I hate KK and QQ

[Sabrazack]

What are the chances that someone has AA when im holding KK? And what are the chances that someone has AA OR KK when i have QQ. Take a 10handed table for example..

[AaronBrown]

When you hold KK, there are 50 unknown cards and 4 of them are Aces. So the chance that one opponent has AA is (4/50)*(3/49) = 0.49%. With 9 opponents you have to multiply by 9 (4.41%) but subtract out the probability that two opponents have AA because that was double counted. That probability is 36*(4/50)*(3/49)*(2/48)*(1/47) = 0.02%. Your overall answer is 4.39%.

The chance of KK is the same as the chance of AA. So you start by doubling the 4.41% to get 8.82%. But the probability of two opponents both getting AA or KK goes up eight times, to 0.13%. That brings us down to 8.69%. To complete the calculation you add back the chance of three opponents getting AA or KK, and subtract the chance of four opponents. But the answer still rounds to 8.69%.

[Jordan Olsommer]

hmm im getting a much smaller answer than that

I did

(# of ways we can give Hero KK)*(# of ways we can pick someone to get AA)*(# of ways we can give them AA)*[ (# of ways we can deal hands to the remaining 8) - (# of ways we can pick one of the remaining 8 to get the other AA)*(# of ways we can deal hands to the remaining 7 )] all over the number of ways we can deal hands to 10 players.

or

(4c2)(9c1)(4c2)[ (48c2)(46c2)....(34c2) - 8*(46c2)...(34c2)]
------------------------------------------------------------------------
(52c2)(50c2)(48c2).....(34c2)

and I get .000198, or 0.02%

Any idea where I'm going wrong here?

[AaronBrown]

We interpreted the question differently. You answered what's the chance on a given deal that I hold KK and someone else holds AA? I answered, given that I hold KK, what is the chance someone else has AA? So if you multiply my answer by the chance of holding KK, you get your answer.

[Jordan Olsommer]

So I'm not going out of my mind? Sweet!

tyvm

[ellipse_87]

Can you expand a little on how you derived the probability of two holding AA, and that of two holding either AA or KK?

[BruceZ]

[ QUOTE ]
(4c2)(9c1)(4c2)[ (48c2)(46c2)....(34c2) - 8*(46c2)...(34c2)]
------------------------------------------------------------------------
(52c2)(50c2)(48c2).....(34c2)

[/ QUOTE ]

As Aaron said, you just included the probability that hero gets KK = 4c2/52c2 = 1/221. Also, as we discussed in that other problem, your first term double counts the times that 2 players have AA, and your second term that you subtract also double counts these cases, for the same reason that we discussed before. So in effect you are not counting the cases of 2 AA at all. If you only wanted the probability that exactly 1 AA was out, then this would be correct, and it is a valid use of the inclusion-exclusion principle (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&Board=&Number=417383&page=&v iew=&sb=5&o=&fpart=). However, this problem asked for the probability that any AA was out, so you would want to include the cases where there were 2 AA, and for this you would need to only subtract half of your second term.

Do you understand where the double counting is coming from now? If not, this will affect you on many other problems of this type.

Also, you can greatly simplify this if you follow the inclusion-exclusion procedure in my post, as it allows you to eliminate all of the terms for the remaining players. The entire expression can be replaced by:

9*6/C(50,2) - C(9,2)/C(50,4) = 4.39%.

Note that your calculation, once we remove the terms for KK, gives 4.37% due to the aforementioned double counting. If we change the -8 to -4, it then gives the correct answer of 4.39%.

[BruceZ]

[ QUOTE ]
Can you expand a little on how you derived the probability of two holding AA, and that of two holding either AA or KK?

[/ QUOTE ]

For two holding AA, the easiest method is to note that the probability that 2 particular players hold all 4 aces is 1/C(50,4), and since only 1 pair of players can hold 4 aces, we can just multiply this by the number of ways to choose the 2 players to get C(9,2)/C(50,4). That is, the pairs of players holding 4 aces are mutually exclusive, so we can just add the C(9,2) probabilities. The C(50,4) is a trick that can be used when all the cards are the same.

Now Aaron did 36*(4/50)*(3/49)*(2/48)*(1/47). The 36 is the number of ways to choose the 2 players C(9,2) = 9*8/2 = 36. Then the first player has probability 4/50 of getting an ace, the second player has probability 3/49 of getting one of the remaining aces, etc.

For AA and KK, this would become 36*(8/50)*(3/49)*(4/48)*(3/47). The 4/50 has been replaced by 8/50 since the first player can get an A or a K. Then, whichever he gets on the first card, he has 3 ways to get the same rank on the second card. Then the second player has a probability of 4/48 of getting the other rank, times 3/47 for the last card of that rank. This appears to be 12 times the value for AA alone, not 8 times. Plus we can still get two AA or two KK, so all together this would be 14 times that of two AA. Now for the exact answer, we would actually need to add back the probability of 3 players having these hands, and then subtract for 4 players, but in practice these terms are so small that they can be ignored.