Re: 4 Pocket Pair...
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Saw this on party a couple of days ago. Just wondering the odds of this happening. The last 4 positions all had pocket pair, JJ, QQ, KK, AA. Two were low stacks and the kings and aces had about 100 each. They all ended up going all in. flop was KQ482r. After the hand was over and everyone was bitching at how party sets up hands for action the guy to the left of the jacks said he had folded pocket 10's...Whats the odds of all the pocket pair being dealt inline or just even in the same hand? First time ive seen every high pocket pair in a pot
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well, for the last 5 players specifically to have TT-AA in that order: the probability of it happening is 12^5 / (52!/42!) = 4.3*10^-12, pretty darn improbable. there are 12 ways to get a specific pocket pair in a particular order, so there are 12^5 ways to deal a particular player TT AND a particular player JJ AND a particular.... AA where order counts. the number of total ways of dealing those players two cards apiece, again where order matters, are (52 * 51 * ... * 43). counting the ways of doing it where order does not matter should give you the same answer but i think this is one of those uncommon times where considering order is easier.
for any players to have them in any of the ten positions, multiply that figure by 10 * 9 * 8 * 7 * 6 to get a probability of 1.3*10^-7, again a pretty small figure, but now we are down to the times-per-ten-million-hands order of magnitude.
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01-03-2005, 04:44 PM
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