Re: Simple Math Question
there was just a thread that would answer this. i don't feel like searching for it. i did the calculation "the hard way", and someone (almost undoubtedly gm) pointed out an easier way which is almost certainly inclusion/exclusion.
i think we all agreed that just using the chance of each player getting aces was close enough to correct to not worry about it since the chance of two people getting aces is so rediculously remote.
since two Ks are out, the chance of getting dealt two aces is 4/50 * 3/49 = .00489795918367346938
call that value a.
the chance of someone getting dealt AA against our hero all four times is approximately a^4 * 5 * 5 * 9 * 9 = .00000116542860715575
pretty unlikely.
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