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  #1  
Old 04-05-2005, 01:07 AM
DoctorWard DoctorWard is offline
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Default Texas Holdem Hands Probability

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?!
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  #2  
Old 04-05-2005, 02:09 AM
gaming_mouse gaming_mouse is offline
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Default Re: Texas Holdem Hands Probability

[ 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
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  #3  
Old 04-05-2005, 03:50 AM
DoctorWard DoctorWard is offline
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Default Re: Texas Holdem Hands Probability

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?
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  #4  
Old 04-05-2005, 04:21 AM
gaming_mouse gaming_mouse is offline
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Default Re: Texas Holdem Hands Probability

[ 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.
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  #5  
Old 04-05-2005, 04:35 AM
DoctorWard DoctorWard is offline
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Default Re: Texas Holdem Hands Probability

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
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  #6  
Old 04-05-2005, 04:44 AM
gaming_mouse gaming_mouse is offline
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Default Re: Texas Holdem Hands Probability

[ 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?
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  #7  
Old 04-05-2005, 05:00 AM
DoctorWard DoctorWard is offline
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Default Re: Texas Holdem Hands Probability

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
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  #8  
Old 04-05-2005, 05:22 AM
gaming_mouse gaming_mouse is offline
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Default Re: Texas Holdem Hands Probability

[ 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?
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  #9  
Old 04-05-2005, 09:40 PM
DoctorWard DoctorWard is offline
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Default Re: Texas Holdem Hands Probability

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
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  #10  
Old 04-05-2005, 09:56 PM
gaming_mouse gaming_mouse is offline
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Default Re: Texas Holdem Hands Probability

[ 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....
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