[cwes]

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i guess i can stand to get burned once every 2,814,000 hands [img]/images/graemlins/grin.gif[/img]

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Dangerous conclusion. I gave the probability of this happening ex ante. I did not give the propability of you being beat by two other sets if you flop a set and two higher cards appear on board.

So if you have got a pair of threes before the flop there obviously are 60 combinations (higher pairs), that can beat your pair. the probability of two other players having two unique valued higher pairs is '(5 choose 2) x (60/(50 choose 2)) x (54/(48 choose 2))'.

The probability of you all flopping a set then stays the same as above (i.e. 8/901080).

So this happens every 485,549 hands in a six handed game to you if you play a pair of threes. Remember this is only about flopping a three if you play a pair of threes and then being beat by two higher sets flopped. It is not about being beat by two higher sets gained on the flop and/or turn and/or the river nor being beat by any other hand.

Given there are three pairs before the flop the probability they all flop a set is just 8/91080 = 11,385^-1.

Given there are three pairs before the flop, your threes and two higher ones, the probability of you and at least one of them flopping a set (and hence beating you unless you get a fourth three and they do not get a fourth whatever) is '2/46 * 4/45 = 258.75^-1'. So starting with a pair against two higher pairs, flopping a set and already being beat by a higher flopped set will happen every 258.75 times you see the flop against two higher pairs.

Different story, huh?

But, hey! It gets even worse. Given you start against two higher pairs with your pair and you flop a set, the probability of at least one of them also flopping a set is just 4/45 = 11.25^-1. From 2.8 millon hands to 11 hands... creepy.