chances a pair will fall on flop

[jonnyreno]

What are the chances a pair will fall on the flop?

[Lost Wages]

Given that you hold 2 cards in your hand, there are C(50,2) = 19,600 possible flops.

Case 1
You hold an unpaired hand. There are 11 ranks that are not in your hand. Each of those can form C(4,2) = 6 combinations, so that's 6*11 = 66 pairs. Each pair can combine with the 46 remaining cards (52) - (your 2) - (the 4 cards of that rank). That gives us 66*46 = 3036 flops (so far). Note that we are only looking at paired flops, we are not counting flops like 777. Now there are 3 cards left of each of the two ranks in your hand. Each can make C(3,2) = 3 pair combinations. Each of those can combine with 47 remaining cards. So that's an additional 2*3*47 = 282 flops (that means that 282 flops will give you trips or a full house). So there are 3036+282 = 3318 paired flops, or a probability of 3318/19,600 = 16.9%.

Case 2
You hold a pair. Then there are 12*C(4,2)*46 + 1*C(2,2)*48 = 3360 paired flops (48 of which give you quads). 3360/19,600 = 17.1%.

Somebody please check me.

Lost Wages

[bigpooch]

As far as your calculations are concerned, they seem right.

Now, if we're just standing on the sidelines and watching
the flop hit the table, there will be C(52,3)=22100
possible flops of which 13x6x48= 3744 are paired or there is
a probability of 3744/22100 = 72/425 of exactly a pair.
This is a 16.94118% chance which compares with your two
cases.

[Lost Wages]

Thanks. I make a typo and it should have been C(52,3) = 19,600 flops (instead of C(50,2)).


Lost Wages

[bigpooch]

Of course you mean C(50,3) = 19600. I wasn't gonna say
anything, but you stubbed your toe! [img]/images/graemlins/smile.gif[/img]

[Lost Wages]

[img]/images/graemlins/blush.gif[/img] I did worse than that, I meant to reply to your message instead of my own. The thanks was for you checking my work.

Lost Wages