# Paradox with probability becoming $1$ from $\frac{1}{2}$

There are two players $$A$$ and $$B$$ that play the following game. Each of them is given a random positive integer number by a fair judge, and the player with the biggest number wins. The judge tells player $$A$$ his number then waits some time(let's say 1 minute) and then tells player $$B$$ his number.

Obviously the probabilities of winning are $$50-50$$ (equal). But after the judge tells player $$A$$ his number (let's name it $$x$$), one could assume (no matter what value the $$x$$ has) that the probability of $$B$$ winning is $$1$$, since there are infinite integers greater than $$x$$ but only $$x-1$$ positive integers smaller than $$x$$(and the probability of draw is $$0$$).

I know that the game must be fair, but why does the information we take from hearing player's $$A$$ number condemn player $$A$$ to lose? Why do we(me at least) think for 1 minute that player $$B$$ will surely win?

It seems like a paradox to me and I would like an explanation/solution for it. I thought of this while reading about another paradox (Necktie paradox) but I can't give myself an explanation for it. What is the flaw in the "logic" I used?

• I strongly doubt the existence of that fair judge. – Michael Hoppe Jul 11 '19 at 6:08
• There is no uniform distribution on the set of positive integers. – Sangchul Lee Jul 11 '19 at 6:09
• Nevertheless, if we assume the numbers $x$ and $y$ are distinct (with no distribution assigned to them) but unknown to player A, and we randomly give one of the numbers to player A, equally likely giving $x$ or $y$, then A can guess whether he was given the higher or lower number, and be correct with probability strictly larger than $1/2$. – Michael Jul 11 '19 at 13:05

The key to the issue is in appreciating that there is no way of picking uniformly from all positive integers.

If we knew that there was a maximum number $$N$$ that could be chosen, then the uniform distribution on the set $$\{0,\ldots, N\}$$ would give each individual number, $$n$$, a probability of $$p_n =1/(N+1)$$.

In the extreme case $$N \equiv \infty$$ this is no longer well defined, so the idea that the judge can be fair doesn't make sense.

Instead we have to turn to a different non-uniform distribution on the integers. Eg. some assignment of probabilities $$p_n$$ (the probability of choosing $$n$$, for $$n \geq 0$$) such that

$$\sum_{n \geq 0} p_n = 1.$$

One classic example would be the Poisson distribution (with, for example, parameter $$\lambda = 1$$) which would give weights

$$p_n =\frac{1}{n!}e^{-1}.$$

With such a weighting it is clear that the judge is not considered fair, and supposing that the first person was given the value $$m$$, then the probability that they win is

$$\sum_{n=0}^{m-1} \frac{1}{n!}e^{-1}$$ which one can show is greater than $$1/2$$ so long as $$m > 1$$.

• Thanks! The key(what persuaded me) was that there can not be a uniform distribution on positive integers. For anyone interested, I googled it to find why and found an explanation here – michalis vazaios Jul 11 '19 at 6:20