The Ticket Swap Paradox

Suppose that ticket A has the value x. Whatever the value of A, ticket B will be worth 2x in 3/7 of cases and x/2 in 4/7 of cases. So the gain of switching from A to B is 1/7( 3(2x) + 4(x)/2 - 7x ) = x/7.

Once the value, y, of ticket B is decided, ticket C will be worth 2y in 4/7 of cases, and y/2 in 3/7 of cases. So the gain of switching from B to C is 1/7( 4(2y) + 3y/2 - 7y ) = 5y/14. Since y > 0 the gain in each case is positive for any value of x. Therefore, if given the chance, we should swap A for B and then B for C.

The problem is that ticket C is none other than ticket A. We have a paradox because after switching twice with positive gain each time we end up with the original value, ie x.

Solution

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