[binions]

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Backdoor flush draws are 23:1 to hit, which is equal to 1 out with 2 cards to come.

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Can you show your derivation?

I used:
10/47 you hit a flush draw on the Turn.
if you hit the flush draw, then you will have 9 outs.

10/47 x 9 outs = 1.91 expected outs

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Do you agree a backdoor flush draw is 23:1 to hit? 10/47 * 9/46 = 4.16% which is 23:1 against.

If so, look at any standard odds chart. 1 out with one card to come is 45:1 assuming 46 unseen cards. 1 out with 2 cards to come is 22.5:1 with 47 unseen cards (or 21.5:1 if there are 45 unseen cards).

Bingo. See Ed Miller's SSHE at p. 102 for confirmation on the math.

Anyway, I have seen Ed Miller (and I think Abdul) say a backdoor flush draw is worth 1.5 outs. But this has to do with:

1. The fact you can fold on the turn when you don't pick up a draw;

2. The fact that you get paid off relatively more often than a flopped 4 flush when you hit.

So the math "out" is inflated by 50% to 1.5 outs given these implied odds/game considerations.

But the math is a backdoor flush draw is 23:1 against which is equivalent to 1 out.

To say you have 1.91 outs on the flop means you have an 8% chance to hit your hand by the river. That's simply false with respect to backdoor flush draws. You don't have an 8% chance. You have a 4.16% chance.