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Drug addict

  • Thread starter Thread starter radosr
  • Start date Start date

radosr

Baruch MFE Faculty
Joined
7/18/07
Messages
696
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A drug addict has a bottle of n pills and initially wishes to take one pill per day. However, he is a bit short on money lately, so he plans to take only half a pill per day. Every day he picks a pill from the bottle at random. If it is a whole pill, he cuts it in half, takes half a pill, and puts the other half back in the bottle. If it is half a pill that he picked, he just takes it. He does this until the bottle is empty. What is the expected maximum number of half pills in the bottle?
 
Let E(n, m) be the expected maximum number of half pills after starting with n whole pills and m half pills. The answer we're looking for is E(n, 0). I tried to solve for it in terms of E(n-1, 0).

This is as far as I got:
E(n, 0) = 2/(n^2 (n-2)!) + (1 - (n-1)(n-2)/n^2)E(n-1, 0) + (n-1)(n-2)/n^2 E(n-3, 3)

I couldn't find a simple expression for E(n-3, 3), but
E(0, 3) = 3 for n = 3
E(1, 3) = 13/4 for n = 4
E(2, 3) = 711/200 for n = 5

Using this formula, and starting with the trivial E(1, 0) = 1,
E(2, 0) = 1/2 + E(1, 0) = 3/2
E(3, 0) = 2/9 + 7/9 E(2, 0) + 2/9 E(0, 3) = 37/18
E(4, 0) = 1/16 + 5/8 E(3, 0) + 3/8 E(1, 3) = 739/288
E(5, 0) = 1/75 + 13/25 E(4, 0) + 12/25 E(2, 3) = 3.0540

It wouldn't be difficult to compute the expected maximum recursively, but I'm not sure how to find a closed-from solution.
 
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