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JeffSpicoli
09-24-2005, 08:04 PM
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

KJL
09-24-2005, 11:11 PM
[ 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.

JeffSpicoli
09-25-2005, 12:14 AM
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

KJL
09-25-2005, 12:27 AM
Sorry that was a typo, the odds in my OP are for 1 pair only.

Cobra
09-25-2005, 01:08 PM
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

Cobra
09-25-2005, 01:11 PM
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

KJL
09-25-2005, 07:17 PM
I apolagize, the number in my original post represents the chance that 1 or more ace or king flop.