Question for the whizes.

[VivaLaViking]

When you count your outs you count single card outs. The relevent percentages can easily be calculated mentally within a fraction of a percent. But commonly other outs may also exist, multiple card outs, when two card values are needed as for a straight. Not that you would bet on runner-runner but it must increase your overall probability. Could it be calculated as a fraction of an out (eg 6 1/2 outs)? How could this be incorporated in calculating the probability of improving yor hand?

[Tom1975]

IIRC, SSHE says a backdoor flush and a backdoor straight with no gaps are worth about 1.5 outs on the flop. I think a 1-gap BD straight is worth 1 out, and a 2 gapper is .5 outs.

[VivaLaViking]

I don't know what IIRC, SSHE is. A one-gap straight is 4 outs assuming the filling card is live (generally unless dealer mishandles). An open ender is 8 outs. So the qustion resolves to the outs for a straight needing two cards, either a 2-gap or 1-gap with needing either end and completing a 3-flush. Thank you for responding so I must keep podering.

[Tom1975]

For a real example, say you have Q /images/graemlins/heart.gifJ /images/graemlins/heart.gif, and the flop comes 10 /images/graemlins/spade.gif6 /images/graemlins/club.gif2 /images/graemlins/heart.gif. When calculating your outs, you would give yourself 6 outs for the overcards, 1.5 outs for the backdoor flush and another 1.5 out for the backdoor straight, for a total of nine outs. Of course pairing one of your overs won't always win, so you should partially discount those outs.

[Tom1975]

IIRC= If I recall correctly
SSHE= Small Stakes Hold 'em (by Ed Miller, et al)

[VivaLaViking]

I think I'll do some math on your example and see how well it correlates as soon as I have time. I'll post my results if you are interested.

[VivaLaViking]

I want to see if I can calculate these fractional outs and correlate the percentages with known probabilities of outs and/or see if they are significant enough to even consider.

First I will start when a 3-flush flops. You will need runner-runner suited. Is this the correct math?

(10 C 1) * (9 C 1) / (52 C 5)

[Tom1975]

No, it's simply (10/47)*(9/46). On the turn there are 47 unknown cards, 10 of which help your flush draw. If you hit one, there are 9 out of 46 cards on the river which complete your flush.

[VivaLaViking]

Tom, if that math is correct your chances of getting a back door flush is ~ .4% This sounds about right but I don't know. Improving a hand with 2 outs by the river from the flop is ~8.4%. so that infers a back door flush is only about .2 outs. Does that sound correct to you?

[Tom1975]

You're off by a decimal place, it's 4%, not .4%

[VivaLaViking]

Yes Tom, your are right. This result indicates that as 2 outs can be calculated:

3-flush probability: (10/47)*(9/46) ~ .042
2 out probability: 1 - (45/47)(44/46) ~.084

.042
---- * 2<--Outs ~ 1 Out
.084

So a 3 flush on the flop could be counted a single out. Does the math look right to you and would you agree?

I just read this section of SSHE last night, conincidentally, and there's no better resource on the subject of backdoor draws (draws that require help from the turn and river).

Your backdoor flush calculations are accurate. While the value of a 4.2% is right about one out, Sklansky adds that you will be able to fold on the turn if you don't hit, and bring in quite a number of bets if you make your flush, as it would be unexpected. For that reason, Sklansky values the backdoor flush as 1.5 outs.

Now, I didn't pay very close attention to Sklansky's math for the backdoor straights, so hopefully I'll be corrected if I'm wrong, but here's how I reason it. For no-gappers, there are two ways to complete your backdoor straight: either you hit one of eight outs on the turn to make a four-straight, and then one of eight outs for the straight on the river; or you hit one of eight outs on the turn to leave you the inside draw, which you then fill in on the river.

For example, the first case is if you have 6-7, and the board has an 8. You could get a 5 or a 9, and then a 4 or a 10. Eight outs both times. In the second case, with the 6-7-8, you could hit a 4 or a 10 on the turn, then requiring only the 5 or 9 on the river, for eight outs, then four outs.

So here's the math:

Case 1: (8/47) * (8/46) = 2.96%
Case 2: (8/47) * (4/46) = 1.48%

The two possibilities together equal 4.44%, also just a little more than one out. Sklansky values the no-gap straight draw at 1.5 outs, likely juicing the value a little for the same reasons as with the backdoor flush.

Bothering with the math for the one- and two-gappers seems a little pedantic (though it can't be much different than Case 2 above), so I'll just report that Sklansky values the one-gap straight draw at 1 out, and the two-gap straight draw at .5 outs.

So there it is. As I said though, you could do no better than SSHE for the subjects of outs, odds, and drawing.

