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Tue, 17 Sep 2019 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 !!H(N)!! for the expected number of half-pills left, if you start with !!N!!. It's easily seen that !!H(N) = 0, 1, \frac32!! for !!N=0,1,2!!, and it's not hard to calculate that !!H(3) = \frac{11}{6}!!. For larger !!N!! it's easy to use Monte Carlo simulation, and find that !!H(30)!! is almost exactly !!4!!. But it's also easy to use dynamic programming and compute that $$H(30) = \frac{9304682830147}{2329089562800}$$ exactly, which is a bit less than 4, only !!3.994987!!. Similarly, the dynamic programming approach tells us that $$H(100) = \frac{14466636279520351160221518043104131447711}{2788815009188499086581352357412492142272}$$ 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. 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 !!H(N)!! first exceeds 6 around !!N=228!! or so, and indeed !!H(226) < 6 < H(227)!!. So based on purely empirical considerations, we might guess that $$H(N) \approx \frac{\log{\frac{15}{22}N}}{\log 2.75}.$$ (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 !!H(n)!! is simply the harmonic numbers, !!H(n) = \sum_1^n \frac1n!!. I feel a little foolish that I did not notice that the first four terms matched. The appearance of !!H(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 ] [Other articles in category /math] permanent link |