The Ticket Swap Paradox Resolution


Is ticket C really ticket A? It isn't because the possible values of ticket B are 2x and x/2, so that the possible values of ticket C are 4x, x or x/4, ie not just x. If ticket C were the same as ticket A then ticket B would be worth 2 times the value of C in 3/7 of cases and 1/2 the value of C in the remaining 4/7. Yet ticket B has the possible values 2x and x/2, while ticket C has the possible values 4x, x or x/4. It just does not compute.

So why does it seem that C is A? The error arises because we are erroneously trying to extend the valid rule that if ticket X is worth two times ticket Y and ticket Z is worth half of ticket X, then Y and Z are equal. This rule does not apply when the value of each ticket in the chain of three is probabilistically contingent on the one before, as in the paradox. The intuitive explanation is that when two forking processes are linked up then the four paths will not cancel each other to return to the original value, at any rate not in general, though there may be some unusual cases where they do so. (I suspect they never do.)

The value of ticket C is actually 76x/49, which is the product of the individual gains, ie 8/7 and 19/14. Alternatively, we can work it out by adding a 12/49 chance of 4x, a (9/49 + 16/49) chance of x, and a 12/49 chance of x/4. This adds up to 48/49 + 25/49 + 3/49 = 76/49 chance of x.

If you still find this hard to see, have a look at this numerical example. Say A is worth 98. Then B is worth 196 x 3/7 + 49 x 4/7 = 84 + 28 = 112. So C is worth 224 x 4/7 + 56 x 3/7 = 128 + 24 = 152. This is confirmed by the gain calculation ie 76/49 x 98 = 152. So ticket C is indeed worth more than ticket A.


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