probability to be beat here?

[BigBiceps]

What is the probability that AA is beat, when the board shows JT983 against 3 random hands?

[LetYouDown]

No possible flush?

Q-X, 7-X, J-10, J-9, J-8, J-3, T-9, T-8, T-3, 9-8, 9-3, 8-3, J-J, T-T, 9-9, 8-8, 3-3.

I came up with 429 combinations beat you. If that's right, a random hand against you will beat you 43.33% of the time. You'd need inclusion/exclusion to solve this exactly I believe, but that could get really hairy. Against 3 people, it looks like you can safely expect to lose if the hands are indeed random.

[danq]

Any Q or any 7 beat you, so 8 of the 45 cards you can't see can't be in any of the three hands if you're good. The probability that nobody has a straight is therefore

37/45 * 36/44 * 35/43 * 34/42 * 33/41 * 32/40 = 28.5%

so it's just over 71% that someone has a straight.

When nobody has a straight, there are 15 cards (out of the remaining 38 non-7-or-queens) that pair the board, and anyone with two of them beats you. The probability that at least one of the three hands has two pair or a set, given that nobody has a straight, turns out to be 43.2%. So the chance that you're beat is

.715 + (.285 * .432) = .838

or you lose about 5 times out of 6.

Dan

(Edited cause I can't count)

[LetYouDown]

There's 45 cards left, we're aware of 7.

C(37,6)/C(45,6) = 28.54%

Edit: I tried to rattle this out with inclusion/exclusion and I came to losing 83.25% of the time. Here's to hoping BruceZ can assault this problem.

[danq]

Edited my first post - I can't count.

For the two-pair-or-set probability, rather than inclusion-exclusion, I just calculated cases. There are 15 out of 37 cards that pair the board. If nobody beats you, there are four cases: nobody has a pair, one hand has one pair, two hands have one pair each, or all three hands have one pair each. Each of these is easy to calculate, then add.

Dan

[LetYouDown]

At the very least, we're in the same ballpark and this definitely seems like a bad scenario to be in. Fold preflop.

[pzhon]

[ QUOTE ]
What is the probability that AA is beat, when the board shows JT983 against 3 random hands?

[/ QUOTE ]
PokerStove (http://www.pokerstove.com) says that when no flush is possible, AA has an equity of 16.7075%, with 16.66% wins and 0.04% ties. This was small enough for an exact calculation.

[ QUOTE ]
Any Q or any 7 beat you, so 8 of the 45 cards you can't see can't be in any of the three hands if you're good. The probability that nobody has a straight is therefore

[/ 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% /images/graemlins/smile.gif ), 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. /images/graemlins/tongue.gif

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.

[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% /images/graemlins/smile.gif ), 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. /images/graemlins/tongue.gif

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

[hansdebelge]

not gonna argue the math, but is it logical that there is a 71% chance someone has 8 cards of 45 cards???? somethings gotta be wrong.

[danq]

[ QUOTE ]
not gonna argue the math, but is it logical that there is a 71% chance someone has 8 cards of 45 cards???? somethings gotta be wrong.

[/ QUOTE ]

Well, 8 out of 45 cards are queens or sevens, and the opponents have a total of 6 cards, so the expected number of queens and sevens in their hands is 6 * 8/45 = 48/45, a little bit more than 1. This doesn't answer the question, but it does reinforce the fact that the probability is pretty high... (The 48/45 double-counts the probability of more than one queen-or-seven, which you don't care about.)

Beyond that, all I can tell you is that the math is right. (Or, to put it another way: remove seven blanks from a deck of cards, shuffle, and deal off the top six; I'll bet you two-to-one there's a queen or a seven in there, as many times as you like.)

Dan

Forget my post, my mind were playing horrible tricks on me... I was mixing up 2 point of views. And I came to very strange conclusions. Im really sorry.