Odds of fullhouse in 7 Cards

[mchilger]

Can someone show how to do this calculation?

tx, Matthew

[ciaran]

There are 35 ways to pull 3 cards from 7. In order to get a FH the first thing that must happen is that those three cards must match without matching any of the 4 remaining cards. 1*(3/51)*(2/50)*(48/49)*(47/48)*(46/47)*(45/46)*35

There are then 6 ways of pulling 2 cards from the remaining 4. For the FH to occur, those two cards must match without matching both of the remaining cards. Note that we've excluded the 4 cards that form our triplet and it's quad-making match already. 3/47*44/46*43/45*6

Multiply those two and you get 2.647% chance of a FH in 7 cards which matches (roughly) what I find here, http://www.pokerhands.com/poker_odds.html (they used a sim to get their result).

If that's not right, I'm sure Bugs will be along shortly to point out any errors.

[ciaran]

Bah, I did all that by hand and it looks like I'm overlooking something.

See, http://en.wikipedia.org/wiki/Poker_probability, which gives it as 2.60% (2.5961%, actually, the value in the table is rounded).

[Bugsbunny]

I'll try to get something up by tomorrow. I need to think how to approach this since there's a number of ways this can happen. I assume we're talking strictly Hold'em (since things would be different for Stud or Omaha).

Also - do you want strictly a FH, or do you want quads included as well? (I think it'll be easier if we include the quads - but not sure yet. I'll probably try to split it out in either case though).

[the_hawk]

I would take the probability of a full house in 5 cards (easily calculated) and multiply by 7_choose_5, although that's probably not quite right either.

[mchilger]

only full house, no quads.

tx Matthew

[janeg]

Al Spach shows how to do this.

3 ways to make a full house with 7 cards:

(a) 2 sets and a singleton.
Two Set ranks: C(13,2) = 78
Set: C(4,3) = 4
Singleton: C(44,1) = 44

78 * (4*4) * 44 = 54,912

(b) A set and 2 pairs
Set Rank: C(13,1) = 13
Pairs Rank: C(12,2) = 66
Set: C(4,3) = 4
Pair: C(4,2) = 6

13 * 66 * 4 * (6 * 6) = 123,552

(c) A set, a pair, and two singletons of separate rank
Set Rank: C(13,1) = 13
Pair Rank: C(12,1) = 12
Singleton Rank: C(11,2) = 55
Set: C(4,3) = 4
Pair: C(4,2) = 6
Singleton: C(4,1) = 4

13 * 12 * 55 * 4 * 6 * (4 * 4) = 3,294,720

Add all three results:
54,912 + 123,552 + 3,294,720 = 3,473,184 possible full house hands.

Total possible 7 card hands = C(52,7) = 133,784,560

3,473,184 / 133,784,560 = 0.025961022 or 2.6%

And I could never have worked that out myself in a million years

[Bugsbunny]

I get a slightly different result. Someone find my mistake:

Total 5 card combinations from 50 cards: 50 choose 2 = 2 118 760

First: If you hold a pair (AA): You need a board of:

BBB + (any 2 not A or B) (This would include BBBCC)
4*12 * (44 choose 2) = 48*946=45408 combinations

ABB + (any 2 not A or B)
2 * 6*12 * (44 choose 2) = 144 * 946 = 136224

ABBBC (This is not included in the above)
2 * 4*12 * 44 = 4224

4224 * 136224 + 45408 = 185856 combinations
185856/2118760= 0.087719232 = chance of getting FH if holding a pair

------
Next if we hold AB:
AAB + (any 2 not A or B) (This includes AABCC)
3 * 3 * (44 choose 2) = 9 * 946 =8514 *2 (BBA) = 17028

AABBC
3 * 3 * 44 = 396

AACCD
3 * 6*11 * 40 = 7920 * 2 (BBCCD) = 15840

AACCC
3 * 4*11 = 132 * 2 (BBCCC) = 264

ACCCD
3 * 4*11 * 40 = 5280 * 2 (BCCCD) = 10560

CCCDD
4*11 * 6*10 = 2640

2640 + 10560 + 264 + 15840 + 396 + 17028 = 46728

46728/2118760=0.0220544092 = chance of getting a FH with a non-pair hand


----
Possible 2 card hands (52 choose 2) = 1326
Possible pairs = 6*13 = 78
Possible non-pair hands = 1326-78 = 1248

Weighting the above results:

((185856/2118760)*78 + (46728/2118760)*1248)/1326=0.0259170458

[janeg]