View Full Version : Full House By the River

Dave H.
11-17-2004, 02:42 PM

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!

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:


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.


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

11-18-2004, 10:35 AM
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.


Dave H.
11-18-2004, 11:04 AM
Good grief!
Nice job!!!...thanks much!