Odds of pockets aces being dealt twice in the same round

[Scott_Baio]

What are the odds of pocket aces being dealt to two people in the same hand at a 10 person table?

Could you show me the formula or equation you used to get the number? Thanks a lot

[SaintAces]

[BruceZ]

C(10,2)/C(52,4) = 6015-to-1.

1/C(52,4) is the proability of 2 specific players having the 4 aces, and we multiply this by the number of ways to choose 2 players out of 10, or C(10,2). Since no more than 2 players can have 4 aces, this method is exact.

[MarkD]

Not to mention extrememly concise. I sure wish I was better at using combinations - I never have been able to use them as well as I should.

[Zetack]

[ QUOTE ]
C(10,2)/C(52,4) = 6015-to-1.

1/C(52,4) is the proability of 2 specific players having the 4 aces, and we multiply this by the number of ways to choose 2 players out of 10, or C(10,2). Since no more than 2 players can have 4 aces, this method is exact.

[/ QUOTE ]

It could be he's asking, if * I * have aces, what are the odds anyother player has aces as well. Is that a different probability?

[BruceZ]

It could be he's asking, if * I * have aces, what are the odds anyother player has aces as well. Is that a different probability?

Yes, that is a very different probability than what he asked for. Of course it is much more likely that 2 players have aces if you already have aces. That's like the guy who takes his own bomb on a plane because he figures the odds of there being two bombs is astronomical.

If you have aces, the probabilty that one of your 9 opponents has has aces with you is 9/C(50,2) = 9/1225 or 135-to-1.

[wacki]

What is C(52,4) and how do I calculate this?

[BruceZ]

What is C(52,4) and how do I calculate this?

Number of ways to choose 4 things out of 52 irrespective of order, or "combinations of 52 things taken 4 at a time". It is equal to 52*51*50*49/(4*3*2*1) or 52!/48!/4!. There are 52 ways to choose the first, 51 ways to choose the second, 50 ways to choose the third, and 49 ways to choose the 4th, and then we divide by 4! different orders they can be chosen. In general, C(n,k) = n!/(n-k)!/k!. This can also be evaluated in Excel with the function COMBIN, or on some calculators.