[danq]

[ QUOTE ]
That cant be right. You say its 8 of 45 unseen that beats him. If so then there would be one less unseen card when we calculate the probability of our opponent getting an X card (which is 100% [img]/images/graemlins/smile.gif[/img] ), 44 unseen cards. 46 cards of the 44 unseen cards gives him any card (the X-card). Even though we dont remove one from the unseen cards he still has a 46/45 chance of getting ANY card. You cant be more than 100% sure to get a second card. [img]/images/graemlins/tongue.gif[/img]

Im thinking there is 8 of 47 unseen cards that gives one opponent a straight. And 46 of 46 unseen cards that gives him any kicker.

[/ QUOTE ]

Not quite sure where you're going with "46 out of 44" and kickers. I stand by my calculation; here's how it goes.

Remove a jack, a ten, a nine, an eight, a three, and two aces from a deck of cards. This leaves the 45 cards you haven't yet seen. Now deal six cards off the top of the deck, and calculate the probability that none of them is a queen or a seven. (Do you agree that's analogous to the first part of this problem, working out the probability that nobody has a straight?)

There are 45 cards left in the deck, including all 8 of the 7s and Qs, so the probability that the first cards is neither a 7 nor a Q is 37/45. IF the first card is not a 7 or a Q, then there are 44 cards left in the deck, including all 8 of the 7s and Qs, so the probability the second card is neither a 7 nor a Q is 36/44. If neither of the first two are 7s or Qs, then the probability the third card is neither a 7 nor a Q is 35/43. And so on, three more times.

Therefore, the probability that none of the six cards are 7s or Qs is
37/45 * 36/44 * 35/43 * 34/42 * 33/41 * 32/40
as in my original (edited) post. This, then, is the probability that three random hands do not make straights on this board.

Dan