a priori probability

[BobS1001]

Suppose that there are five coins, four heads and one tail. They are divided into two piles, three on the left, and two on the right. Now you would say that it was 3:2 against the tail’s being included in the smaller pile. Now take two coins away from the larger pile, with the proviso that neither of them be the tail. (That is what happens in bridge, where the discarding is selective and a player who has the critical honor, a king or queen, does not play it wantonly.) At this stage there is only one coin on the left and, as before, two on the right; but it remains 3:2 against the tail’s being on the right.

It is just hard for me to believe that with the one coin on the left and two on the right, that a gambler can get rich by betting even money that the tail is on the left. (You could do that if the odds are 3:2 against the tail being on the right.)

[PairTheBoard]

Bet even money the tail is in the pile of 3 at the beginning. Then remove 2 heads from the pile of 3 before declaring the winner. Does that feel better?

PairTheBoard

[dansalmo]

This is just a different way of looking at a classic problem known as the "Monte Hall" problem. You can searchthis forum or google and find many examples and explainations of it.