## Monday, March 31, 2014

### Another absurdity about infinite fair lotteries

It is easy to generate a method for choosing a natural number in such a way that the probability of choosing n is 2n. For instance, toss a fair coin and let n be the number of the first toss on which you get heads. Thus, n=1 if you get heads on your first toss, n=2 if you get tails on the first and heads on the second, n=3 if you get tails on the first two and heads on the third, and so on. (And if you never get heads, count that as just another way of choosing 1—after all, the probability of never getting heads is zero.)

Could one have a way of choosing a natural number, guaranteed to return some natural number, but such that the probability of choosing n is between 4n and 3n? Surely not! That would be absurd: the probabilities would be too small.

But now suppose you can have an infinite fair lottery. Follow the following procedure. Toss a fair coin until you get heads. If it took an even number m of tosses to get to heads, let your chosen number be n=m/2. If it took an odd number of tosses to get to heads, or if you never got to heads, then choose the natural number n via an infinite fair lottery.

What's the probability of getting n in this new process? Well, the probability of getting heads for the first time on the 2nth toss is 2−2n=4n. But if you didn't get heads for the first time on an even-numbered toss, you still have a chance of getting n, namely via the infinite fair lottery. The latter chance is either zero or infinitesimal. So, the probability of getting n is 4n or 4n plus an infinitesimal. But it's absurd to have a random choice of natural number where the probability of getting n is 4n. And if it's 4n plus an infinitesimal, then it's between 4n and 3n, which we agreed was absurd.

So infinite fair lotteries lead to absurdity.