Probability of flopping a straight draw with a connector

[chalimac]

What is the probability of flopping a straight draw with an unsuited 0-gap connector like 76?
Does it differ MUCH from other connectors like 65 or T9?

Thank you. Can't find this anywhere.

[chapstick]

For 67 in the pocket I get 26.9%. It seems high.
But I do know that given a connected pocket like
67, you're about 9.4% to make a straight if you
see every community card, so 26.9% doesn't seem
too unreasonable.

Here's how I got it 26.9%:

There are C(50,3) possible flops. C(50,3)=19600.
Many of these possible flops produce a straight
draw (4 cards that could make a straight with the
turn or river). I'm assuming a straight draw does
not mean 4 cards in a row. So, 3467 is a straight
draw. If that's incorrect, I'll edit this post.
Anyway, here ar all the possible flops that could
make a straight draw:

3, 4, not(5)
3, 5, not(4)
4, 8, not(5)
5, 9, not(8)
8, T, not(9)
9, T, not(8)
4, 5, not(3 or 8)
5, 8, not(4 or 9)
8, 9, not(5 or T).

The first six in the list have only one "not" card, so
they represent six possible 4*4*38 combinations. For
exmaple, for {3, 4, not(5)}, there are 4 possible threes,
4 possible fours, and 38 possible cards that are not
threes, fours, or fives.
The last four in the list have two "not" cards, so
there are four 4*4*34 combinations.

In total, this gives
6*4*4*38 + 3*4*4*34 = 5280
possible combinations of flops that give a straight draw.

Dividing this by all the possible flops gives
5280/19600 = 26.9%.

Any errors? Anything unclear?

For 9T, or 56, the chances are equally likely.
For the end connectors like 23 or QK, the odds
decrease substantially since you have less
possible flops.

[HelloGoodbye]

"Does it differ MUCH from other connectors like 65 or T9?"

Doesn't differ at all.

[chalimac]

Thanks for the calculation.

If I get you right, then the probability of flopping an open-ended straight draw is only in these instances:

4, 5, not(3 or 8)
5, 8, not(4 or 9)
8, 9, not(5 or T)

That would be: 3*4*4*34 = 1632

1632/19600= 8.32% or 11 to 1

Is this right for the open-ended?

Thanks again.

[chapstick]

Yes, I think that is right. These scenarios are
in the class of probabalities that may be useful
to know, but are almost impossible to calculate on
the fly.

A small typo in my previous post. In the text
where I wrote "four 4*4*34 combinations", I
should have wrote "four 4*4*34 combinations".

[chalimac]

[ QUOTE ]

A small typo in my previous post. In the text
where I wrote "four 4*4*34 combinations", I
should have wrote "four 4*4*34 combinations".

[/ QUOTE ]

Can't see the difference.
I wanted that info to calculate odds preflop when facing a raise in No-limit holdem.
Thanks.

[chapstick]

Doh. I wrote "four" but I should have written "three.
Sorry.

[BruceZ]

[ QUOTE ]
Thanks for the calculation.

If I get you right, then the probability of flopping an open-ended straight draw is only in these instances:

4, 5, not(3 or 8)
5, 8, not(4 or 9)
8, 9, not(5 or T)

That would be: 3*4*4*34 = 1632

1632/19600= 8.32% or 11 to 1

Is this right for the open-ended?

Thanks again.

[/ QUOTE ]

There are problems with the way you are handling pairs (which is a common problem with this type of calculation). You are over counting draws which pair the hole cards, and you are not counting any draws which pair the board. I computed the probability of open-ended straight draws plus double gut shot draws in this archived post (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&Number=355216&page=&view=&sb =5&o=&vc=1). In this same post I also compute the probability of flush draws, both with and without the straight draws, and the numbers were collaborated with different methods. Here is the calculation from that post of the probability of open ended straight draws and double gut shots, including straight draws which are also flush draws:

[ QUOTE ]
[ 3*(15*34 + 1*27 + 2*6*4) + 4*4*4*2 - 2 ] / C(50,3) = 9.60%.


