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Optimal mix of deposits and withdrawals over 100 turns

Question

A solo game uses two opaque boxes and lasts exactly 100100 turns. On each turn you must choose one action: place, which puts $1 (supplied by a third party) into one of the two boxes at random, or take, which empties one of the boxes at random into your pocket. You learn nothing about the boxes' contents until the game ends. With optimal play, what is your expected total winnings?

Short answer

2961/32≈92.532961/32\approx92.53. It is best to do all places first. With pp places followed by t=100−pt=100-p takes, each dollar is collected unless its box is never picked, so the payoff is p (1−2−t)p\,(1-2^{-t}). This peaks at p=94p=94: 94⋅6364=29613294\cdot\tfrac{63}{64}=\tfrac{2961}{32}.

Adapted from QuantQuestions.io; question and answer rewritten by SelfMadeQuant.com.

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