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View Full Version : The AA vs KK. Is it really so trivial ?

10-14-2005, 01:39 PM
Say you are holding KK or some other hand (not important as long as an ace isn't part of it)

Then I have read on the forum that the probability of EXACTLY one holding AA is:

9*C(4,2)/C(50,2) - C(9,2)/C(50,4)

Now, this should be read like the probability that atleast one player has AA minus the probability that exactly two players has AA. So far so god. But I dont feel convinced that the first part is calculated correctly.

lets say that A to I represent players getting AA with P(A) = P(B) = ... = P(I) = the probability that player x gets AA = C(4,2)/C(50,2)

Then the probability of atleast one player getting AA should be P(A U B U...U I) = P(A)+P(B)+...+P(I) - P(AB) - P(AC) - ... - P(HI) and they are the only ones we have to care about since the probability for all with 3 or more will be zero sice only two person can have AA.
We then get P(A U B U...U I) = 9*C(4,2)/C(50,2)- 36 * (the probability that two have AA) = 9*C(4,2)/C(50,2) - 36*1/C(50,4)

Now we have compensated for calculating the intersections twice and we have the answer for atleast one player holds AA. To get Exactly one we need to remove 36*1/C(50,4) one more time.

So in total we get:
9*C(4,2)/C(50,2) - 2*C(9,2)/C(50,4) ~4,38%

AaronBrown
10-14-2005, 06:23 PM
You are correct. The first formula is the probability that at least one player has AA. Your formula is the probability for exactly one player.

BruceZ
10-14-2005, 06:51 PM
[ QUOTE ]
Say you are holding KK or some other hand (not important as long as an ace isn't part of it)

Then I have read on the forum that the probability of EXACTLY one holding AA is:

9*C(4,2)/C(50,2) - C(9,2)/C(50,4)

[/ QUOTE ]

This is the probabilty of at least 1 player having AA, which is the the case for which I have always given this probabilty.

10-15-2005, 07:27 AM
Yea. I see now. I think I've been confusing myself whether it was atleast one or exactly one. But everything makes sence now, thanks guys