Re: Odds of OESD
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
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