heads up % someone flops pr or better

[J LU]

Let's say I'm on BB w/ K 7 (or whatever) and the SB completes and I check. Now the flop comes.

What are the odds that 1 of us will hit at least 1 pair? Lets assume that neither of us have pocket pair already.

How does this change with a full 9 or 10 player table?

This may have been done before so any links would also be appreciated.

[gaming_mouse]

Let's assume the flop comes with three different cards:

That means there are 9 cards out there (that either of you could hold) that would give at least one of you a pair. So
we calculate the change that neither of you has a pair. Well, assuming each of you has 2 random cards (not completely true, because SB CHOSE to complete and might have throw away some hands -- but the assumption is close enough) we calculate as follows:

1 - (40/49)*(39/48)*(38/47)*(37/46)=56%

The calculation is similar for more players, although the assumption of randomness becomes less and less valid. That is, with many players, the particular cards on the flop affect the calculation greatly. If the flop is QKA it becomes much more likely that someone has a pair than if the flop is 234. However, no matter what the flop is, it is highly likely that there is at least one pair with more than 3 players.

gm

[J LU]

Can you also show me the odds that both SB and BB would each flop a pair?

Thanks for your answer.

[gaming_mouse]

Can you also show me the odds that both SB and BB would each flop a pair?

Assuming each of them does not already have a pair and also assuming that none of their cards overlap, ie, the two players hold a total of 4 distinct cards. First we calculate the ordered probability of the first card pairing player 1, the second card pairing player 2, and allowing the third card to be anything. Then we multiply that by 6 to account for all possible orderings.

(6/48)*(6/47)*6=.0954, about 10%

We can remove the above assumptions but it's a bit of a pain and I don't feel like doing it. Also, note that if you have a pair, this DOES NOT MEAN there is a 10% chance that your opponent also has a pair. That would be a differenct calculation entirely, and the answer would be closer to 30%.

gm

[Cobra]

I am not an expert on proboabilities but here is an interesting web site I came accross.

http://wilsonsoftware.com/besthand.html

Cobra