HE: Backdoor flush = + 2 outs?

[RockLobster]

Hi Guys--

I seem to remember reading that, when counting your outs in hold'em, a backdoor flush counts as 2 outs. Does this sound right to anyone? If I remember right, a backdoor flush is about 23-1 against, which would make sense.

Any thoughts? Thanks in advance...

[BruceZ]

It is 23-1 against. This is (10/47)(9/46) = 4.16%. This is approximately the same as having 1 out on the flop since it takes 2 cards to make your hand. You count it as 1 out for the purpose of determining if the pot is big enough to see the turn. Only if you were all-in on the flop, or if you knew there would be no turn bet, would it be the same as 2 outs.

[RockLobster]

[ACPlayer]

Why is it equal to 2 outs if you are all in on the flop? The chance of making the backdoor draw on the flop is the same whether u have more money or not.

[BruceZ]

Actually I see no use for thinking of it as 2 outs even when you are all-in, but it has that effect mathematically. The probability of making your hand is the same when you go all-in on the flop, but you don't need the pot odds to be as high to go all-in since you now get to see both cards for the same price. The pot odds you need are still determined by the odds of hitting 1 out with 2 cards to come (23-1), and these are approximately the same odds you would need to see the turn if you had 2 outs and you were not all-in. I was just trying to make sense of the statement about it being worth 2 outs.

For example, if you had middle pair along with your backdoor flush draw, and you thought 2-pair would be good, then if you could not go all-in, you would count this as 6 outs for the purpose of calling a bet on the flop (5+1), and you would need pot odds of 41-6 or 6.8-1 to call. But if you could go all-in on the flop, then this is 6 outs with 2 cards to come, and you would only need pot odds of 1-(41/47)(40/46) = 4.1-1. You should memorize that 6 outs with 2 cards to come was about 4-1, you should not say that you have 6 outs twice for 12 outs which requires pot odds of 35-12 or 2.9-1. This is inaccurate, and it would be even more inaccurate for a larger number of outs, though it would be fairly accurate if all you had were the backdoor draw.

[ACPlayer]

I understand the point about how to account for odds when there are two cards to come. However, my point is that the backdoor draw is one out whether you are allin or not, on the flop

In your note you state:
"But if you could go all-in on the flop, then this is 6 outs with 2 cards to come, and you would only need pot odds of 1-(41/47)(40/46) = 4.1-1. "

This calculation is actually only true if the 6 outs were six outs on the turn and on the flop. The backdoor out is a pseudo out as it requires the first card to be one of the backdoor out cards. THat is if you are not all-in the number of outs on the turn could reduce to five if the allin outs dont hit.

So, because on the flop we are talking about the backdoor out being about 4% and knowing that one out twice is about 4% we are converting the backdoor chance into approx one out on the flop (this by itself assumes that 2 cards are to come). It does not matter whether you are allin or not. On the turn we either have 5 outs or 14 outs. Note that for accurate calcualation we would need to confirm that the backdoor outs are not already counted in the other outs (for example the 2 pair outs are not counted) but that is a secondary effect.

[BruceZ]

However, my point is that the backdoor draw is one out whether you are allin or not, on the flop

I agree it is always worth 1 out, but since you can see 2 cards for the same price when you can go all-in, you only need 23-1 pot odds to call all-in since these are the odds of making your hand in 2 cards, and these are the same pot odds you would need to hit a 2-outer on the turn. When we can't go all-in, we determine based on 2 cards to come that we have 1 out, but we use that pseudo-out as if we can make our hand in one card, that is, we would not call to see the turn unless the pot odds were 46-1. Actually you would need more than this if all you had was the backdoor draw since you would be getting less than 46-1 on the turn, but this problem is diluted in real cases where you have more outs. You don't need as high a price to call a bet all-in.

In your note you state:
"But if you could go all-in on the flop, then this is 6 outs with 2 cards to come, and you would only need pot odds of 1-(41/47)(40/46) = 4.1-1. "
This calculation is actually only true if the 6 outs were six outs on the turn and on the flop. The backdoor out is a pseudo out as it requires the first card to be one of the backdoor out cards. THat is if you are not all-in the number of outs on the turn could reduce to five if the allin outs dont hit.


The above actually comes out to 3.1-1, not 4.1-1, and it comes out virtually the same if we do an exact calculation taking into account the effect you mention:

1 - [ (32/47)(41/46) + (10/47)(32/46) ] = 3.1-1.

Here I am separating out the case where you catch a flush card on the turn from the case where you don't. So you can treat it the same as if you had 6 outs with 2 cards to come.

Since this comes out to 3.1-1 instead of the 4.1-1 that I mistyped earlier, it is actually very close to what you would get if you just doubled your outs to 12 and said the pot odds should be 35-12 or 2.9-1. So I can't use this example to teach that lesson, but this becomes less accurate with a larger number of outs.