[bigpooch]

First of all, the game is symmetrical with respect to all
players, i.e., the optimal strategy that each player uses is
identical with each other and it looks like this: bet $y
when h>z(x,y,N) where z is the minimal hand to show a profit
which is obviously a function of x, y and N.

Let's assume that the higher number wins and that the
probability that anyone gets a hand less than r is just r
(or that each real number anyone is dealt is an independent
identically distributed random variable with uniform
distribution over (0,1] ).

For the real number z which is minimal for which you bet $y,
the chances that nobody else gets a hand better than z (and
hence won't play) is z^(N-1) since there are (N-1) other
hands and the chances that one is worse is z. Here, you
"pick up the antes" which nets you +$(N-1)x.

On the other hand, with a probability of 1-z^(N-1), someone
does pick up a hand better than z and since that person is
playing optimally and bets, you end up losing $(x+y).

Thus, you need the critical value of z to satisfy that the
expectation

z^(N-1)*[+(N-1)x] + (1-z^(N-1))*[-x-y] is zero.

(for a hand higher than z, then the expectation >0 since the
LHS is an increasing function in z with N>=2, x>0, y>0).

This reduces to

z^(N-1) = (x+y)/(Nx+y)

or z = [(x+y)/(Nx+y)]^(1/(N-1))


The more interesting question is this: what if position is
important, i.e., the first player to the left of the button
acts first, etc. It's clear that if everyone is playing
optimally, because of the "gap principle", the first player
doesn't quite need as strong a hand since the other players
"know" that it is very unlikely that they can win if they
hold a hand slightly better than the utg player's minimal
opening hand (as long as x,y and N are reasonable).