The Universe of Discourse

Wed, 18 Sep 2019

Breaking pills

Suppose you have a bottle that contains !!N!! whole pills. Each day you select a pill at random from the bottle. If it is a whole pill you eat half and put the other half back in the bottle. If it is a half pill, you eat it. How many half-pills can you expect to have in the bottle the day after you break the last whole pill?

Let's write !!E(N)!! for the expected number of half-pills. It's easily seen that !!E(N) = 0, 1, \frac32!! for !!N=0,1,2!!, and it's not hard to calculate that !!E(3) = \frac{11}{6}!!.

For larger !!N!! it's easy to use Monte Carlo simulation, and find that !!E(30)!! is almost exactly !!4!!. But it's also easy to use dynamic programming and compute that $$E(30) = \frac{9304682830147}{2329089562800}$$ exactly, which is a bit less than 4, only !!3.994987!!. Similarly, the dynamic programming approach tells us that $$E(100) = \frac{14466636279520351160221518043104131447711}{2788815009188499086581352357412492142272}$$ which is about !!5.187!!.

(I hate the term “dynamic programming”. It sounds so cool, but then you find out that all it means is “I memoized the results in a table”. Ho hum.)

As you'd expect for a distribution with a small mean, you're much more likely to end with a small number of half-pills than a large number. In this graph, the red line shows the probability of ending with various numbers of half-pills for an initial bottle of 100 whole pills; the blue line for an initial bottle of 30 whole pills, and the orange line for an initial bottle of 5 whole pills. The data were generated by this program.

screenshot of the graph
described above.  In each case, the probability starts relatively
high, then drops rapidly to nearly zero.

The !!E!! function appears to increase approximately logarithmically. It first exceeds !!2!! at !!N=4!!, !!3!! at !!N=11!!, !!4!! at !!N=31!!, and !!5!! at !!N=83!!. The successive ratios of these !!N!!-values are !!2.75, 2.81,!! and !!2.68!!. So we might guess that !!E(N)!! first exceeds 6 around !!N=228!! or so, and indeed !!E(226) < 6 < E(227)!!. So based on purely empirical considerations, we might guess that $$E(N) \approx \frac{\log{\frac{15}{22}N}}{\log 2.75}.$$

(The !!\frac{15}{22}!! is a fudge factor to get the curves to line up properly.)

I don't have any theoretical justification for this, but I think it might be possible to get a bound.

I don't think modeling this as a Markov process would work well. There are too many states, and it is not ergodic.

[ Addendum 20190919: Ben Handley informs me that !!E(n)!! is simply the harmonic numbers, !!E(n) = \sum_1^n \frac1n!!. I feel a little foolish that I did not notice that the first four terms matched. The appearance of !!E(3)=\frac{11}6!! should have tipped me off. Thank you, M. Handley. ]

[ Addendum 20190920: I was so fried when I wrote this that I also didn't notice that the denominator I guessed, !!2.75!!, is almost exactly !!e!!. (The correct value is !!e!!.) I also suspect that the !!\frac{15}{22}!! is just plain wrong, and ought to be !!e^\gamma!! or something like that, but I need to take a closer look. ]

[ Addendum 20191004: More about this ]

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