Texas Holdem Hands Probability

[DoctorWard]

OK here are my workings...
a) Suits = 13 different cards with 4 choose 2 suits = 13 x 4c2 = 78 hands
b) Suited connector = 13 combinations with 4 suits = 13 x 4 = 52 hands
c) All paint = 16 paint cards choose 2 = 16c2 = 120 hands
d) A-x suited = A with 12 other cards by 4 suits = 12 x 4 = 48 hands

Total = 298 hands

Agree now with all except (c), am confident of these workings!

Then the overlaps of A-x suited agree it is 12 overlaps
Then the overlaps of Suited connectors with All paint is 12 overlaps (A-K, K-Q & Q-J by the 4 suits)
Then the overlaps of Suits with All paint is 24 overlaps (A-A, K-K, Q-Q & J-J by 4 suits choose 2, i.e. 4 x 6 = 24)

Therefore we have
= 298 - 12 - 12 -24
= 250 hands!

[gaming_mouse]

[ QUOTE ]

c) All paint = 16 paint cards choose 2 = 16c2 = 120 hands


Agree now with all except (c), am confident of these workings!


[/ QUOTE ]

Your calculation includes things like AA, KK, etc, which are already counted in the pairs.

[gaming_mouse]

[ QUOTE ]


(78 + 52 + 86 + 48) - 12 - 8 -12 = 232



[/ QUOTE ]

In my orig post I made an arithmetic error. 86 above should be 96. The correct final answer is 242.

[jason1990]

1. any pair -- 6*13 = 78 hands
2. any suited connector -- 13*4 = 52 hands
3. all paint hole cards -- 16C2 = 120 hands
4. A-x suited -- 12*4 = 48 hands

1&2 -- 0
1&3 -- 6*4 = 24
1&4 -- 0
2&3 -- 3*4 = 12
2&4 -- 2*4 = 8
3&4 -- 3*4 = 12

1&2&3 -- 0
1&2&4 -- 0
1&3&4 -- 0
2&3&4 -- 4

1&2&3&4 -- 0

Total = 78 + 52 + 120 + 48 - 24 - 12 - 8 - 12 + 4
= 246.

Does this agree with any of your answers?

[jason1990]

Did you subtract AKs one too many times?

[gaming_mouse]

[ QUOTE ]
Did you subtract AKs one too many times?

[/ QUOTE ]

Yep. I forgot the third term in inclusion-exclusion. Nice catch.

[DoctorWard]

Does everyone agree?

ANSWER:
P(any pair) = 13 different cards with 4 suits to choose from, with only 2 cards available = 13 x 4C2 = 13 x 6 = 78 hands;
P(suited connector) = 13 combinations within one suit by 4 suits = 13 x 4 = 52 hands;
P(all paint) = there are 6 ways to combine A, K, Q & J, 4 choose 2. For each, there are 16 possible hands = 16 x 4C2 = 16 x 6 = 96 hands; and
P(A-x suited) = A goes with 12 other cards in that suit and there is 4 suits = 12 x 4 = 48 hands.

These hands are not all mutually exclusive therefore remove:
a) A-x suited overlaps with All Paint in 12 situations (A-K, A-Q & A-J by the 4 suits);
b) All paint overlap with Suited Connectors in 12 situations (A-K suited, K-Q suited & Q-J suited by the 4 suits);
c) A-x suited overlaps with Suited Connects in 8 situations (A-K and A-2 by the 4 suits);and
d) All paint overlap with Pairs in 24 situations (A-A, K-K, Q-Q & J-J by 4 suits choosing 2, i.e. 4 x 4C2 = 4 x 6 = 24).

Then since we have removed all A-K suited so need to add back A-K by the 4 suits = 1 x 4 = 4

Therefore the total hands are equal to:
= 78 + 52 + 120 + 48 - 12 - 12 - 8 - 24 + 4
= 246 hands

The total possible combination of hands is 52 choose 2 = 52C2 = 1,326

Therefore you will play 246 in 1,326 hands or 18.195 % of the time. Therefore, you come out betting approximately 2 hands per round.

