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The prisoner and the two urns

Chance and uncertaintyLevel 2/5

One of the canonical pieces of distribution under uncertainty: the optimal strategy is not to balance, but to create a small island of sure success.

A prisoner receives 50 white balls and 50 black balls. You must distribute them between two urns, however you want, but no urn can be left empty.

Then the jailer will do this:

  1. will choose one of the two urns at random;
  2. will draw a ball at random from the chosen urn. If the ball is white, the prisoner is free.

If it is black, it dies. How should you distribute the balls to maximize your chance of saving?

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Hints

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  1. Distributing them almost equally is not the best.
  2. It is advisable to create at least a possibility of certain success.
  3. The optimal strategy is to isolate a single cue ball.

Solution

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The best strategy is:

  • urn 1: a single white ball
  • urn 2: the other 49 white balls and the 50 black ones So, if the jailer chooses urn 1—which happens with probability 1/2—the prisoner is safely saved. If you choose urn 2, the probability of drawing white is 49/99. The total probability of being saved is then:

(1/2) × 1 + (1/2) × (49/99) = 74/99 that is, approximately 74.75%.

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