Suited Connectors...

[DoctorWard]

Hello All

A question related to texas hold'em... If you hold any suited connector (eg. Ac-2c or 9h-Th) what is the probability of flopping a straight, flush or straight flush. Please express the result as a percentage to four decimal places and 1 in X to two decimal places. Since it is harder to flop a straight for A-2, 2-3 & 3-4 over middle suited connectors I think you need to weight the chance of getting each suited connector by the probability of flopping a straight, flush or straight flush for each suited connector.

Thanks for your help

[LetYouDown]

Grinded this out kinda quickly so if someone notices anything wrong, please correct.

19600 possible flops with any suited connector.

For 4-5 suited through J-10 suited:
252 combinations give you a straight and only a straight.
161 combinations give you a flush and only a flush.
4 combinations give you a straight flush.

For 3-4 suited and Q-J suited:
189 combinations give you a straight and only a straight.
162 combinations give you a flush and only a flush.
3 combinations give you a straight flush.

For 2-3 suited and K-Q suited:
124 combinations give you a straight and only a straight.
163 combinations give you a flush and only a flush.
2 combinations give you a straight flush.

For A-2 suited and A-K suited:
63 combinations give you a straight and only a straight.
164 combinations give you a flush and only a flush.
1 combination gives you a straight flush.

The math to combine the odds based on "any" suited connector should be trivial from here. Obviously the probability of each hand on the flop is the (number of combinations)/19600.

[DoctorWard]

thanks for the post...

I would agree with the basic methodology however, I would disagree with the odds for "2-3 suited and Q-K suited". Here I would propose the following:

For 2-3 suited and K-Q suited:
126 combinations give you a straight and only a straight (not 124 as previous)
163 combinations give you a flush and only a flush.
2 combinations give you a straight flush.

Assuming that I am correct with the above statement then the odds would be

A-2: 1/13 x 228/19600 = 228/254800
2-3: 1/13 x 291/19600 = 291/254800
3-4: 1/13 x 354/19600 = 354/254800
4-5: 1/13 x 417/19600 = 417/254800
5-6: 1/13 x 417/19600 = 417/254800
6-7: 1/13 x 417/19600 = 417/254800
7-8: 1/13 x 417/19600 = 417/254800
8-9: 1/13 x 417/19600 = 417/254800
9-T: 1/13 x 417/19600 = 417/254800
T-J: 1/13 x 417/19600 = 417/254800
J-Q: 1/13 x 354/19600 = 354/254800
Q-K: 1/13 x 291/19600 = 291/254800
K-A: 1/13 x 228/19600 = 228/254800

Summing to 4665/254800, or 1.8308%, or 1 in 54.62 times.

Agreed?