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  #1  
Old 09-15-2005, 03:59 AM
davet davet is offline
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Default trips to full houses


Scenario. I flop trips, or improve to trips by the river, what is the probability of some other player having a full house?

Conditions: 9 player game with no pre- flop raising, even with AA.
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  #2  
Old 09-15-2005, 07:25 PM
KJL KJL is offline
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Default Re: trips to full houses

Do you have a pocket pair?
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  #3  
Old 09-15-2005, 09:07 PM
davet davet is offline
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Default Re: trips to full houses

[ QUOTE ]
Do you have a pocket pair?

[/ QUOTE ]

no.

quick note: Trips and Set have specific definitions.

Trips: one card in your hand and two on the board.

Set: Two cards in your hand and one on the board.

I figure you already know this, and are just checking, no insult intended here.

I am implying, however, that my opponent has a pocket pair.
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  #4  
Old 09-15-2005, 09:21 PM
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Default Re: trips to full houses

what is it when u have a 55 in your hand and 55 is on the flop.. trips right
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  #5  
Old 09-15-2005, 10:53 PM
KJL KJL is offline
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Default Re: trips to full houses

Yea, I was just checking.
Ok so you hold A7 and the board reads KQJ77. The probability that villian has either KK, QQ, or JJ is:
3*C(3,2)/C(45,2)=1/110 or .9%
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  #6  
Old 09-16-2005, 12:51 AM
elitegimp elitegimp is offline
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Default Re: trips to full houses

[ QUOTE ]
[ QUOTE ]
Do you have a pocket pair?

[/ QUOTE ]

no.

...

I am implying, however, that my opponent has a pocket pair.

[/ QUOTE ]

Why? What if you have 23 and your opponent has 34 on a board of

3347J

edit: by my count, there are 9 ways your opponent could have 3x for a boat, and 9 ways (s)he could have a pocket pair for a boat, so there are 18 possible hands he could hold. Since you are looking at 7 of the 52 cards in the deck (your hand and the board), there are only 45 cards left to deal your opponent's hand with => 990 possible hands he could have.

18/990 = 2/110, or about 0.0181818 or 1.8%

So, IF YOU ARE HEADS UP, the probability that your opponent has a full house when you have trips is 1.8%

The probability one of 9 players have it can be approximated by 1-(972/990)^9, which is approximately 0.152 or 15.2% However, this over counts times when 2 or more opponents have boats, so you'd want to use inclusion-exclusion to get an exact answer.

If I had to guess, though, I'd say around 14% if you assume that everyone goes to show down. A computer simulation would give a much better approximation, but I'm not feelin' it tonight. Sorry [img]/images/graemlins/smile.gif[/img]
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  #7  
Old 09-16-2005, 02:33 AM
danzasmack danzasmack is offline
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Default Re: trips to full houses

[ QUOTE ]

Scenario. I flop trips, or improve to trips by the river, what is the probability of some other player having a full house?

Conditions: 9 player game with no pre- flop raising, even with AA.

[/ QUOTE ]

Pretty good if he is loose/passive and capped.
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