We play a game as follows:

I place one dollar on the table. I repeatedly flip a coin. Each time the coin comes up heads, I double the money on the table. The first time the coin comes up tails, you take the money and the game is over.

What's a fair admission price for the game?

Would you play the game with me for $100?

I think I outsmarted myself by bypassing simple odds.

Consider:

There is 1/2 chance that your first toss will be tails and I will win $1. For that, I should pay you $.50

There is 1/4 chance that your first toss will be heads and your second will be tails, and I will win $2. So I should pay you another $.50

For a streak of any length, the probability of reaching and stopping at that particular length, multiplied by the payoff, means I should pay you another $.50.

So, unless you set a maximum number of tosses (the last one worth $1, instead of $.50), the "fair price" of the game is Aleph Null (countable infinity).

I believe that this is another case, however, where the practical and the theoretical are at tenterhooks. I suspect that in real games you will not have to pay out the higher sums often enough, and will usually profit from my $100. (after all, the probability of a streak exceedind 6 heads is only 1/128)

Posted by TomM on 2002-10-09 19:03:32