Can You Find Your Pills?

I’ve been prescribed to take $1.5$ pills of a certain medication every day for $10$ days, so I have a bottle with $15$ pills. Each morning, I take two pills out of the bottle at random.

On the first morning, these are guaranteed to be two full pills. I consume one of them, split the other in half using a precision blade, consume half of that second pill, and place the remaining half back into the bottle.

On subsequent mornings when I take out two pills, there are three possibilities:

I get two full pills. As on the first morning, I split one and place the unused half back into the bottle. I get one full pill and one half-pill, both of which I consume. I get two half-pills. In this case, I take out another pill at random. If it’s a half-pill, then I consume all three halves. But if it’s a full pill, I split it and place the unused half back in the bottle. Assume that each pill, whether it is a full pill or a half-pill, is equally likely to be taken out of the bottle.

On the $10$th day, I again take out two pills and consume them. In a rush, I immediately throw the bottle in the trash before bothering to check whether I had just consumed full pills or half-pills. What’s the probability that I took the full dosage, meaning I don’t have to dig through the trash for a remaining half-pill?

Solution

Track the bottle by its contents $(f,h)$, the number of full pills and half-pills. The whole process is a Markov chain on these states. Drawing two of the $t=f+h$ pills, the three immediate outcomes and their probabilities are $$\begin{aligned} \text{two full}\ (f{-}2,h{+}1) &:\ \binom{f}{2}\big/\binom{t}{2},\\ \text{one of each}\ (f{-}1,h{-}1) &:\ fh\big/\binom{t}{2},\\ \text{two halves} &:\ \binom{h}{2}\big/\binom{t}{2}. \end{aligned}$$ In the two-halves case a third pill is drawn from the remaining $t-2$: it is full with probability $f/(t-2)$, sending $(f,h)\to(f-1,h-1)$ (split it, return a half), or a half with probability $(h-2)/(t-2)$, sending $(f,h)\to(f,h-3)$.

I have taken the full dosage exactly when the bottle empties on the tenth draw, that is, when exactly two pills remain after the first nine mornings. Pushing the distribution from $(15,0)$ forward nine days and summing the weight on states with $f+h=2$ gives the exact $$P = \frac{80529}{101920} \approx \boxed{0.7901}.$$

The computation

Evolve the exact distribution over $(f,h)$ states with rational arithmetic: each day, redistribute every state’s probability across the four outcomes above, then read off the total probability on the two-pill states after nine days.

from fractions import Fraction as F
from math import comb
from collections import defaultdict
dist = {(15, 0): F(1)}
for _ in range(9):
    nd = defaultdict(F)
    for (f, h), pr in dist.items():
        t = f + h
        if t < 2: nd[(f, h)] += pr; continue              # fewer than two pills: carry
        c2 = F(1, comb(t, 2))
        if f >= 2: nd[(f-2, h+1)] += pr*comb(f, 2)*c2          # two full
        if f and h: nd[(f-1, h-1)] += pr*(f*h)*c2              # one full, one half
        if h >= 2:                                            # two halves, draw a third
            p2h = comb(h, 2)*c2
            if t > 2:
                nd[(f-1, h-1)] += pr*p2h*F(f, t-2)            # third is full: split it
                nd[(f, h-3)]   += pr*p2h*F(h-2, t-2)          # third is half: consume all
            else:
                nd[(f, h-3)]   += pr*p2h                      # no third pill to draw
    dist = nd
print(sum(pr for (f, h), pr in dist.items() if f + h == 2))   # 80529/101920