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There are 30 red marbles and 30 blue marbles. Your opponent may arrange these marbles in any way he/she chooses into 2 urns. You then pick one of these 2 urns. You get 10 dollars if you draw red and 0 dollars if you draw blue. How much would you be willing to pay to play this game?

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  • $\begingroup$ How much you would be willing to bet is far too subjective. You might want to simplify your question to "what is the probability of winning?" $\endgroup$ – Ataraxia Jul 28 '13 at 0:19
  • $\begingroup$ This a recurring convention in this sort of problem that you would be willing to pay up to the expected value of the game. $\endgroup$ – MJD Jul 28 '13 at 0:24
  • $\begingroup$ @MJD Didn't know that, thanks for the info. $\endgroup$ – Ataraxia Jul 28 '13 at 0:26
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Assuming your opponent is trying to minimize your expected value, you should be willing to pay about $\$2.54$.

Your opponent should place one blue marble in one urn, and the rest of the marbles in the other urn. This leaves you with a $15/59$ chance of winning the game, which makes your expected value $\$10\cdot 15/59=150/59$, which is a bit over $\$2.54$.

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