View Full Version : odds of flopping quads

11-11-2002, 11:50 PM
What is the proper formula for figuring the odds
on flopping quads when I have a pocket pair in HE? Any help would be appreciated. I keep getting different answers. Thanks.

11-12-2002, 12:06 AM
You have 2 cards in your hand, leaving 50 unseen cards. There are (50 choose 3)=19600 different flops that can come up. (2 choose 2) * (48 choose 1)=48 of these give you quads. Probability of flopping quads is 48/19600.
Would you believe that once I flopped quads with a pocket pair 4 straight times? Well, you shouldn't.

11-12-2002, 12:13 AM
Thanks for reply. However, I figure the possible flops to be (50*49*48)=117600. I agree on the 48. So, I get 48/117600. Am I miscounting the total number of flop combos?

11-12-2002, 12:24 AM
50*49*48 gives you every possible combination, but it makes it order dependent. Dividing by 6 takes out the order dependency, and gives you the correct total of 19600 combinations.


11-12-2002, 12:26 AM
That's the issue! Thanks to both for the help. Got it now. Much appreciated.

11-14-2002, 03:36 PM
I tried this. Here is my results using Aces.

50*49*48=117600 total possible flops.

AAx = happens 48 times
AxA = happens 48 times
xAA = happens 48 times

144 ways two Aces can come up in 117600 flops.
144/117600 I get 3/2450. /forums/images/icons/confused.gif

Where does my logic fail? And what does (50 choose 3) and (2 choose 2) and (48 choose 1) mean?

11-14-2002, 06:34 PM
(N Choose K) is equal to the number of ways that one can choose K differentiable objects from N total differentiable objects (without repetition).

The formula is:

(N Choose K) = N! / [ (N-K)! K! ]

N! = 1 * 2 * 3 * ... * N

Hope that helps!


11-14-2002, 07:04 PM
So (48 choose 47) = (48 choose 1) = 48?
but (48 choose 46) = 1028?

I knew I should have studied math.

11-14-2002, 09:08 PM
"AAx = happens 48 times"

"Where does my logic fail? And what does (50 choose 3) and (2 choose 2) and (48 choose 1) mean? "

Your logic fails because there are two ways to get each outcome. Say you have AcAs, then board could come AhAdx or AdAhx. So you get 288/117600=3/1225=48/19600 as given earlier.


11-14-2002, 11:40 PM
Yes, that is precisely correct.

(48 choose 47) = (48 choose 1) = 48

This is a combinatorial identity, that is, (N choose K) = (N choose (N-K)).

Do you see why this must be true? Say you've got 48 balls, numbered from 1 to 48. If you choose 47 of them to take away, you can see what you did from a different perspective... that you chose 1 from 48 to leave. It follows then that if you have N differentiable objects, and you choose K of them, it's the same as if you "chose" (N-K) of them to not be chosen.

If you still don't see why this is true, please ask again, and I will try to provide a different example.


11-14-2002, 11:44 PM

11-15-2002, 03:13 PM
Now I see where my logic failed. Thank you Craig. AAx can happen 96 times, AxA also 96 times, and xAA 96. Muy bueno.

RMJ, your example was great. I even got it on the first read. /forums/images/icons/smile.gif