A second red ball from the same urn

There are nn urns; urn ii (1in1 \le i \le n) holds 11 red ball and 2i12^i - 1 blue balls (so 2i2^i balls total). You pick an urn uniformly at random, draw a ball, and it is red. You replace it and draw again from the same urn. Let p(n)p(n) be the probability the second draw is also red. Compute limnp(n)\displaystyle\lim_{n\to\infty} p(n).

Show hints (2)+
  1. Bayes: posterior on urn ii after a red draw 12i\propto \tfrac1{2^i}. Then average 12i\tfrac1{2^i} over it.
  2. p(n)=4i2i1/31=13p(n)=\tfrac{\sum 4^{-i}}{\sum 2^{-i}}\to \tfrac{1/3}{1}=\tfrac13.

Answer

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0.3333 (± 0.005)

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Asked at: Jane Street, Citadel

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