Re: Odds of OESD
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)
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