computing expected entries based on duplicates

[rufus]

It would make more sense to indicate how many duplicate winners there were on each day.

Let's assume, for a moment, that the chance to win is fixed at p, and that everyone who wins plays again.

Then the chance of winning exactly twice in four days is going to be:
p^2(1-p)^2*6=6k

And the chance of winning exactly once in four days is:
p*(1-p)^3*4=296k

So you can solve:
p^2(1-p)^2*6 / (p*(1-p)^3*4) =6/296
p=1/75
so roughly 75*77=5775 people play per day.