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how to calculate these probabilites

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how to calculate these probabilites - Sat Feb 05, 2011, 04:03 AM
(#1)
gonchi1207's Avatar
Since: Dec 2010
Posts: 17
Hi there !,,,


well i have this doubt,..



one of my weakest areas is when i have to weigh whether my opponent is bluffing me or not... for example... if i hold K-K and the flop goes


A-7-2

and someone makes a raise of half size of the pot, i usually get trouble how to tell if hes holding an ace,, bluffing, or semi bluffing with a draw...

besides that, i am a person that like to know concrete numbers and probabilities, so lets see if someone can help me with these 2 situations , telling me the probabilities and explaining to me how they calculate them.


1 )

Me holding K-K raise to 3 times BB preflop and got 4 callers


FLOP : A-7-2

What is the probability that at least one opponent holds an ace?
what is the probability that one opponent holds pocket aces?



2)
Me holding A-5 other 2 players call
FLOP : A-7-7
What is the probability at least one opponent holds a 7?
what is the probability that at least one opponent holds an ace with a higher kicker?
what is the probability that at least one opponent holds either a 7, an ace with a higher kicker, or both ( thats it a-7 ) ?






thanks in advantage for the help
 
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Sat Feb 05, 2011, 09:22 AM
(#2)
TrumpinJoe's Avatar
Since: Jun 2010
Posts: 4,557
The answer to each of your scenarios is "It depends!". And what it depends on is your opponrnts range in each situation.
 
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Sat Feb 05, 2011, 01:28 PM
(#3)
gonchi1207's Avatar
Since: Dec 2010
Posts: 17
nah what do you mean by it depends?.. if you mean you cant be sure because someone would have folded an ace so theres less probability that some opponent remaining in the pot holds an ace too, then the same uncertanty apply when calculating outs.


i do have an idea on how to calculate these probabilites but i wanted to be sure... thats why i asked.. for example, for my first question my approach would be:


Me holding K-K raise to 3 times BB preflop and got 4 callers


FLOP : A-7-2

What is the probability that at least one opponent holds an ace?

theres 4 opponents on the pot so theres 8 hole cards, 2 for each of them,, then there are 8 possibilities of drawing an ace, and 3 aces remaining in the deck ( cause 1 is on the board )
sooooo...

since there are 47 cards in the deck to calculate all combinations of these 47 hands drawing 8 cards its comb(47,8)

47 ! / (47-8)!*8!

now lets calculate the drawings that does NOT include an ace , since there are 3 aces remaining in a 47 card deck 47-3 = 44

44! /(44-8)!8!

so, to calculate the probability of neither of them being dealt an ace its

(44! / (44-8)!8!) / (47 ! / (47-8)!*8!) = 0.56 = 56 %

so if i want to calculate the probability of one of them holding at least an ace i just have to from the total drawings extract those cases in which neither of them is holding and ace


100 - 56 = 44 % that at least one opponent is at least holding one ace.



can someone tell me if it is correct this approach and the calculations? can someone tell me the answer to the other scenarios i posted which i find them more difficult than this one?

thanks again
 
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Sat Feb 05, 2011, 05:56 PM
(#4)
TrumpinJoe's Avatar
Since: Jun 2010
Posts: 4,557
Your calculation assumes that you are getting 3 callers with any two cards, which is unreasonable. A better approach is to assign hand ranges to each caller and then calculate the probability of an ace being present in at least one of their hands.
 
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Sat Feb 05, 2011, 09:44 PM
(#5)
gonchi1207's Avatar
Since: Dec 2010
Posts: 17
aham, no i get why u said it depends... so i could narrow down the hand range for example if someone from early position called my bet i wont put him in 6-4 thats what youre saying....


though just to be sure my mathematical reasoning is right, suppose you have never played a single hand against these opponents and you know that they are super-agressive and willing to play any 2 cards. then,, is this calculation valid ?
 
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Sat Feb 05, 2011, 10:56 PM
(#6)
TrumpinJoe's Avatar
Since: Jun 2010
Posts: 4,557
Your methodolgy is appropriate for NE2 callers.
 
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Tue Feb 08, 2011, 10:27 AM
(#7)
JoBaloo's Avatar
Since: Jun 2010
Posts: 129
BronzeStar
Quote:
Originally Posted by gonchi1207 View Post
...
so, to calculate the probability of neither of them being dealt an ace its

(44! / (44-8)!8!) / (47 ! / (47-8)!*8!) = 0.56 = 56 %

so if i want to calculate the probability of one of them holding at least an ace i just have to from the total drawings extract those cases in which neither of them is holding and ace

100 - 56 = 44 % that at least one opponent is at least holding one ace.
...
It's probably more useful to use a tool like PokerStove to evaluate this situation first with 4 opponents and random cards (you win 41%), then giving 'em some brains (they all called your raise so at least top quarter of all PF hands) where your chances plummet to 14%. BTW, I told PokerStove the flop was a rainbow.

--
JoBaloo

Last edited by JoBaloo; Tue Feb 08, 2011 at 01:42 PM.. Reason: ...
 

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