simple Q for you math geniuses

[AH-TitanFan]

For the odds of 1 person making a pocket pair wouldn't it just be 3/51 ~ .0588, since the first card doesn't really matter, but the second card must pair with the first card. Thus there is a 3/51 chance a player will be dealt a pocket pair. (or approx. ~ 6%)

[BrassGuru]

Yes, that is another way to look at it. From a "counting" perspective, I get the likelihood of a person being a pocket pair as 13 * 4C2 /52C2 = 0.588 = 5.88%. Yours has a more intuitive feel, but we get to the same place.

[CrazyN8]

[ QUOTE ]
The chance of 3 particular people all having a pocket pair is (using a very slight and negligible cheat by assuming that all 4 are of different ranks):

(13*6/(52 choose 2))*((12*6+1)/(50 choose 2))*((11*6+2)/(48 choose 2)) = 0.00021131857

And for 4:

0.00021131857*(10*6+3)/(46 choose 2) = .0000128628695

The second term of inclusion exclusion is therefore:

378*.0000128628695 + (630 -378)*0.00021131857= 0.0581144443

Which means that the 1 term approximation is not very accurate after all. The 2nd term approximation is:

12.6 - 5.8 = 6.8

but is only guarenteed to be accurate within 5.8%. I don't feel like going after the 3rd term, as it only get messier. But it looks like the binomial answer will be pretty close after we add back on the 3rd term.

[/ QUOTE ]

do i read that right...the chance of 3 people having pocket prs is .02% ?

what about all 3 flopping a set?

had this happen once. I had middle set.

[gaming_mouse]

[ QUOTE ]
..the chance of 3 people having pocket prs is .02% ?

[/ QUOTE ]

No, that was for 3 particular people. Like, seat 1, 2, and 3. The chance of ANY three people is much higher (see binimial approximations in previous posts)

[ QUOTE ]
what about all 3 flopping a set?

[/ QUOTE ]

I havn't done this yet, cause I'm lazy. Also, you need to clarify what you mean. GIVEN that those three people already all have pocket pairs, or just the chance that on the next hand, 3 people will flop a set?

[seemee]

So to clarify the question, as Im awful at maths:

If 2 people hold a different wired pair, what are the odds of them both making trips by river?

[gaming_mouse]

[ QUOTE ]
So to clarify the question, as Im awful at maths:

If 2 people hold a different wired pair, what are the odds of them both making trips by river?

[/ QUOTE ]

2*2*(44 choose 3)/(48 choose 5), or about 3%