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  #1  
Old 07-26-2004, 02:47 AM
getfunky getfunky is offline
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Default Trips Probability Quetstion

I was trying to figure out how to calculate the odds of getting trips after the flop.

Calculating the odds of getting the third card of my pair is simple, but I was thinking I still need to calculate the odds of an unpaired card, either in my hand or on the board, hitting runner, runner to make trips by the river.

Thanks for the help.
-ST
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  #2  
Old 07-26-2004, 03:58 AM
Precision1C Precision1C is offline
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Default Re: Trips Probability Quetstion

Assuming that you have AB in your hand and CDE flop all of differnt ranks then the odds of AAA, BBB, CCC, DDD or EEE by the river are identical and equal to the inverse of 47 choose 2 times 3.

So each case has a .27% chance of occurring so the total chance is 1.39%.
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  #3  
Old 07-26-2004, 01:45 PM
Shawsy Shawsy is offline
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Default Re: Trips Probability Quetstion

[ QUOTE ]
I was trying to figure out how to calculate the odds of getting trips after the flop.

Calculating the odds of getting the third card of my pair is simple, but I was thinking I still need to calculate the odds of an unpaired card, either in my hand or on the board, hitting runner, runner to make trips by the river.

Thanks for the help.
-ST

[/ QUOTE ]

I think you have implied that you have a pair on the flop, and you want to know the odds of having trips on the river. The thing is, if two cards come runner, runner to make trips with a rank other than your pair, then you have made a full house, not just trips. So the way you have posed the question is open to a little interpretation. The way I read it, you are interested in the probability of making trips, a full house, or quads by the river when you have a pair on the flop. Correct me if I am wrong in this assumption.
You can make quads when two of your paired cards come on the turn and river. This will happen in C(2,2) or 1 way.
You can make a full house when one of your paired card comes and one of the unpaired cards comes. This will happen in C(2,1)xC(9,1) or 18 ways.
You can make trips when one of your paired cards comes, and the other card is not one of the unpaired cards. This can happen in C(2,1)x(36,1) or 72 ways.
You can make a full house when the turn and river are runner, runner with one of the unpaired cards. This will happen in 3xC(3,2) or 9 ways.
The probability of making trips, a full house or quads by the river when you have a pair on the flop is therefore 100/C(47,2) or 100/1081 = 9.25%.
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  #4  
Old 07-27-2004, 03:45 AM
getfunky getfunky is offline
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Default Re: Trips Probability Quetstion

I appologize for the confustion. I am trying to calculate the chance of getting three of a kind and no better by the river. Assuming that after the flop you have a pair. The pair can be anthing from a pocket pair, to the board pairing to one of your hole cards pairing up.

Thanks,
-ST
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  #5  
Old 07-27-2004, 10:36 AM
Shawsy Shawsy is offline
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Default Re: Trips Probability Quetstion

This should be easy then. If you have a pair on the flop, then the only way to make trips (and not a full house or quads) is to have exactly one card from your paired rank come, and a "blank" come.
In other words, your hand at the flop looks like BBCDE, and you need BX to come. (In your original post you were asking about runner, runner, but this would give you a full house, C D or E full of B.)
BX can come in C(2,1)xC(36,1) or 72 ways out of C(47,2) or 1081 ways, for a probability of about 6.7%.
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  #6  
Old 07-27-2004, 02:13 PM
getfunky getfunky is offline
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Default Re: Trips Probability Quetstion

Sometimes you don't see the forest for the trees. I guess my question was flawed from the start. Thanks for turning the lightbulb on.

-ST
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