Odds of OESD

[jpg7n16]

I know this has probably been asked before but shoot I'm tired...

what are the odds in hold 'em with two connected (no gapper) cards flopping an OESD?
what about 1-gappers?

then more importantly....

what calculations do you use to find those numbers?

[LetYouDown]

Say you have 6-7 offsuit.

The flops that give you an OESD are:

4-5-X
5-8-X
8-9-X

Seems like (16 * 40) combinations of each type of draw are possible, or 1920 total possible combinations out of 19600. So my guess is about 9.8% or 9.2-to-1.

Feel free to correct, it's early for me =).

[AaronBrown]

I will expand a bit on LetYouDown's excellent post.

Let's say you have 6-7. 8-9-x gets you an open-ended straight draw. There are 4 8's and 4 9's. If you get an 8 and a 9, there are 48 remaining cards that could fill out the flop (everything except your hole cards and the 8 and 9). This includes other 6's, 7's, 8's and 9's. Of those 48 cards, 8 (the 5's and the 10's) give you a straight instead of an OESD. So there are 4*4*40 = 640 ways to get an OESD and 4*4*8 = 128 ways to get a straight.

The same analysis applies with 4 and 5 or 5 and 8 in the flop. So there are 3*640 = 1,920 ways to flop an OESD. There are 19,600 ways to flop (50 unseen cards not counting your hole card, so 50*49*48 = 117,600 ways for the flop to come down, but you divide by 3*2*1 = 6 because you don't care about the order the cards fall in; you can get this in Excel with COMBIN(50,3), the number of ways to draw three cards from a pack of 50).

You might think there are 3*128 ways to get a straight, but there's some double counting in there. You could get 3, 4, 5; 4, 5, 8; 5, 8, 9; or 8, 9, 10. Each of those can come in 4*4*4 = 64 ways. That's 4*64 = 256 ways. Divide that by 19,600 to get the probability of a straight flop.

The odds change if your hole cards include 2, 3, Q, K or A. Then not all the straights are possible, some of them are round the corner straights (like Q, K, A, 2, 3).

[Orpheus]

I actually have a slightly different answer for the probability of a OESD from ungapped connectors, but let me address the 1-gapped connectors first

FIRST APPROXIMATION
you can a) fill the gap [4 candidates] then cap one of the two ends [2x4 candidates], or b) do it the other way around. The DC has some surprising effects on the stats. let's write the possibilities the long way to demonstrate that we aren't double counting. For each case, we have three terms, representing the DC dealt last, middle or first, respectively.

Case A (fill gap first): (8/50)*(4/49)*(48/48) + (4/50)*(49/49)*(8/48) + (50/50)*(4/49)*(8/48) = (1536 + 1568 + 1600)/(50*49*48) = 4704/(50*49*48)
Case B (cap end first): (8/50)*(4/49)*(48/48) + (8/50)*(49/49)*(4/48) + (50/50)*(4/49)*(8/48) = (1536 + 1568 + 1600)/(50*49*48) = 4704/(50*49*48)

Alas, this lovely symmetry must be broken - which only makes sense, since you are obviously more likely to "hit the big target" (the eight cards that will cap one end) before you hit the small target (the four cards that will fill the gap)

SECOND APPROXIMATION
As AaronBrown noted, this can double-count the same flop. As an example, let's assume we're holding 8T. The flop 9sJcJd would be counted twice depending on which J we treat as a "don't care" -- but it's obviously only one single possible flop, and putting it in two different classes won't make it any more likely. Therefore, we should change the numberators, disqualify cards that would fulfill our requirements from being counted as "don't cares" (unless, of course, the OESD is already made, in which case we readlly don't care [Well, we *would* in a real game of poker, but that's another story]

To adjust the allowed don't-cares in this approximation, subtract the remaining "outs" from the total cards remaining in the "don't care" position.

Case A (fill gap first): (4/50)*(8/49)*(48/48) + (4/50)*(41/49)*(8/48) + (38/50)*(4/49)*(8/48) = (1536 + 1312 + 1216)/(50*49*48) = 4064/(50*49*48)
Case B (cap end first): (8/50)*(4/49)*(48/48) + (8/50)*(45/49)*(4/48) + (38/50)*(8/49)*(4/48) = (1536 + 1440 + 1216)/(50*49*48) = 4448/(50*49*48)
Total= (4064 + 4448)/(50*49*48) = 7.24%

I was surprised that the difference between Case A and Case B was so small, but having the "big target" (the caps) still open decreases the allowable don't-cares considerably.

