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  #1  
Old 06-03-2005, 10:06 AM
PokerProdigy PokerProdigy is offline
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Default Ace-Queen

If you have ace-queen in your hand, what are the chances/probability that a king (and NO ace) will appear on the flop?
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  #2  
Old 06-03-2005, 10:19 AM
querulous querulous is offline
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Default Re: Ace-Queen

You want at least one king with no ace on the board?

4 * 46 * 45 = 8280 flops (four kings * 46 non-A cards left in deck * 45 non-A cards left in deck)

So 8280:11320 or about 42.2% of the time
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  #3  
Old 06-03-2005, 10:22 AM
LetYouDown LetYouDown is offline
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Default Re: Ace-Queen

I come up with 3612/19600 or about 18.4% or 5.43 to 1 for exactly one king and no ace.
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  #4  
Old 06-03-2005, 10:31 AM
LetYouDown LetYouDown is offline
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Default Re: Ace-Queen

A number of questions. 8280:11320 = 42%? If the odds of having at least one King on board are 42%, that should be the same for any non-Ace card. So every denomination has a 42% chance of appearing on the flop? Doesn't seem very likely.
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  #5  
Old 06-03-2005, 03:07 PM
querulous querulous is offline
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Default Re: Ace-Queen

I overcounted somewhere there, I get 4140:15460 now, which is 21.1%

Four ways to choose a king times 46c2 = 4140 out of 19600 possible flops that contain at least one K and zero A's.
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  #6  
Old 06-03-2005, 03:26 PM
LetYouDown LetYouDown is offline
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Default Re: Ace-Queen

I think you're still forgetting to take out the remaining Aces. 4 * 43 C2 is the number I came up with a few posts above.
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  #7  
Old 06-03-2005, 04:09 PM
AaronBrown AaronBrown is offline
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Default Re: Ace-Queen

I agree with LetYouDown. Out of 19,600 possible flops, 16,215 (82.7%) have no Ace if you hold AQ. There are 47 possible cards, 4 cards from each of 11 ranks plus 3 Queens. C(47,3) = 16,215. 4,140 (21.1%) of the flops have exactly one King. There are four Kings, plus 46 other cards (4 from each of 10 ranks, 3 Aces and 3 Queens), 4*C(46,2) = 4,140.

There are 3,612 (18.4%) flops with exactly one King and no Ace. There are four Kings, and 43 other cards (4 each from 10 ranks, plus three Queens). 4*C(43,2) = 3,612.
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  #8  
Old 06-03-2005, 05:12 PM
querulous querulous is offline
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Default Re: Ace-Queen

I'm not removing the remaining kings as you seem to be. You're not counting K-K-x or K-K-K.
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  #9  
Old 06-03-2005, 09:02 PM
LetYouDown LetYouDown is offline
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Default Re: Ace-Queen

But if you're saying 4 Kings * the remaining cards...you're gonna end up double counting your K-K combinations.
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  #10  
Old 06-03-2005, 11:03 PM
Cobra Cobra is offline
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Default Re: Ace-Queen

There are 19600 unique flops if you hold AQ. (50c3).

Out of the 50 remaining cards you have 3 aces, 4 kings and 43 other cards. The probability of getting One or more kings and no ace is as follows:

4*(43c2)+(4c2)*43+(4c3) = 3612+258+4 = 3874 flops.

The first term is the number of flops with one king and no ace, the second term is the number of flops with two kings and no ace, and the third is the number of flops with three kings.

Therefore the probability is 3874/19600 = 19.77%

Cobra
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