A,A or A,Ks or A,Ko at a full table, what are the odds

[rybones]

does anyone know where to find or have the odds of someone having A,A at a table of 10 during a given hand?

I also have the same question for A,Ks and A,Ko. and would like to know how those odds change as the table gets smaller?

In addition, what are the odds of two or more players having pp 10 handed. Again, how does this change as the number of players goes down?

Any thoughts on where to find this info would be great.

Any info on how to do the math would be fun to look at, but for me it would be like seeing the opera: looks and sounds great, but I don't get it.

Thanks,

Ryan

[rybones]

I have tried the search function and can't find anything. does anyone have a link?

Ryan

[A_C_Slater]

There are 6 ways to make AA and 16 ways to make AK. The odds of getting AA are 220 to 1.

[rybones]

sorry, I must have phrased my question wrong. I know my odds for any specific pocket pair is 220 to 1. What I want to know is what are the odds of anyone at a full table having A,A? do I simply devide 220/10 and the answer is 22 to 1 that someone will have A,A? or is it more complex?

Also, would that simply mean that the odd of two or more pocket pairs in a given hand are 44 to 1? and 66 to 1 for three sets of pp?

Thanks to anyone who can answer

Ryan

[A_C_Slater]

That I do not know. But I do know that if you have KK the odds of someone holding AA is 19.5 to 1 at a full table.

[MortalWombatDotCom]

[ QUOTE ]
does anyone know where to find or have the odds of someone having A,A at a table of 10 during a given hand?

I also have the same question for A,Ks and A,Ko. and would like to know how those odds change as the table gets smaller?

In addition, what are the odds of two or more players having pp 10 handed. Again, how does this change as the number of players goes down?

Any thoughts on where to find this info would be great.

Any info on how to do the math would be fun to look at, but for me it would be like seeing the opera: looks and sounds great, but I don't get it.

Thanks,

Ryan

[/ QUOTE ]

so, i figured it out to be ~ 4.5%, about what you'd expect if you could just multiply 10 * (1/221).

you can't just multiply 10*(1/221) because the probabilities of players getting AA are not independent, but with highly improbable events and small numbers of trials, the approximation is close.

anyway, for the combinatorics savvy, here's my methodology (let me know if you see an error):

let p2a be the probability that 2 aces total are dealt to 10 players. let p3a and p4a be the probabilities that 3 or 4 aces total are dealt to 10 players, respectively.

p2a = choose(4, 2) * choose(48, 18) / choose (52, 20)
pick two of the 4 aces, pick 18 of the non-aces, divide by the total number of ways to pick 20 cards out of 52. similarly:
p3a = choose(4, 3) * choose(48, 17) / choose (52, 20)
p4a = choose(4, 4) * choose(48, 16) / choose (52, 20)

let s2 be the probability that at least one player has AA given 2 aces are dealt. let s3 and s4 be defined analogously for 3 and 4 total aces dealt.

if 2 aces are dealt to the players, assume without loss of generality that one of them is AS. whoever is dealt the AS has a 1/19 chance of also gettting the other A. s2 = 1/19

if 3 aces are dealt to the players, assume without loss of generality that two of them are AS and AH. whoever is dealt the AS has a 1/19 chance of also getting the AH. if that person does not get the AH, then there is a 2/18 chance that the remaining A is dealt to one of the two ace holders. so the chance of someone holding 2 aces is 1/19 + (1-1/19)(2/18) = 3/19. therefore s3 = 3/19.

similarly, with 4 aces dealt, repeat the above experiment for the first 3 aces. if nobody has been dealt 2 of them, there is still a 3/17 chance that one of the three players with AS, AH, or AD gets the AC. s4 = 3/19 + (1-3/19)(3/17)

so the probability that at least one player is dealt AA is p2a * s2 + p3a * s3 + p4a * s4 ~ .045

[gaming_mouse]

Mortal,

You are right that just multiplying by 10 gives a good approximation.

I think your method for the exact answer is correct, but there is definitely an easier way. Just use the inclusion-exclusion principle. Here is a longer than necessary explanation of it:

Let A1 = P(Player 1 has aces)
Let A2 = P(Player 2 has aces)
etc.

Then:

P(A1 or A2 or .. or A10) = P(A1) + P(A2) + ... + P(A10) - P(A1 & A2) - P(A1 & A3) - ... - P(A9 & A10)

The subtraction is done over all possible pairs of two players. But we can just solve for P(A1 & A2) and then multiply by (9 choose 2):

(9 choose 2)*(6/1326)*(1/1325)= 0.000122940323

So the above tiny number is what you subtract from the approximation

10*(6/1326)

to get the exact answer. As you noted, the approximation is almost exact anyway.

You might want to check if this agrees with your method, but I'm 95% sure it will.

gm