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  Poker - Probability calculations.
 
  #1  
19-01-2007, 2:55 AM
Shoestringx
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Location: Waterloo
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Probability calculations.

Ok so this came up in the Test your Small Stakes Hold'em ....... threads in the Hand analysis section, so I decided to post about it in case anyone was interested.

I'll start with somthing very simple, the probability of drawing a specific pocket pair (for this example I'll show with A's).

p(A,A) = p(A1) * p(A2) (where p(A1) and p(A2) are probabilities of getting an ace on the first and second card respectively.

P(A1) is very simple, there are 52 cards in the deck and 4 aces therefore 4/52.
P(A2) is a also very simple, there are 51 remaining cards in the deck and 3 aces, therefore 3/51

p(A,A) = (4/52)*(3/51) = 0.45%

That calculation was easy as there is only one possible outcome of cards, and they both need to be Aces.

However lets say you wanted to find the probability of finding any A,K then you need to do:

p(A,K) = p(A1)*p(K2) + p(K1)*p(A2)
= (4/52)*(3/51) + (4/52)*(3/51)
= 0.90%

Now lets say you are dealt those two A,A to start, what are the odds of flopping a set (or maybe better, but no quads)?

p(Aset) = p(A3)*p(X1)*p(X2) + p(X1)*p(A3)*p(X2) + p(X1)*p(X2)*p(A3)
=[(2/50)*(48/49)*(47/48)] + [(48/50)*(2/49)*(47/48)] +
[(48/50)*(47/49)*(2/48)]
= 11.51%

I'll take some time later tonight and work out a couple more complicated examples if there is any interest in that (I think Zebranky mentioned an example, I'll probably try that one later).
 

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  #2  
19-01-2007, 3:00 AM
mrsnake3695
I'm confused
 
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OK, this is way to complicated for me.
  #3  
19-01-2007, 3:04 AM
pokerrqueenn
CardsChat Elite
 
Location: virginia
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Quote:
Originally Posted by mrsnake3695
OK, this is way to complicated for me.
rofl. so it isn't just me.
  #4  
19-01-2007, 3:44 AM
Welly
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Quote:
Originally Posted by Shoestringx

p(Aset) = p(A3)*p(X1)*p(X2) + p(X1)*p(A3)*p(X2) + p(X1)*p(X2)*p(A3)
=[(2/50)*(48/49)*(47/48)] + [(48/50)*(2/49)*(47/48)] +
[(48/50)*(47/49)*(2/48)]
= 11.51%
All 3 parts of the equation are the same.

Multiple by 3 instead.
  #5  
19-01-2007, 3:46 AM
Shoestringx
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Location: Waterloo
Plays at: PokerStars
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Posts: 222
Quote:
Originally Posted by Welly
All 3 parts of the equation are the same.

Multiple by 3 instead.
I know. Ijust wanted to show the full equation for the purpose of explanation.
  #6  
19-01-2007, 5:25 AM
ChuckTs
whitebread
 
Location: lopping off my C-game
Posts: 11,567
Could you do me a favour, shoestring, and list the permutation and combination equations out for me? Like I said before, can't remember much from high school; this was actually the class (statistics + probability) that got me thinking to applying it to poker, as most of the example questions were perfect for it.
  #7  
19-01-2007, 7:12 AM
zowzza
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From your example, "However lets say you wanted to find the probability of finding any A,K then you need to do:
p(A,K) = p(A1)*p(K2) + p(K1)*p(A2)= (4/52)*(3/51) + (4/52)*(3/51)= 0.90%"

shouldn't it really be p(A,K) = p(A1)*p(K2) + p(K1)*p(A2)= (4/52)*(4/51) + (4/52)*(4/51)= 1.2% since you have still have 4 cards left in the deck to hit. you've only taken out one card from the deck but none of the hit cards for the second card you need, just dimished the deck by one card. Or even the simplest formula should be [(8/52)*(4/51)] if I'm correct, since the first pocket card can be one of 8 cards and the second can be one of 4 cards still in the deck but the deck has dimished by one card. and that would give you the net result of .012 or 1.2%
  #8  
19-01-2007, 7:33 AM
Shoestringx
Expert Member
 
Location: Waterloo
Plays at: PokerStars
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Posts: 222
Quote:
Originally Posted by zowzza
From your example, "However lets say you wanted to find the probability of finding any A,K then you need to do:
p(A,K) = p(A1)*p(K2) + p(K1)*p(A2)= (4/52)*(3/51) + (4/52)*(3/51)= 0.90%"

shouldn't it really be p(A,K) = p(A1)*p(K2) + p(K1)*p(A2)= (4/52)*(4/51) + (4/52)*(4/51)= 1.2% since you have still have 4 cards left in the deck to hit. you've only taken out one card from the deck but none of the hit cards for the second card you need, just dimished the deck by one card. Or even the simplest formula should be [(8/52)*(4/51)] if I'm correct, since the first pocket card can be one of 8 cards and the second can be one of 4 cards still in the deck but the deck has dimished by one card. and that would give you the net result of .012 or 1.2%
Yes it should, sorry my fault, thanks for pointing that out The second equation you used : [(8/52)*(4/51)] is obviously also correct, I am trying to show the entire equation for the sake of it being easier to understand, but it can be simplified the way you have:
(a*b) + (a*b) = 2a*b

Chuck for these calculations, the only equation you need is the basic equation of probability:

p = n/N

Where n is the # of favorable cases and N is the number of total cases.

The equations listed in my book for Permutations and Combinations are:

P(n,r) = n!/(n-r)! The # of permutations of n things taken r at a time.

C(n,r) = n!/r!(n-r)! The # of combinations of n things taken r at a time.

Not sure if these are what you are looking for?

Last edited by Shoestringx : 19-01-2007 at 7:51 AM.
  #9  
19-01-2007, 9:10 AM
ChuckTs
whitebread
 
Location: lopping off my C-game
Posts: 11,567
Exactly what I was lookin for, shoestring - ty sir
  #10  
19-01-2007, 4:08 PM
millarski
Advanced Member
 
Location: Scotland
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Posts: 153
I found these two articles for poker probability on wikipedia. They are quite lengthy but are an interesting read.

Hold'em

http://en.wikipedia.org/wiki/Poker_p..._hold_%27em%29

Omaha

http://en.wikipedia.org/wiki/Poker_p...ty_%28Omaha%29
 



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