[DoctorWard]

Another way to look at it

Probability working is as follows:

P(pair) = 1 x 3/51
P(suited connector) = 1 x 2/51
P(A-x suited) = 1/13 x 12/51 x 2 but we remove A-2 and A-K which was covered by suited connector,
= 1/13 x 10/51 x 2
P(all paint) = (1/13 x 3/51 x 2) x 6 (for AK, AQ, AJ, KQ, KJ, QJ)

= 18.25%

[DoctorWard]

ANSWER:
P(Pair) = 13 different cards with 4 suits to choose from, with only 2 cards available = 13 x 4C2 = 13 x 6 = 78 hands;
P(Suited Connector) = 13 combinations within one suit by 4 suits = 13 x 4 = 52 hands;
P(All Paint) = 16 paint cards in the deck, with only 2 to choose from = 16C2 = 120 hands; and
P(A-X Suited) = Ace goes with 12 other cards in that suit and there are 4 suits = 12 x 4 = 48 hands.

Therefore, a total of:
= 78 + 52 + 120 + 48
= 298 hands

However, some of these hands are not mutually exclusive therefore we remove:
a) A-x Suited overlaps with All Paint in 12 situations (A-K, A-Q & A-J by the 4 suits);
b) A-x Suited overlaps with Suited Connectors in 4 situations (A-2 by the 4 suits) (n.b. it may seem you should remove A-K in this scenario too however, doing this will remove A-K entirely in which case you will need to add it back);
c) All Paint overlap with Suited Connectors in 12 situations (A-K suited, K-Q suited & Q-J suited by the 4 suits); and
d) All Paint overlap with Pairs in 24 situations (A-A, K-K, Q-Q & J-J by 4 suits choosing 2, i.e. 4 x 4C2 = 4 x 6 = 24).

Therefore, we remove a total of:
= 12 + 4 + 12 + 24
= 52 hands

Therefore, the total hands that meet the criteria are:
= 298 - 52
= 246 hands

The total possible combination of hands is 52 choose 2 = 52C2 = 1,326

Therefore you will play 246 in 1,326 hands or 18.55% of the time.

OR
Using probabilities...

P(Pair) = 52/52 x 3/51
P(Suited Connector) = 52/52 x 2/51
P(All Paint) = (16/52 x 15/51) - (16/52 x 3/51) {to remove Pairs} - (4 suits x 2! order x 3 cards (A-K, K-Q & Q-J) = (4 x 3 x 2) / (52 x 51)
i.e. P(All Paint) = (16/52 x 15/51) - (16/52 x 3/51) - (4 x 3 x 2)/(52 x 51)
P(A-x Suited) = 1/13 x 12/51 x 2 but we remove A-2 which was covered by suited connector and we remove A-K, A-Q & A-J suited which was covered by All Paint
= 1/13 x 8/51 x 2

= 1/17 + 2/51 + 42/663 + 16/663
= 123/663
= 18.55%

Therefore, you come out betting approximately 2 hands per round.

[pzhon]

As a side note, many people consider 32 to be a connector, but it is no more connected than 52. In fact, 32 makes fewer straights than 52 as there are fewer one-card straights, just as many potential two-card straights, but more two-card straights by 32 are counterfeited.

http://twodimes.net/h/?z=872198
5[img]/images/graemlins/heart.gif[/img] 2[img]/images/graemlins/heart.gif[/img] beats A[img]/images/graemlins/spade.gif[/img] A[img]/images/graemlins/club.gif[/img] 6.94% with all other hearts, 2s and 5s removed from the deck. The only way to win is with a straight.

http://twodimes.net/h/?z=872201
3[img]/images/graemlins/heart.gif[/img] 2[img]/images/graemlins/heart.gif[/img] beats A[img]/images/graemlins/spade.gif[/img] A[img]/images/graemlins/club.gif[/img] 5.19% with all other hearts, 2s and 3s removed from the deck. The only way to win is with a straight.

I consider 32 (with 42, KQ, and KJ) to be a two-gapper and 43 (or QJ) to be a one-gap connector.