Texas Holdem Hands Probability

[DoctorWard]

You sit down at a $5-$10 limit table where the blinds are $2.50 (SB) and $5.00 (BB). There are nine others at the table. Your starting criteria regardless of whether you post a blind or the pot has been raised is that you play every time with:

any pair;
any suited connector;
two all paint hole cards (i.e. A-K, A-Q, A-J, K-Q, K-J, Q-J); and
A-x suited

How many times a round (i.e. a round being when the dealer button moves on full rotation around) would you "come out" betting?

Show workings please?!

[gaming_mouse]

[ QUOTE ]
You sit down at a $5-$10 limit table where the blinds are $2.50 (SB) and $5.00 (BB). There are nine others at the table. Your starting criteria regardless of whether you post a blind or the pot has been raised is that you play every time with:

any pair;
any suited connector;
two all paint hole cards (i.e. A-K, A-Q, A-J, K-Q, K-J, Q-J); and
A-x suited

How many times a round (i.e. a round being when the dealer button moves on full rotation around) would you "come out" betting?

Show workings please?!

[/ QUOTE ]

any pair -- 6*13 = 78 hands
any suited connector -- 13*4 = 52 hands
all paint hole cards -- 16*6 = 86 hands
A-x suited -- 12*4 = 48 hands

There are 52 choose 2 = 1326 total hands

(78 + 52 + 86 + 48)/1326 = 0.199095023

Just under 20% of the time

[DoctorWard]

Gaming Mouse

Thanks for that however, aren't there a number of those outcomes which are double ups? Eg. Ac & Kc is both a suited connector and all paint. Therefore, I come up with a mutually exclusive result of 210 different hands out of 1,326 which is 15.84%.

Thoughts?

[gaming_mouse]

[ QUOTE ]
Gaming Mouse

Thanks for that however, aren't there a number of those outcomes which are double ups? Eg. Ac & Kc is both a suited connector and all paint. Therefore, I come up with a mutually exclusive result of 210 different hands out of 1,326 which is 15.84%.

Thoughts?

[/ QUOTE ]

Yep. Careless mistake on my part. I didn't double check your math, but as long as you removed the doubles from Ax, suited connectors, and paint i'm sure you got it.

[DoctorWard]

Sorry to ask such a pointless question but can you please confirm there are 210 mutually exclusive combinations. I think I have removed them all however, I am not 100%. No rush ~ your assistance is appreciated.

Doctor_Ward

[gaming_mouse]

[ QUOTE ]
Sorry to ask such a pointless question but can you please confirm there are 210 mutually exclusive combinations. I think I have removed them all however, I am not 100%. No rush ~ your assistance is appreciated.

Doctor_Ward

[/ QUOTE ]

There are 3*4 = 12 double counted paint.
A2s and AKs represent another 8 double counted cards.

So we have:
(78 + 52 + 86 + 48) - 12 - 8 = 244

So that is different from your count. Not sure why? Do you see anything I missed?

[DoctorWard]

I count at least 20 double ups, however some might be triple ups?

For example, Ah Kh is a suited connector, all paint AND A-x suited.

I still get 210

[gaming_mouse]

[ QUOTE ]
I count at least 20 double ups, however some might be triple ups?

For example, Ah Kh is a suited connector, all paint AND A-x suited.


[/ QUOTE ]

I took this into account.

Why don't you show me your work, so I can double check that?

[DoctorWard]

Gaming Mouse

I have set up a spreadsheet for this so potentially it's easier for me to email it to you?

However, in the attempt of getting this resolved here are my calculations...
a) Pairs = 78 hands
b) Suited Connectors = 55 hands
c) All Paint (A, K, Q & J) = 120 hands
d) A-x suited = 48 hands

Firstly this differs from your workings in both (b) and (c)

Now going in order you look at pairs, so you get 78 hands. Then you add in the suited connectors of 55 hands and nothing here overlaps, therefore we are at 133 hands.

Next add in the all paint of 120 hands however, there are 36 overlaps here, so now we are at 217 hands.

Next add in the A-x suited of 48 hands however, there are a further 16 overlaps, so we get to 249 hands.

This is my result!

78 + 55 + 120 + 48 - 36 - 16 = 249 hands

[gaming_mouse]

[ QUOTE ]

a) Pairs = 78 hands
b) Suited Connectors = 55 hands
c) All Paint (A, K, Q & J) = 120 hands
d) A-x suited = 48 hands

Firstly this differs from your workings in both (b) and (c)

[/ QUOTE ]

How did you get 55 for suited connectors? There are 13 possible: A2-TJ is 10, JQ,QK,KA is 3 more. For each, there are 4 hands, one for each suit. 13*4 = 52.

There are 6 ways to combine A,K,Q & J. For each, there are 16 possible hands. 16*6 = 96.

[ QUOTE ]

Next add in the all paint of 120 hands however, there are 36 overlaps here, so now we are at 217 hands.

[/ QUOTE ]

There are 3 connected hands among the 6 combinations of paint cards: JQ, QK, KA. For each one, 4 hands are also suited. Thus 4*3 = 12 overlaps.


[ QUOTE ]
Next add in the A-x suited of 48 hands however, there are a further 16 overlaps, so we get to 249 hands.

[/ QUOTE ]

A2 and AK represent 4 overlaps each with suited connectors, for a total of 8 overlaps with suited connecters. In my original response, I forgot to take into account the Ax suited overlaps with paint: AKs, AQs, AJs -- gives us 12 more overlaps.

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

So I'm still getting a different answer. As you're still leaving out your work, I can't see where you (or I) am going wrong....

[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/images/graemlins/heart.gif 2/images/graemlins/heart.gif beats A/images/graemlins/spade.gif A/images/graemlins/club.gif 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/images/graemlins/heart.gif 2/images/graemlins/heart.gif beats A/images/graemlins/spade.gif A/images/graemlins/club.gif 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.