Imagine a pack of cards, each of which has one number on one side and the number directly above on the other.

There is one card with 1 on one side and 2 on the other,

two cards with 2 on one side and 3 on the other,

four cards with 3 and 4,

eight cards with 4 and 5,

and so on, ad infinitum.

So that there are 2^{n+1} cards with n on one side and n+1 on the other.

Let us now use that pack to play a game of chance.

Two people, Alice and Bob, stand facing each other,

while the host draws one card from the pack and puts it between them,

so that each can see the side facing him or her, but not the other side.

The winner is the one who sees the lowest number.

What is Alice’s probability of winning?

If she sees a 1, the other side must be a 2, and she has won.

If she sees a number n>1, the hidden number,

on the other side of the card, is either n-1 or n+1.

In the first case she loses, in the other case she wins.

Since there are twice as many cards of the second type as there are cards of the first,

she has a probability 2/3 of winning.

Unfortunately, the same argument holds for Bob :

he also has a probability 2/3 of winning.

Since one must win, but not both,

the two probabilities should sum to 1,

so that 2/3 + 2/3 = 1,

Quite a remarkable equality !!

The argument is perfectly correct, but there's a problem with it.

Find it.

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