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View Full Version : Probability of at 7-8-10 (2 suited) or 7-8-9 (suited) type of flop?

Dan Mezick
10-12-2004, 03:18 PM
If you are a Prob-Wonk, can you help me answer this? How many flops per 100 will have one type (1-gap 3-straight, 2-suited) or the other type (3-running straight, 2-suited) textures?

In all cases the flop is expected to have 2 or more suited cards.

Thanks in advance for any help along these lines.

For the 1-gapper straight type, the gap can be in both places, bottom (5-x-6-7) or the top (5-6-x-7)

Also, if you can display the approach to solve, I can reduce my frequency of questions of this type in the future.

I suspect it's 15 to 21 per 100; that's a guess.

Is it actually higher? Rememeber every flop will have at least 2 suited cards.

Lost Wages
10-12-2004, 05:21 PM
I have no idea what a Prob-Wonk is but I'll try to answer your question.

There are 12 3-straights from A23 to QKA, each can be formed 4*4*4 = 64 ways. If we subtract the rainbow ways then we will know the 2 suited and single suited ways. For rainbow, the first card can be one of four suits, the second one of three suits and the third one of two suits, so 4*3*2 = 24 of the 64 ways are rainbow leaving 40 that are not. Therefore, there are 12*40 = 480 non-rainbow 3-straight flops.

There are 11 X-XX flops from A34 to JKA and 11 XX-X flops from A24 to JQA. The number of non-rainbow ways will be the same for each as it was for the 3-straights, so 22*40 = 880 non-rainbow one-gap flops.

There are a total of 52*51*50/(3*2) = 22,100 flops, so your conditions will be met (480+880)/22,100 = 6.15% of the time (~6 times per 100 flops).

Lost Wages

Dan Mezick
10-12-2004, 11:46 PM
Thanks for this outstanding post which answers my question completely !!

ddubois
10-18-2004, 08:23 PM
I am hijacking this thread because it is very similar to a question I wanted to ask.

Problem: Identify how much more likely it is that a 0-gap hand will flop a OESD relative to the likelihood of a 1-gap hand flopping a OESD.

I will now attempt to answer my own question by LostWages' method of enumerating flops.

If I have 89...
x67 6x7 67x x76 7x6 76x
xT7 Tx7 T7x x7T 7xT 7Tx
xTJ TxJ TJx xJT JxT JTx
All of these flops have an equal number of ways of being generated. There are (48x4x4) ways for x67 to be dealt (in that order), and the same for the other combinations. With 18 combinations, that is 48x4x4x18 = 13824 flops, of 48x47x46 = 117600 possible flops (again, if you consider the cards to be ordered). Thus a 11.76% chance of flopping an OESD.

If I have 8T...
x79 7x9 79x x97 9x7 97x
x9J 9xJ 9Jx xJ9 Jx9 J9x
All of these flops have an equal number of ways of being generated. There are (48x4x4) ways for x79 to be dealt (in that order), and the same for the other combinations. With 18 combinations, that is 48x4x4x12 = 9216 flops, of 48x47x46 = 117600 possible flops (again, if you consider the cards to be ordered). Thus a 7.84% chance of flopping an OESD.

I assume these numbers will also be true for every holding between 45 thru TJ for connectors, and 35 thru TQ for 1-gappers, since they all have the same "range of motion" to make their straights.

Correct so far? It's OK to assume ordering right, as long as you do it for both subsets of flops and total flops, yes?

Oh, one caveat, these numbers are not just OESDs only, they include the flopped straights as well. Without doing the homework, I think 1/6ths of these flops will be an actual straight (48x4x4 - 40x4x4), not an OESD, and similarly, the connectors flop a straight precisely 50% more often than the 1-gappers, 1.96% versus 1.31%.

Lost Wages
10-19-2004, 11:07 AM
The connectors 45 thru JT are commonly referred to as "max stretch" and, yes, the solution is the same for each.

For simplicity, let's assume that you have an unsuited max stretch connector (e.g. 89o). Otherwise you would have to subtract those times that you flopped a flush.

Given 2 starting cards, there are (50*49*48)/(3*2) = 19,600 possible flops.

There are 4 ways to flop a straight (567, 67T, 7TJ &amp; TJQ). Each can be formed 4*4*4 = 64 ways for a total of 256 straight making flops.

The probability of flopping a straight is 256/19,600 = 1.31%

There are 3 ways to flop an open end straight draw as you noted (67X, 7TX and TJX). 67 can make 16 combinations which can combine with 52-(2 in your hand)-(4 tens)-(4 fives)-(4 sixes)-(4 sevens) = 34 other cards for a total of 16*34 = 544 ways. In addition, 667 and 677 can each make ((4*3)/2)*4 = 24 ways. So 67X, where "X" does not make a straight, has a total of 544+24+24 = 592 ways. The same is true for 7TX and TJX so there are a total of 3*592 = 1776 open end straight draw flops.

The probability of flopping an open end straight draw is 1776/19,600 = 9.06%

Everyone who considers this problem for the first time forgets about the 8 out double gutshot straight draws. All DGSD's are of the form X-XXX-X. Max stretch connectors can flop two different DGSD's. For 89 they are 6TQ and 57J. Each can be formed 4*4*4 = 64 ways for a total of 128 ways.

The probability of flopping double gutshot straight draw is 128/19,600 = 0.65%

<font color="red">The probability of flopping a straight or 8 out straight draw with a max stretch offsuit connector is 1.31% + 9.06% + 0.65% = 11.02% </font>

Armed with this knowledge we can make quick work of the max stretch 1-gappers. We add the condition that all outs on the draws must include both hole cards or things can get icky (e.g. 68 with a flop of 9TJ). Note that given this the max stretch 1-gappers are 35 thru TQ.

They can flop 3 straights 64 ways each for a total of 192 ways.

The probability of flopping a straight is 192/19,600 = 0.98%

They can flop 2 OESD's 592 ways each for a total of 1184 ways.

The probability of flopping an open end straight draw is 1184/19,600 = 6.04%

They can make 1 DGSD 64 ways.

The probability of flopping double gutshot straight draw is 64/19,600 = 0.33%

<font color="red">The probability of flopping a straight or 8 out straight draw with a max stretch offsuit 1-gapper is 0.98% + 6.04% + 0.33% = 7.35% </font>

I'm tired now.

Lost Wages