THIRD APPROXIMATION
There is another special case to consider: the "don't care" might complete the straight. I assume you don't want to count completed straights as OESDs. Interestingly, we can know *how many* cards to disallowed without knowing which cards they are: i.e. if the first flopped card is a 6 or Q, it may or may not end up completing a streight with out 8T -- but we do know that *either* the sixes or the Queens will complete our OESD (depending on whether our OESD ends up containing a 7 or a J), so we know there will be four more "danger cards, even if we don't know their identity until later in the flop. (the other four cards that would complete a straight are already deducted: they are the "unused cap" that could have made a OESD)

Case A (fill gap first): (4/50)*(8/49)*(40/48) + (4/50)*(37/49)*(8/48) + (34/50)*(4/49)*(8/48) = (1280 + 1184 + 1088)/(50*49*48) = 3552/(50*49*48)
Case B (cap end first): (8/50)*(4/49)*(40/48) + (8/50)*(41/49)*(4/48) + (34/50)*(8/49)*(4/48) = (1280 + 1312 + 1088)/(50*49*48) = 3680/(50*49*48)
total= (3552 + 3680)/(50*49*48) = 6.15%

It's interesting that you're only slightly more likely to cap an end before you fill the gap, but this is an artifact: as you know from playing, the sooner you fill the gap, the more likely you are to complete the straight (which disqualifies a hand in this calculation)

[Orpheus]

It should be noted that there really isn't a single meaningful value for all connectors. AK and A2 are obviously ruled out, because they can only create single-ended straight draws. 23 and QK are only half as likely to become OESD because flopping an A forces a single-ended straight draw (if you get a straight draw at all). Therefore, almost one-third of all connectors (4/13) are hobbled!

For the remaining "normal" connectors, let me create some terms to reduce my typing:
There are 16 'candidate cards' that are "in range" to create an OESD: 4x -2, -1, +1 and +2.
if two cards are both above or below the CC, they "match", if one is above and the other is below, they "oppose"
The other 34 remaining cards are DCs ("don't cares"). Candidate cards may become DCs during the course of the flop
The original connectors are CC and the three flop cards F1, F2 and F3, in order.
To simplify the equations, I'm only going to show the numerator; the denominator is, of course, (50*49*48)

FIRST APPROXIMATION
if F1 is a DC, there are 16 allowable F2, if F2 is a ±2, then there are only 4 allowable F3. If it is a ±1, there are 8 allowable F3 cards (i.e. if F2 is CC+1, F3 can be CC+2 or CC-1)

If F1 is a ±2, We must either draw a matching 1 and a DC (the opposing 2 is a DC in this case); or an opposing 1 AND 2 in either order (creating an OESD "on the other side" of the CC, but not completing a straight), The DC can be F2 or F3.

IF F1 is a ±1, we must draw a DC and a matching 2 (the opposing 2 becomes a DC) OR an opposing 1. The DC can be F2 or F3.


Case 1 (F1=DC): 34 * [8*4 + 8*8]
Case 2 (F1=±2): 8 * [4*38 + 38*4 + 8*4]
Case 3 (F1=±1): 8 * [4*38 + 38*4 + 4*34 +34*4]

Case 1 + Case 2 + Case 3 = 3296 + 2688 + 4608 = 10592 OESD flops

10592 OESD flops / (50*49*48) possible flops = 9% (okay, 9.006800272...)

I wouldn't take this "normal connector -> OESD" probability at its face. It's too late for me to be doing math reliably

[Cobra]

LetYouDown

Your answer is slightly off. By doing it the way you did you are double counting certain flops. These things are tricky.

For example if the flop comes 4-5-X you should originally say x cannot be a 8,3,4,5. This leaves 34 cases for X, you then go back in and add the times that x are a second 4 or 5. This eliminates the double count of the fours and fives. So the answer should be:

= 4*4*34 = 544 + 2*(4c2)*4 = 544 + 48 = 592 flops. The second term is either flopping two fours and a five or two fives and a four.

So for the three subcases that give you an OESD you have a total of 592*3 = 1776 flops / 19600 = 9.1% or 1 in 11.04.

To the original poster remember there are an additional 128 flops that give you a double gut buster(still an eight out draw). This brings the total flops that give you eight outs from a max stretch connector to 1176+128=1904 flops or 9.7% or 1 in 10.29.

Cobra

[LetYouDown]

Touche. I've been known to overlook the obvious...thanks for pointing it out.