Probability calculations.

Shoestringx

Shoestringx

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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).
 
mrsnake3695

mrsnake3695

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OK, this is way to complicated for me.
 
Welly

Welly

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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.
 
ChuckTs

ChuckTs

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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.
 
zowzza

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%
 
Shoestringx

Shoestringx

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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%

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:
ChuckTs

ChuckTs

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Exactly what I was lookin for, shoestring - ty sir :)
 
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