PDA

View Full Version : Back door etc

08-21-2002, 04:51 PM
2 q's

1 What is the probability of flopping a straight w/ JT or an open ender?

2. What is the probability of a backdoor stright if Q-7-2 or 9-2-3 flops to that same hand? I tried to calculate it but my answer didn't seem right as it just about equalled the prob of a backdoor flush.

Thanx

Joe

08-21-2002, 05:23 PM
P(flopping a straight if you hold JT)=

P( AKQ)+ P(KQ9) +P( Q98) +P(987)=

4(64)/ 50C3

P(making a straight if you hold JT and the board is Q-7-2)=

P(AK) +P(K9) +P(98)

= 3(16)/47C2

Note P(back door flush)= 10C2/47C2= 45/47C2 so it should be close.

Ill do openended later (its more of a hassle).

08-21-2002, 07:46 PM
what is C?

I hope this isn't a retarded question.

Joe

08-21-2002, 11:12 PM
like... 50 choose 3...

number of ways to choose 3 things from 50.

formula is something like n!/k!(n-k)!... I believe... it's been a long time since I did any math and didn't just look up an answer.

k is the number you are choosing, 3 in this case, and n is the total number... 50 in this case.

~D

08-22-2002, 12:51 AM
Right, or combinations.

C(50,3) = 50!/[(47!)(3!)] = 50*49*48/3! That is, there are 50 ways to choose the first, 49 ways to choose the 2nd, and 48 ways to choose the 3rd, then we divide by 3! = 6 since there are 3! ways to get the same 3 in a different order. If we didn't divide by 3! we would have

P(50,3) or "permutations of 50 things taken 3 at a time".

08-22-2002, 05:49 AM
P(openend w/a paired board)=

P(KKQ) +P(QQK) +P( QQ 9) + P99Q) +P(998) +P(889) =48(3)/50C3

P(openended w/o a paired board)=

P(KQ X (Xnot K, Q, A or 9)+ P(Q9X) +(P98X)

=16(34) + 16(34) +16(34)/50C3= 48(34)/50C3

Finally we can also compute the probability of a double gutter ball.

P(K97) +P(AQ8) =2(64)/50C3.

SO P(openended or double gutter)=48(3) +37(48)+ 2(64) / 50C3= about 10%

and also P(straight or openended)= 48(40)+2(64)+ 6(64)/ 50C3=