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View Full Version : Aces!!!.........to aces

HardCory
04-18-2005, 05:26 PM
what are the odds that I get aces and some one else has them also. Its probably already posted but the search isnt working

Beerfund
04-18-2005, 06:29 PM
[ QUOTE ]
Its probably already posted but the search isnt working

[/ QUOTE ]

No, never been discussed.

Mano
04-18-2005, 06:31 PM
Not sure if this is right, but I'll give it a shot:

Assuming a 10 player table, and you have two of the aces, there are C(50,18) ways to choose 18 cards, 18! different permutations, and 2^9 of each of those should be equivalent (switching each players first card with the second) so the total number of ways to deal two cards to the remaining nine players should be C(50,18)*18!/2^9 .

Using same method as above, if another player has the other two aces, nine possibilities for whose got the aces, there are C(48,16) ways to pick the 16 cards for the other 8 players, 16! orderings and 2^8 equivalent orderings. So C(48,16)*9*16!/2^8 different ways for another player to have two aces.

Dividing the bottom amount by the top, I get that there is about a .73% chance that another player has aces, or about 135:1 against.

BruceZ
04-18-2005, 07:09 PM
[ QUOTE ]
Not sure if this is right, but I'll give it a shot:

Assuming a 10 player table, and you have two of the aces, there are C(50,18) ways to choose 18 cards, 18! different permutations, and 2^9 of each of those should be equivalent (switching each players first card with the second) so the total number of ways to deal two cards to the remaining nine players should be C(50,18)*18!/2^9 .

Using same method as above, if another player has the other two aces, nine possibilities for whose got the aces, there are C(48,16) ways to pick the 16 cards for the other 8 players, 16! orderings and 2^8 equivalent orderings. So C(48,16)*9*16!/2^8 different ways for another player to have two aces.

Dividing the bottom amount by the top, I get that there is about a .73% chance that another player has aces, or about 135:1 against.

[/ QUOTE ]

If you just want to know the odds against another player having aces when you have aces, the answer is exactly 9*1/C(50,2) =~ 135:1. However, the question asked for the odds that you get aces and someone else has aces, and the odds against that are (1/221)*9/C(50,2) = 30,080:1.

Mano
04-19-2005, 12:15 PM
I did some cancelling above, and my answer does indeed reduce to 9/C(50,2). For some reason I thought that just multiplying the probability of getting the aces by nine would only be approximate. Bruces answer is obviously much simpler.

BruceZ
04-19-2005, 03:07 PM
[ QUOTE ]
I did some cancelling above, and my answer does indeed reduce to 9/C(50,2). For some reason I thought that just multiplying the probability of getting the aces by nine would only be approximate. Bruces answer is obviously much simpler.

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The reason it is exact is that only one other player can have AA, so the events of each player having AA are mutually exclusive.

This would not be the case if we wanted the probability of someone having AA when we hold KK, for example. In that case, multiplying by 9 would double count the times that 2 players have AA, so we must subtract that off, to get 9*6/C(50,2) - C(9,2)*1/C(50,4). In this case, the events of 2 players having AA are mutually exclusive, so we can multiply 1/C(50,4) by C(9,2). This is the inclusion-exclusion principle (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&amp;Number=417383&amp;page=&amp;view=&amp;sb =5&amp;o=&amp;vc=1).