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Old 12-13-2004, 12:09 PM
NoPepperJack NoPepperJack is offline
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Default A pair in the flop

I have done the math in my head but doubt my understanding of how to get there. Here is the problem..
Assuming that no one has a matching card, what are the chances the board will pair on the flop? Then what is the chance, if the flop has a pair.. that someone at the table has made a set... and does the number of players at the table make a difference?

This came up because yesterday I was plaqued with catching top pair... but facing the flop with a pair... making me lay down my top pair when someone bet hard at the pot.

Please explain how you came to your numbers?

Thanks,

Jack
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Old 12-14-2004, 02:10 AM
Cobra Cobra is offline
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Default Re: A pair in the flop

I will give this a shot. I am new to the probability area so hopefully someone else will check me.

1. The probability the flop has a pair in it. The flop must come in the form of (x,x,z), therefore,

13*3*48/c(52,3)=.0847%

Thats 13 ways to choose the first x, 3 ways to choose the second and 48 other cards for the third.

2. Assuming 10 handed and you do not have an x, therefore you have two z's. This leaves 2x's and 45 z's left.

Probability breaking the possible hands into vectors.

(z,z) No one has a set. c(2,0)*c(45,18)/c(47,18)=.376%

(x,z) One person has a set. c(2,1)*c(45,17)/c(47,18)=.483%

(x,x) Two people have a set or someone has 4 of a kind.

c(2,2)*c(45,16)/c(47,18)=.142%

Now these percentages are not that they are still there after the flop, just that they were dealt the hands prior to the flop.

If you go down to five handed the percentages are as follows.

(z,z) No one was dealt a set. c(2,0)*c(45,8)/c(47,8)=.685%

(x,z)=.289%

(x,x)=.026%

So the answer is yes it does depend on how many people are dealt cards.

Cobra
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  #3  
Old 12-14-2004, 02:28 AM
gaming_mouse gaming_mouse is offline
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Default Re: A pair in the flop

[ QUOTE ]
1. The probability the flop has a pair in it. The flop must come in the form of (x,x,z), therefore,

13*3*48/c(52,3)=.0847%

Thats 13 ways to choose the first x, 3 ways to choose the second and 48 other cards for the third.

[/ QUOTE ]

Almost. For each rank, there are (4 choose 2) = 6 possible ways to make a pair. Each of those goes with 48 possible 3rd cards. So it should be 13*6*48/(52 choose 3) = .1694.

Don't have time now to check the other part.

gm
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Old 12-14-2004, 02:51 PM
NoPepperJack NoPepperJack is offline
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Default Re: A pair in the flop

Thank you both for the answers. I am having some trouble understanding parts of the answer. I hope you don't mind giving me a little math lesson.

So it should be 13*6*48/(52 choose 3) = .1694

The "c" or choose is confusing to me. I understand +-/* but not choose. Can you explain the process?
Also, since the percentage is .1694 that translates to about a 5.9/1 dog. Meaning one in 6 flops will probably contain a pair. If that is true, I have indeed learned something.

Thanks,
Jack


And does
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  #5  
Old 12-14-2004, 03:10 PM
Dave H. Dave H. is offline
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Default Re: A pair in the flop

[ QUOTE ]
So it should be 13*6*48/(52 choose 3) = .1694

The "c" or choose is confusing to me. I understand +-/* but not choose. Can you explain the process?
Also, since the percentage is .1694 that translates to about a 5.9/1 dog. Meaning one in 6 flops will probably contain a pair. If that is true, I have indeed learned something.


[/ QUOTE ]

52 choose 3 represents the number of ways you can choose 3 cards (the flop) from 52 cards (the entire deck). Order does NOT matter, i.e. A,2,3 is the same as 3,A,2

The calculation for this is 52!/(3!*(52-3)!)

Note: 52! = 52*51*50*...*1
3! = 3*2*1
(52-3)! = 49! = 49*48*47*46*...*1

If you write down the numbers, you see that most of them cancel out.

That is: 52!/(3!*(52-3)!) =

(52*51*50)/(3*2) = 22,100

Hope that helps.
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