting the number of permutations is impossible for me extrapolation from a 1 to 2 to 3 to 4 tablet simplification of the 15 tablet scenario above.

Any ideas/solutions welcome. Preferably the solutions would be hidden in a spoiler box, as for the time being I’m still trying to work it out myself.

Many thanks!

Answer 1

It turns out that on the 14th (edited from 13th - I was using zero index before) day, you are more likely to pick a half pill than a whole pill. On that day, the expected number of whole pills is ~5.597 and the expected number of half pills is ~5.807.

To do this,

I just wrote a Python program that enumerates all the possible outcomes each day, along with the probability that each outcome would occur. To compute tomorrow's possible outcomes, you take each possible situation S from today and check each "branch" that can happen - selecting a whole or half pill. Each branch has a probability of occurring (based on how many of each are in S), and a resulting situation (one less half pill, or one less whole pill and one more half pill). You multiply the probability of being in S by the probability of taking the branch, and add that to tomorrow's probability for S', the new situation that we just created.

Finally,

For each day, you can compute the expected number of whole pills (just multiply the number of whole pills in each situation by the probability of the situation). The total expected whole pills for the day will be the sum of all of these values. The same can be done for the half pills (and if you want to check your work, 2*expected_whole + expected_half should equal (30 - day), since that's how many total half-pills you still have left).

I can give you the Python code if you want. Not sure it can be spoilered (I've never learned how to have multi-line spoilers).

Python Code

Answer 2

Starting at $(0,0)$, we do the followings:

• If a whole pill is taken, we move one unit to the right.
• If a half pill is taken, we move one unit upwards.

Therefore, if we are at point $(x,y)$ we have:

• $(15-x)$ whole pills left
• $(x-y)$ half pills left
• $(15-y)$ total pills left

Now we define $f(x,y)$ as the probability that we reach point $(x,y)$. Obviously,

$f(0,0)=1$ and $f(-1,y)=0$

We reach point $(x,y)$:

• From $(x-1,y)$ if we take a whole pill
• From $(x,y-1)$ if we take a half pill

Therefore,

$f(x,y)=\frac{15-(x-1)}{15-y}f(x-1,y)+\frac{x-(y-1)}{15-(y-1)}f(x,y-1)$

I use microsoft excel to calculate $f(x,y)$ for $0\leq x \leq 15$ and $0\leq y \leq 15$

Now, the expected value of # of half pills - # of whole pills after $k$ days is calculate by summing

$((x-y)-(15-x))f(x,y)$

for all $(x,y)$ which satisfy

$x+y=k-1$

This expected value becomes positive (we have more half pills than whole pills) for $k\geq 14$.

Thus the answer is 14.

Answer 3

If I understand the question correctly, at the end of the 15th day has consumed 7.5 pills because he is allowed half a pill a day. Now, for the remaining 15 days he needs another 7.5 pills. On the 16th day, assuming in package are 5 whole pills and 5 half pills, on that day he has an equal chance of picking a whole or half pill.