KK vs. AA

[avisco01]

Just curious...when you hold KK at a 9 handed table, what are the odds that someone has AA? What I mean is, for all the times you are dealt KK, how many times will you be up against AA? I thought I read somewhere (don't remember where) that for every 25 times you hold KK, you will be up against AA once, thus its roughly 25:1. Is this true, and if so, how is this figured? As the number of players decreases, all the way down to heads up, how less likely is it for this horrible scenario to take place? Any help would be greatly appreciated, thanks.

Yeah. If I'm not mistaken, the odds of a specific hand is 212 to 1. Multiply that by 8, and your at approx 8/200 or 1/25.

[BruceZ]

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Just curious...when you hold KK at a 9 handed table, what are the odds that someone has AA? What I mean is, for all the times you are dealt KK, how many times will you be up against AA? I thought I read somewhere (don't remember where) that for every 25 times you hold KK, you will be up against AA once, thus its roughly 25:1. Is this true, and if so, how is this figured? As the number of players decreases, all the way down to heads up, how less likely is it for this horrible scenario to take place? Any help would be greatly appreciated, thanks.

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8*6/C(50,2) - C(8,2)/C(50,4) =~ 24.6-to-1

See details in this post.

Next time use the search function.

[AaronBrown]

Not quite. The odds of getting a specific pair dealt to you, like AA, is 1/221 or 220 to 1. There's one chance in 13 (4 in 52) of getting an Ace on the first card, and if you do, one chance in 17 (3 in 51) on the second. 13*17 = 221.

Once you have KK, or any other two non-Aces, the chances of another player getting AA increases to 4/50 * 3/49 = 12/2,450 or about 1/204. That's the appropriate number to use for this problem.

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Not quite. The odds of getting a specific pair dealt to you, like AA, is 1/221 or 220 to 1. There's one chance in 13 (4 in 52) of getting an Ace on the first card, and if you do, one chance in 17 (3 in 51) on the second. 13*17 = 221.

Once you have KK, or any other two non-Aces, the chances of another player getting AA increases to 4/50 * 3/49 = 12/2,450 or about 1/204. That's the appropriate number to use for this problem.

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wouldn't that be heads up?

[BruceZ]

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Not quite. The odds of getting a specific pair dealt to you, like AA, is 1/221 or 220 to 1. There's one chance in 13 (4 in 52) of getting an Ace on the first card, and if you do, one chance in 17 (3 in 51) on the second. 13*17 = 221.

Once you have KK, or any other two non-Aces, the chances of another player getting AA increases to 4/50 * 3/49 = 12/2,450 or about 1/204. That's the appropriate number to use for this problem.

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wouldn't that be heads up?

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He's talking about the chance of one particular player having AA when you have KK. That's the same whether it's heads-up or a full table. Then if you want the chance that any of your N opponents have it, you would multiply that number by N for a very good approximation. To be exact, you would then have to subtract the probability that 2 opponents have it, which is C(N,2)/C(50,4), but this is a small number.

This is the exact result, hopefully ... [img]/images/graemlins/wink.gif[/img]

There are 50 cards (without the two kings) 4 of which are aces. We randomly draw 18 cards without replacement and randomly place those 18 cards into 9 hands, 2 cards each. What is the probability that at least one hand contains two aces?

First we draw 18 out of 50, that gives us a sample space of C(50,18) possible ways to draw 18 from 50. Now there are three cases in which at least one hand could contain two aces.


1. Among the 18 cards drawn there are exactly 4 aces.
The prob. for this event is C(46,14)C(4,4). The prob. that at least two of those are in one hand is 9*C(16,2)/C(18,4). From the 18 "free spaces" we pick four for the aces, 9 of those cases put two aces together in one hand and for the other aces there are C(16,2) free spaces left.

2. Among the 18 cards drawn there are exactly 3 aces.
The prob. for this event is C(46,15)C(4,3). The prob. that two of those are in one hand is 9*C(16,1)/C(18,3). From the 18 "free spaces" we pick three for the aces, 9 of those cases put two aces together in one hand and for the other aces there are C(16,1) = 16 free spaces left.

3. Among the 18 cards drawn there are exactly 2 aces.
The prob. for this event is C(46,16)C(4,2). The prob. that those are in one hand is 9*C(16,0)/C(18,2).

All together we get:

(C(46,14)*C(4,4)*9*(16,2)/C(18,4)+
C(46,15)*C(4,3)*9*(16,1)/C(18,3)+
C(46,16)*C(4,2)*9*(16,0)/C(18,2)) / C(50,18)

= 0.0440816

[BruceZ]

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This is the exact result, hopefully ... [img]/images/graemlins/wink.gif[/img]

There are 50 cards (without the two kings) 4 of which are aces. We randomly draw 18 cards without replacement and randomly place those 18 cards into 9 hands, 2 cards each. What is the probability that at least one hand contains two aces?

First we draw 18 out of 50, that gives us a sample space of C(50,18) possible ways to draw 18 from 50. Now there are three cases in which at least one hand could contain two aces.


1. Among the 18 cards drawn there are exactly 4 aces.
The prob. for this event is C(46,14)C(4,4). The prob. that at least two of those are in one hand is 9*C(16,2)/C(18,4). From the 18 "free spaces" we pick four for the aces, 9 of those cases put two aces together in one hand and for the other aces there are C(16,2) free spaces left.

2. Among the 18 cards drawn there are exactly 3 aces.
The prob. for this event is C(46,15)C(4,3). The prob. that two of those are in one hand is 9*C(16,1)/C(18,3). From the 18 "free spaces" we pick three for the aces, 9 of those cases put two aces together in one hand and for the other aces there are C(16,1) = 16 free spaces left.

3. Among the 18 cards drawn there are exactly 2 aces.
The prob. for this event is C(46,16)C(4,2). The prob. that those are in one hand is 9*C(16,0)/C(18,2).

All together we get:

(C(46,14)*C(4,4)*9*(16,2)/C(18,4)+
C(46,15)*C(4,3)*9*(16,1)/C(18,3)+
C(46,16)*C(4,2)*9*(16,0)/C(18,2)) / C(50,18)

= 0.0440816

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Congratulations, you just found the world's most complicated way to compute 9*6/C(50,2), which is just 9 times the probability of 1 player being dealt AA. This is what your expression reduces to.

As I already explained, this double counts the cases where 2 players have AA, so these must be subtracted off. In your case, this double counting comes from the term 9*C(16,2).

Why did you do all of this? Didn't you understand that I already posted the exact solution in this thread? The exact answer for 9 opponents is simply:

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

The probability of 2 players getting AA is subtracted by the second term. I even posted a link to this post which explains this in detail with the inclusion-exclusion principle.

Why don't you use the search function? [img]/images/graemlins/confused.gif[/img]

[LetYouDown]

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Congratulations, you just found the world's most complicated way to compute 9*6/C(50,2)

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Probability POTD for sure.

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Why don't you use the search function?

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Say what?