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?

  • $\begingroup$ I strongly doubt the existence of that fair judge. $\endgroup$ – Michael Hoppe Jul 11 '19 at 6:08
  • 5
    $\begingroup$ There is no uniform distribution on the set of positive integers. $\endgroup$ – Sangchul Lee Jul 11 '19 at 6:09
  • 1
    $\begingroup$ 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 $1/2$. $\endgroup$ – 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 $m > 1$.

  • 1
    $\begingroup$ 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 $\endgroup$ – michalis vazaios Jul 11 '19 at 6:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.