What the heck:

One-gap straight:

You need to hit one of eight outs on the outside, and one of your four inside outs. So that's your 1.48% from the previous example, though you can hit the inside first or the outside first, so you double up the 1.48% for 2.96%. Juice it for value, and Sklansky's value of one out makes sense.

Two-gap straight:

Fewer options here: you must hit one of your eight inside outs, and then one of your four remaining inside outs. So that's just 1.48%; no two ways to do that one. Bump it up a little, and there's your half an out.

[Tom1975]

Yes, the math looks good. I think Sklansky actually says a backdoor flush draw is worth more like 1.5 outs in practicality, because if the turn does not improve you to a four flush or otherwise help you, you can fold and save a big bet.

[VivaLaViking]

Thanks, I wont disagree that there may be an extra value assigned to this. especially at a short table or with a few callers but my thoughts are that you may find youself in a reverse implied odds situation.

Also, I prefer to do the math myself for verification purposes and so I'll see the raw numbers. On to other problems, the straight. I prefer to tackle straights as a function of solutions not merely gaps as A, 3, 4, 5 and 2, 3, 5, 6 differ in the probabilities. In the first example your solution is only 2, 6; but in the second the solution is 4, 7 or 4, A. They are both one gap straights. I will start working on this somtime today.

[VivaLaViking]

Sorry for the bad examle but you know what I'm trying to say. When one card is needed to fill the inside and then either 1 or 2 outside possibilties exist to complete the straight.

Well, I think I've covered just about all the math you'll need in my posts, but of course I'd be curious to see what you'd come up with.

The two-gap probabilites are correct for any straight, and the one gap for everything except the wheel and Ace-high. A one-gap wheel or Ace-high backdoor straight draw (e.g., A-2-4) has the same probability as a regular two-gapper, as I see it: you've got eight outs, then four, and you must hit both.

The no-gaps on the high/low end also have the same probabilities as regular one-gap or two-gap backdoor straight draws. If you've got A-2-3, for instance, and are looking for the wheel, that's just like a two-gapper. 2-3-4 looking for the wheel or 6-high is a one-gapper. I think that about covers it.

[VivaLaViking]

The genesis of my thought process began with the probability of flopping a straight. I used the number of solutions as the base quantity for my calculation because of the ease of seeing solutions. Technically a solution is a unique combination that completes a straight, as opposed to permeations. To make sure that we are on the same wavelength; given down cards (3 5) there are only 3 solutions (A 2 4), (2 4 6) and (4 6 7). Straight flushes are not discounted so:
.
3 [Solutions] x (4 C 1)[Suits/Card] ^3 [Cards/Solution]
-------------------------------------------------------
Total Flops
.
3 * 64
= --- = 3* .0033
(50 C 3)

3 *.0033 ~ .01
.
This implies the number of flopped straight, which includes straight flushes, may be deduced based solely on the number of solutions, .0033 per solution.

Next I consider flopped near straights; as my ultimate goal is to deduce the number of equivalent outs; and the outs of a flopped near straight requiring only a single card may simply be counted, no other analysis is needed. Only flopped near straights requiring two cards will be considered, the 3-straight. I would like to continue my calculations though solutions and a two gap can only have 1 solution, a one gap may have a maximum of 2, and a 0 gap a max of 3 solutions. The probability reduces to counting the solutions and multiplying by the probability of choosing two distinct card values.

(8 C 1)(4 C 1)
--------------- ~.014 per solution
(47 C 1)(46 C 1)

Again, I will base the number of outs on the known value of 2 outs. 2 out probability: 1 - (45/47)(44/46) ~.084

.014 [per solution] .014 Solutions
------------------- * 2 = ---- = --------- [Outs]
.084 .042 3

This implies 1/3 out per solution for straights requiring 2 card.

Would someone please check this math and comment so I can move to the next item.

[VivaLaViking]

3-flush probability: (10/47)*(9/46) ~ .042
2 out probability: 1 - (45/47)(44/46) ~.084

.042
---- * 2<--Outs ~ 1 Out
.084

So a 3 flush after the flop could be counted a single out. Some disagreement exists on the actual weight that could be assigned but my thoughts are that it should only be increased, or decreased, depending on the other circumstances of the game and players.

[VivaLaViking]

A solution is defined as a unique combination that completes a straight, as opposed to permeations. To make sure that we are on the same wavelength; given down cards (3 5) there are only 3 solutions (A 2 4), (2 4 6) and (4 6 7). Straight flushes are not discounted.


Single solution probability:

(8 C 1)(4 C 1)
--------------- ~.0148
(47 C 1)(46 C 1)

2 out probability: 1 - (45/47)(44/46) ~.0842

.0148
----- * 2 ~ 1/3
.0842



So straight needing 2 cards after the flop could be counted as 1/3 out per solution. Some disagreement exists on the actual weight that could be assigned but my thoughts are that it should only be increased, or decreased, depending on the other circumstances of the game and players