1 / [(4/52) * (3/51) * (4/50) * (3/49) * (4/48) * (3/47) * (4/46) * (3/45) * (4/44) * (3/43)] = 2.3e11

Would someone kindly confirm this is correct or, if not, provide the correct way to calculate (or estimate) the odds of this?

Thanks in advance to any helpful person regarding this matter.

I'm not sure, but there may be additional division as well.

And then you'd have the odds of it happening on the next hand. I.E. You have to divide by the total number of hands you ever played or will ever play, to get the odds that you'll ever witness such an event.

52C20 * 10C5 / (52C2/6 * 50C2/6 * 48C2/6 * 46C2/6 * 44C2/6)

First term # of cards dealt to pocket

2nd term 5 of 10 Players dealt the hand

Denominator defines each pair as dealt (6 ways for each pair)

Quote:DJTeddyBearNo. You then have to divide by ( 10 * 9 * 8 * 7 * 6 ) since you don't really care who gets which pair.

I think that the number you are looking for here is 10 choose 5, ie, 10!/5!5! (not 10!/5!).

Quote:And then you'd have the odds of it happening on the next hand. I.E. You have to divide by the total number of hands you ever played or will ever play, to get the odds that you'll ever witness such an event.

You don't divide (or, rather, multiply) by the number of hands to get the probability that it happens at least once in that many hands. If that were true, you could get probabilities of over 100%, which is impossible.

The way to calculate this for independent events is (for n hands and an event that happens with probability p on one hand) 1 - (1 - p)^n.

In other words, the probability of it not happening on one hand is 1-p. Since the hands are independent, the probability of it not happening on any of n hands is (1-p)^n. Subtract from one to negate again and get the probability of it happening at least once in n hands, and you get 1 - (1-p)^n

Does it mean that five players had AA, AA, KK, KK, QQ, or five had AA, KK, QQ, JJ, TT (and, in the second case, do you want to assume that none of those were held by two players - e.g. it was not AA, AA, KK, QQ, JJ, TT)?

It's early, and I'm not going to double check my work.Quote:treecattSome years ago I was in a limit Texas Hold'em game in which the top five pocket pairs were dealt. I've always wondered what the odds were of this happening. I've done some calculations but I'm not sure I did it correctly.

1 / [(4/52) * (3/51) * (4/50) * (3/49) * (4/48) * (3/47) * (4/46) * (3/45) * (4/44) * (3/43)] = 2.3e11

Would someone kindly confirm this is correct or, if not, provide the correct way to calculate (or estimate) the odds of this?

Thanks in advance to any helpful person regarding this matter.

First, the answer depends on the number of players at the table. A full table has 10 seats, so we will assume all seats are full. You can see how to do the math 10 seats with this example, and modify it for whatever number of seats the table has.

For 10 seats, choose 5 of them for the premium hands. That's combin(10,5). Then, the hands can be arranged in any fashion in those 5 seats, so that's 5!. So, the number of ways the 5 premium hands can occur at the table is combin(10,5)*5! = 10*9*8*7*6 = 30240, in other words, this event can occur in 30240 ways. Another way to get the same result is that there are 10 locations for AA, then 9 locations for KK, then 8 locations for QQ, and so on.

Once the arrangement is fixed, you can go on to the probability of each particular hand in its location. The probability of AA in the first location is combin(4,2)/combin(52,2). After 2 cards are removed, the probability of KK in the second location is combin(4,2)/combin(50,2). The probability of QQ in the third location is combin(4,2)/combin(48,2). And so on.

The final answer (in a fashion you can copy/paste into Excel) is:

p = combin(10,5)*fact(5)*combin(4,2)^5/(combin(52,2)*combin(50,2)*combin(48,2)*combin(46,2)*combin(44,2)) = 0.00000013107, in other words, 1-in-7629256.

I may have made a mistake. That happens sometimes. My answer is the same as 98clubs, except for his term 52C20, which I don't understand. I have 5! in its place.