Probability

How does one calculate the odds of being dealt, in no-limit poker, two pocket hands in a row (discounting suits), and assuming 9 players? Example: getting AJ in one hand and then AJ in the very next hand.

Being dealt A is 4 of 52...7.6%; then J 4 of 52...7.6%. Being dealt AJ, I would assume, is then 7.6% x 7.6% = .058%.

Assuming the above calculation is correct, would AJ twice in a row to the same player be .058% x .058% = .00033%?

Thanks for the help.

Quote: Chicago

How does one calculate the odds of being dealt, in no-limit poker, two pocket hands in a row (discounting suits), and assuming 9 players? Example: getting AJ in one hand and then AJ in the very next hand.

Being dealt A is 4 of 52...7.6%; then J 4 of 52...7.6%. Being dealt AJ, I would assume, is then 7.6% x 7.6% = .058%.

Assuming the above calculation is correct, would AJ twice in a row to the same player be .058% x .058% = .00033%?

Thanks for the help.

If you get dealt the Ace, it's gone, so the Jack coming out is no longer based on 52 cards, but 51. Secondly, either the Ace or the Jack can be the first card you get, so given that independency, the first card being EITHER an Ace or a Jack is 8/52. Thus:

8/52 * 4/51 = x

0.01206636500754 = x

You want this to happen for two consecutive hands, so:

0.01206636500754 * 0.01206636500754 = x

0.00014559716449518578 = x

or:

1/.00014559716449518578 = 6,868.3 or 1:6,868.3

NOTE: The number of players only becomes relevant to the calculation if you actually know what the players have, lacking this, all calculations would be based upon a complete deck.

Quote: Chicago

How does one calculate the odds of being dealt, in no-limit poker, two pocket hands in a row (discounting suits), and assuming 9 players?

If all cards have no suit, then the answer is 48/51 * (2 * 4/52 * 4/51) + 3/51 * (4/52 * 3/51) = 1.16%.
The first term is for a non-pair hand, catching it again is 2 * 4/52 * 4/51. You get a non-pair hand with probability 48/51.
The second term is for a paired hand, catching it again is 4/52 * 3/51. You get a paired hand with probability 3/51.

I thought he was specifically asking about A-J?

EDIT: The Wiz also seems to indirectly agree with me:

https://wizardofodds.com/ask-the-wizard/texas-hold-em/probability/

Halfway down the page:

Quote:

The probability of receiving ace/king is (8/52)*(4/51) = 0.012066.

The probability of receiving A-J would be no different, thus my formula is correct.

I think the original question wasn't stated clear enough. Does he look for a repetion of any hand, or any specific hand ?

I agree with you. Hopefully he'll come back and let us know!