View Full Version : Full House By the River
Dave H.
11-17-2004, 02:42 PM
Holdem:
As Ac in the pocket. Can someone provide the explanation/formula for determining the probability of a full house by the river?
How about As 9c in the pocket?
Much appreciated!
gaming_mouse
11-18-2004, 09:27 AM
Ignoring cases that produce 4 of a kind, mutually exclusive boards which make a full house with your pocket AA:
1) Trips + Two Pair
2) Trips Alone without A
3) Trips Alone + A
4) Two pair + A
5) One Pair + A
Trips + Two Pair (1584 total)
Consider a specific full house, say TTT33. How many are there?
4*6 = 24
Now, how many specific full houses are there, excluding any that use aces?
nCr(12,2) = 66
So there are 66*24 = 1584 possible boards containing a non-A full house.
Trips Alone without A (84480 total)
For each card rank, there are 4 ways to make trips. Since we ignore aces, there are 12 card ranks, giving us 48 ways to make trips.
Now lets consider how many ways the other 2 cards can come when there are trips on board, and we are not allowed to have aces or pairs. For the first card, there are:
52 - 4 - 4 = 44 possible cards.
There are only 40 for the second card.
So the total for this section is:
48*44*40 = 84480
Trips Alone + A (4224 total)
2 aces left. 44 cards which are neither aces nor part of the trips.
48*44*2 = 4224
Two pair + A (4752 total)
There are 6*6 = 36 ways to make up any specific two pair. There are 66 possible two pair combinations in which neither pair is a pair of aces (see full house calculation above). Since there are two aces left, we have:
36*66*2=4752
One Pair + A (253440 total)
The A is fixed, the pair is fixed, and there are 44*40 ways to put together the remaining two cards.
There are 12 possible non-ace pairs, 6 ways to make up each one. We have:
12*6*44*40*2 = 253440
Putting It Together
We add everything up and divide by the number of possible boards.
1584 + 84480 + 4224 + 4752 + 253440 = 348480
There are nCr(50,5) = 2118760 boards
348480/2118760=.1644, about 16%
I think that's right but it's late and I may have forgotten something or made an arithmetic error.
You'll have to solve the A9 case on your own /images/graemlins/laugh.gif
BTW, knowing these kinds of odds have little practical value at the poker table.
Cheers,
gm
Lost Wages
11-18-2004, 10:17 AM
So there are 66*24 = 1584 possible boards containing a non-A full house.
If he holds AA then he can't make a non-A full house. /images/graemlins/smile.gif
Lost Wages
gaming_mouse
11-18-2004, 10:35 AM
[ QUOTE ]
If he holds AA then he can't make a non-A full house. /images/graemlins/smile.gif
[/ QUOTE ]
Well, yes, I guess his aces would play. But the point is there is a full house on the board in these cases.
gm
Dave H.
11-18-2004, 11:04 AM
Good grief!
Nice job!!!...thanks much!
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