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?
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 $m > 1$.