Re: How do you calculate this?
OK here's what I got:
How I looked at this problem was we've got the six-handed table, and we're going to make three decks of cards: one deck comprised of solely the four kings, one of solely the four aces, and the third deck of the 44 remaining cards plus whichever two aces happen to not get dealt.
Now we pick two people at random of the six at the table to be the lucky (or unlucky, depending on your point of view) people to receive KK. there are 6c2 ways of doing this.
Now we deal out the four kings to the two players - there are 6 ways of doing this (4c2)
Now we pick a lucky player to receive AA. There are four players left, so there are four ways of doing this.
Next, we have to see which two aces he gets out of the four available - again, 4c2 = 6 ways.
now we put the remaining two aces back in our mini-deck of 46 cards and deal out the rest of them to the three remaining poor sods at the table. This can be done (46c2)*(44c2)*(42c2) ways.
OK, so now after all that we have the number of ways we can deal out KK to two players and AA to one.* So now to get the probability that this will happen, we have to divide this number by the total number of ways we can deal 6 two-card holdem hands out of a deck of 52: (52c2)(50c2)(48c2)(46c2)(44c2)(42c2)
And after all that, the answer that I get is approximately .000001, or approximately one in a million.
* note that this probability still includes the slim probability that someone of the remaining three players will be dealt AA as well (making for two KKs and two AAs at the table). Perhaps someone who is more mathematically inclined can correct for this.
btw, thanks for the puzzle [img]/images/graemlins/smile.gif[/img]
-EDIT: Actually, now that I think about it, all you'd have to do to correct for the possibility of one of the remaining three getting dealt the other two aces is subtract three from the numerator, since there are only three ways of giving one of the three remaining players the only remaining AA, so it would have a negligible effect on the odds.
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