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#1
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Searched...no luck...
Three questions...
What are the odds (and the math) of flopping exactly 1 pr from two non-paired cards? I've seen anywhere from 27-30%. What are the odds of making a straight by the river (pre-flop odds, that is) of connectors (87o etc.)? And for 1-gap connectors? If it's been posted in the past, please post the link. I couldn't find it. Thanks in advance. J |
#2
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Re: Searched...no luck...
[ QUOTE ]
What are the odds (and the math) of flopping exactly 1 pr from two non-paired cards? I've seen anywhere from 27-30% [/ QUOTE ] 1-[(44/50)*(43/49)*(42/48)]= a 32.5% chance to make a pair or better, when holding 2 no-paired cards. [ QUOTE ] What are the odds of making a straight by the river (pre-flop odds, that is) of connectors (87o etc.)? [/ QUOTE ] This is actually hard to do because different connectors make a different number of straights. For AK and A2: [4*4*4*C(47,2)]/C(50,5)=~3% For KQ and 23: [2*4*4*4*C(47,2)]/C(50,5)=~6.5% For QJ and 34: [3*4*4*4*C(47,2)]/C(50,5)=~10% For all other connectors: [4*4*4*4*C(47,2)]/C(50,5)=~13% [ QUOTE ] And for 1-gap connectors? [/ QUOTE ] For AQ and A3: [4*4*4*C(47,2)]/C(50,5)=~3% For KJ and 24: [2*4*4*4*C(47,2)]/C(50,5)=~6.5% For all others: [3*4*4*4*C(47,2)]/C(50,5)=~10% Note: The numbers 1-gapers only include when you use both of your hole cards to make the straight. It does not include the times when you hold AQ and the board reads JT98x I will add these calculations later, if you want. |
#3
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Re: Searched...no luck...
Thanks. Really appreciate the assist. How would the calculation change if I wanted to figure out the odds of exactly 1 pr...not 1 pr or better?
J |
#4
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Re: Searched...no luck...
Sorry that was a typo, the odds in my OP are for 1 pair only.
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#5
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Re: Searched...no luck...
I apolagize, the number in my original post represents the chance that 1 or more ace or king flop.
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#6
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Re: Searched...no luck...
The probability of getting a pair and only a pair is
=2*3*(44c2)/(50c3) = 28.96% Assume you have AK there are three aces left and three kings. There are 44 other cards. Let say you want the probablity of hitting a pair of aces and only a pair that would be 3*(44c2)/(50c3) but the probability of hitting either ace or a king would be twice this. Cobra |
#7
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Re: Searched...no luck...
Actually what I gave you would include the hands that have a pair of board cards as well. This would give you two pair. One in your hand and one on the board. I have to go right now but if you want the hands that only have a pair that is yours and the board does not pair, I can do this latter.
Cobra |
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