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Math Problems and Logic Puzzles
Someone shows you two boxes and he tells you that one of these boxes contains two times as much as the other one, but he does not tell you which one this is. He lets you choose one of these boxes, and opens it. It turns out to be filled with $10. Now he gives you the opportunity to choose the other box instead of the current one (and skip the $10 of the first box), because the second box could contain twice as much (i.e. $20). The Question: Should you choose the second box, or should you stick to your first choice to maximize the expected amount of money? A Hint : If you have $10, and you could double this with a chance of 1/2, or half it with a chance of 1/2, one would expect an average of 1/2 * $20 + 1/2 * $5 = $12.5 (so you would expect to gain $2.5)!...
This is the two-box paradox. The tempting reasoning that switching gives $12.50 expected value is flawed. When you see $10, the possible pairs are ($5,$10) or ($10,$20). Since you chose the $10 box, if the pair was ($5,$10) you definitely picked the larger one (probability 1), but if the pair was ($10,$20) you only had a 50% chance of picking the smaller one. This creates selection bias: seeing $10 is more likely from the ($5,$10) pair (2/3) than from ($10,$20) (1/3). So the other box contains $5 with 2/3 probability and $20 with 1/3 probability, giving expected value of $10 - no gain from switching.