very nice, thanks!
And yes I am somewhat familiar with Gauss but it’s a bit rusty now.

Andrew MW, sorry for the wait. About how we can done it, I already explain why must the numbers be taken at random. The problem now is to know what is the distribution of probabilities of those numbers. The easiest way to find them is to think no one can gain advantage against them. This is a work hypothesis, to believe that distribution exist. If it does not exist, we are stuck with distributions which act like the game scissors-paper-stone… if we can gain advantage against each distribution using a certain strategy, the later must also be vulnerable to other strategies (or else we could not gain advantage against any distribution and we would fall in a contradiction).

If he does exist, the fact that no one gains advantage against it means also no one will loose against it playing with the same numbers. This because, two players playing with that same distribution must not loose against each other. To understand that, suppose the distribution covers n numbers and anyone playing the first number of it against it, will loose an average of V points. Now, someone playing the same distribution, will play the first number with probability p1, which means in average will loose due to it, that V*p1. But for the global average be zero, that means that the results of playing the other numbers according to the same distribution against the same distribution, must be V*p1… meaning it is possible to win against the distribution and entering in contradiction with what we asked from it.

So, against the searched distribution, no one can gain or loose in average when playing the same numbers. Suppose the distribution has only 3 numbers… then, what we are searching for is the solution to:

p1+p2+p3=1 (to assure 1, 2, and 3 are the only numbers played)
-2p2+p3=0 (average gain playing 1 must be zero)
2p1-2p3=0 (average gain playing 2 must be zero)
-p1+2p2=0 (average gain playing 3 must be zero)
-p1-p2+2p3<=0 (playing 4 must lead to no gain)
-p1-p2-p3<0 (of course playing 5 or more leads to losses).

So what we do is to solve the first 3 equations for p1, p2 and p3, and see if the numbers are all between 0 and 1, and substituted in the other conditions, they lead to true propositions. If that not happens, we try this time with 4 numbers in the distribution. And 5, and 6, and etc… Note that once found a certain distribution, it is not possible to exist other with more numbers. Do that and discover Andrew's solution when we try for 5 numbers.