odds of full house on board
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odds of full house on board
I have tried to find this using odds calculators but can't seem to find odds on just the board....can anyone tell me a site that will let me calculate a full house coming out on the board? Thanks so much...nice to be here...
- wptworker
- Posts: 5
- Joined: Wed Apr 07, 2004 8:17 pm GMT
Well... I can figure this out using probablilty, but right now I don't have the time. I will try to figure it out tomorrow, but look online for the odds of drawing a full house on your first five cards in draw poker. That will tell you pretty much the same thing.
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thepheonix - Posts: 53
- Joined: Tue Jan 20, 2004 11:28 pm GMT
Okay.... i did some rough calculations and here is what I got:
Chances of a Full house on the board (if either card in your hand does not match the board): 0.1268%
about 1 in every 789 hands
This is a rough calculation and here is how I got it:
The way I figured it would be if the cards were all dealt one by one. Each one calculated as a separate event with separate probability.
Lets call card "X" the card that appears 3 times on the board, and card "Y" the one that appears twice.
There are then only 10 distinguishable permutations for these cards (since suits do not matter):
XXXYY
XXYYX
XXYXY
XYXXY
XYYXX
XYXYX
YYXXX
YXYXX
YXXXY
YXXYX
The probability that one certain permutation of the full house falls is: 0.00162% therefore the probability that any one of the 10 falls makes the number increase by a factor of ten giving us the answer! Yay me!
Please feel free to make what you think is a correction to this.
Chances of a Full house on the board (if either card in your hand does not match the board): 0.1268%
about 1 in every 789 hands
This is a rough calculation and here is how I got it:
The way I figured it would be if the cards were all dealt one by one. Each one calculated as a separate event with separate probability.
Lets call card "X" the card that appears 3 times on the board, and card "Y" the one that appears twice.
There are then only 10 distinguishable permutations for these cards (since suits do not matter):
XXXYY
XXYYX
XXYXY
XYXXY
XYYXX
XYXYX
YYXXX
YXYXX
YXXXY
YXXYX
The probability that one certain permutation of the full house falls is: 0.00162% therefore the probability that any one of the 10 falls makes the number increase by a factor of ten giving us the answer! Yay me!
Please feel free to make what you think is a correction to this.
-

thepheonix - Posts: 53
- Joined: Tue Jan 20, 2004 11:28 pm GMT
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