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Win

:)
Had 88 vs 44 on a board like that once......

WARNING: This contains maths!!

The odds of flopping quads with a PP are about 407 to 1 or 1/408, which we'll call 'p'. The odds of flopping it twice in a single SnG obviously depends on the number of hands that take place, which we'll call 'n'.

If you're after the odds for exactly twice the formula is:

(p^2) *[ (1-p)^n-2 ] * n! / [ 2! * (n-2)! ]

Although this assumes that you have a PP every hand. The probability of being dealt a pocket pair is 16 to 1 or 1/17 and the chance of flopping quads with a non-PP hand is 9799 to 1 or 1/9800. So the overall chance of flopping quads in any one hand is:

(1/17) * (1/408) + (16/17) * (1/9800) = 1/4613 roughly, which we'll redefine as 'p'.

So using the previous formula, assuming a nice round number of hands in the SnG, say 100. The odds of flopping quads exactly twice in a SnG would be approximately 1 in 3584.

If you wanted to know the probability for flopping quads twice or more in a sample you would have to subtract the probability of it happening exactly once and not at all from 1, adjusting the formula in the appropriate manner. I.e.

1 - (p^1) *[ (1-p)^n-1 ] * n! / [ 1! * (n-1)! ] - (p^0) *[ (1-p)^n ] * n! / [ 0! * (n)! ]

That would be the cse without the factorial part of the formula at the end. Without the factorial part it would be calculating the odds of flopping quads in hands 1 and 2, and not flopping them in the subsequent ones. It basically takes into account all the possible combinations of hands which either flop or don't flop quads.