That is, there are 3 2-card combinations, 15 of these are unsuited, and there are 48-8-6 = 34 ways to pick the last card so that it doesn’t complete the straight and doesn’t pair the board. There is 1 2-card combination that is suited in our suit, and there are 34-7=27 ways to pick the last card so it doesn’t complete the straight, doesn’t pair the board, and doesn’t make a flush. We subtract 7 instead of 9 since the 2 flush cards that make a straight were already subtracted. The third term is for the paired boards. There are 2 ranks that can pair, 6 ways to make the pair, and 4 ways to choose the other card. Then we add the 4*4*4*2 double belly busters, and subtract 2 for the double belly buster draws that are flushes.

[/ QUOTE ]

[chapstick]

[ QUOTE ]
[ QUOTE ]
Thanks for the calculation.

If I get you right, then the probability of flopping an open-ended straight draw is only in these instances:

4, 5, not(3 or 8)
5, 8, not(4 or 9)
8, 9, not(5 or T)

That would be: 3*4*4*34 = 1632

1632/19600= 8.32% or 11 to 1

Is this right for the open-ended?

Thanks again.

[/ QUOTE ]

There are problems with the way you are handling pairs (which is a common problem with this type of calculation). You are over counting draws which pair the hole cards, and you are not counting any draws which pair the board.


[/ QUOTE ]

[/ QUOTE ]

BruceZ, I've only been on here a few days, but have noticed your combinatorial-prowess, so I don't want to say you are wrong. But, still, I don't see a flop that can be double counted in the possibilities above.

For example, a flop of [4,5,8] is not wanted since it completes a straight. The flop [4, 5, 6], which pairs the board, cannot be a member of {5, 8, not(4 or 9)} and cannot be a member of {8, 9, not(5 or 7)}. Sorry, I am confused.

[BruceZ]

[ QUOTE ]


BruceZ, I've only been on here a few days,

[/ QUOTE ]

Welcome!

[ QUOTE ]
I don't see a flop that can be double counted in the possibilities above.

[/ QUOTE ]

Neither do I. You didn't double count anything, that was a misstatement on my part; however, you are excluding all cases of the board pairing, such as 4,5,5, which should be counted. You are also counting some made flushes. Here we can clearly see the differences between our counts:

Your count: 3*4*4*34

My count: [ 3*(15*34 + 1*27 + 2*6*4) + 4*4*4*2 - 2 ]

Where you have 3*(16*34), I have 3*(15*34 + 1*27), where I have separated out the 2-card combination that is suited and of the same suit as our suited hole cards. Since we don't want to make a flush, there are only 27 cards that complete these flops instead of 34 (that is 34-7, not 34-9 since 2 make a straight and have already been accounted for). Then I add in the 3*(2*6*4) flops that contain a pair (2 ranks that can pair times 6 ways to make the pair times 4 ways to choose the unpaired card), and finally the 4*4*4*2 double gut shots, minus 2 of those which are flushes.

[chapstick]

[ QUOTE ]

however, you are excluding all cases of the board pairing, such as 4,5,5, which should be counted.


[/ QUOTE ]

Shoot! Yes, of course. I agree. My mistake.

[ QUOTE ]

You are also counting some made flushes.


[/ QUOTE ]

Yes, I am. I guess it doesn't make much sense to include them, but they are still straight draws... at least, that's my excuse! /images/graemlins/wink.gif

[ QUOTE ]

Here we can clearly see the differences between our counts:

Your count: 3*4*4*34

My count: [ 3*(15*34 + 1*27 + 2*6*4) + 4*4*4*2 - 2 ]

Where you have 3*(16*34), I have 3*(15*34 + 1*27)


[/ QUOTE ]

Yes, I understand now. Although, I trust your numbers are correct, I will redo my calculation. To the original poster, my answer only deserves part marks at best.