This interesting problem was told to me by Jim Farned. You have a prescription to take one half of a pill per day for 20 days, but the pharmacist (who is too busy to divide pills for you) gives you 10 whole pills in a bottle. On day 1, you remove a pill from the bottle, break it into two half-pills, take one, and return the other half-pill to the bottle. On all subsequent days you shake the bottle thoroughly and pour something out - whatever comes out first - either a half pill or a whole pill; if it's a half pill you take it and you're done for that day; if it's a whole pill, you split it into two half-pills, take one, and put the other back in the bottle, exactly like you did on day 1. On day 20 there can be only one half pill left in the bottle, but on day 19 there are two possibilities: either there is one whole pill or there are two half-pills left in the bottle. What is the probability that there are two half-pills in the bottle on day 19? Answer : Solutions (listed by author)(in the order devised -- MAG) Michael A. Gottlieb (approximation, with difference equations) (pdf, 74K) (nb, 12K) Michael A. Gottlieb (exact, with recursive functions) (pdf, 108K) (nb, 36K) Michael A. Gottlieb (exact, with Markov chains) (pdf, 95K) (nb, 24K) Michael A. Gottlieb (fast & accurate approximation, with differential equations) (pdf,104K) |
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