Odds of four of a kind holding a pocket pair in holdem

What are the exact odds of making any four of a kind versus any two,three,four or one specific four of a kind, holding a pocket pair in Texas Holdem?

16.473429

I believe you gave me the odds of holding a pocket pair. What are the odds of making four of a kind?

.00327

Thanks, would it be correct to say that .00327* number of players* hands dealt=probability of someone hitting four of a kind?

No it would not. Was hoping someone would challenge my answer. I have absolutely no idea !

I believe you are very close!

You are new here or you would know I just make up numbers. ME, I am not. SIGH

Has anyone else noticed a pattern where a new user, with a handle all in lower case, asks a very specific question on a topic, comments on replies, and then is never heard from again?

Not that I mind. But if it's just one person, there's no need for the multiple identities.

It looks like the probability of being dealt a pocket pair (0.0588) then getting 4 of kind in the next 5 cards (0.00816) is about 0.00048, or 1 in 2083.

You can estimate the probability of anyone getting 4 of kind using your formula, but I don't think it's exactly correct

close enough thank you. I think you have a handle on it.

According to wikipedia the probability of getting 4 of a kind using the best 5-cards out of a 7-card poker hand is:
p = 0.168% or about 1/3 of your value.

Frequency of 7-card poker hands

A better approximation is to assume a Bernoulli trial. Probability P of a specific hand occurring exactly x times in N hands given a probability p of occurring in one hand is a Bernoulli trial:

P(x) = COMB * p ^x * (1 - p) ^ (N - x)

where COMB = N! / x! (N-x)!

The approximation is to assume that each hand dealt to each player is independent when in actuality they are not. Therefore:

N = n * h (approximately)

where n = number of players
h = number of hands dealt

The probability of getting at least one hand (i.e. one or more) is:

P(x>0) = 1 - P(x=0)
= 1 - (N! / 0! (N-0)! ) p ^ 0 (1 - p) ^ (N - 0)
= 1 - (1 - p) ^ (n*h)

So for 10 players playing 100 hands and p = 0.168%:

P( x > 0 ) = 1 - (1 - 0.168%) ^ (10)(100)
= 1 - 18.5%
= 81.5%

Using your approximation:

P( x > 0) = (0.168%)(10)(100)
= 168%

which is clearly nonsensical.

Kinda looks like 10-Player chance.

You have 9-9... there are 18 other cards that cannot be a 9.
That leaves 32 cards eligibile for the remaining 2 nines.
Of the 32 cards remaining, five will be drawn that must contain both nines.
There are 13 possible pocket pairs 2 to Ace.

There is also the chance that TWO players draw quads following the example...

9-9 : K99KX where x is any other card, and also this case
9-9 : K99KK where one player has the case King, again, there are 12 possible "2nd Quads"

Gool Luck