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.

[BruceZ]

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

This isn't right, you could actually call a bet on the flop getting less than 46-1. This is because you will usually catch a non-flush card on the turn and can fold, so it will still only cost 1 bet in most cases. The average cost of going for the backdoor flush is 1 + 2(10/47) = 1.4 small bets since it costs 2 small bets to call on the turn, but you will only call on the turn 10/47 of the time when you catch a flush card. Since out of 24 attempts you will make the flush 1 time, and since it costs an average of 1.4 bets per try, you should have pot odds of 23-1.4 = 32-1 to call on the flop. So for the purpose of deciding whether to call a bet on the flop, the backdoor draw would actually be worth 47/33 = 1.4 outs. Since people usually memorize pot odds for integer numbers of outs, you could round this down to 1, but it could also make sense to round it up to 2. The reason is, you will win more than what is in the pot on the flop when you make the flush. That is, you have implied odds. In some situations you may have to be concerned that you won't have the best hand, such as when there is a pair out. You may also have to be concerned with getting raised either on the flop or the turn. Judgement is always required in these cases, but perhaps it is best to count it as 1 when there is danger. When you can go all in, it is worth exactly 1 out assuming it will make the best hand.

[BruceZ]

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.

This calculation is valid for the case of the backdoor flush plus middle pair too, and it is no coincidence that this came out the same as the exact calculation in the above post. The reason is that when you hold two suited cards, it is impossible to make your backdoor draw AND hit one of your other 5 outs since the two cards of that suit are already on the flop. Since these events are mutually exclusive, the probability of getting one of them is the sum of the probabilities for each one, and this is the same as summing the outs (5+1=6). So it IS the same as if you had 6 outs on both the flop and the turn. The only reason it's even off in the 2nd decimal place is because the backdoor draw isn't precisely 1 out.

[BruceZ]

Since out of 24 attempts you will make the flush 1 time, and since it costs an average of 1.4 bets per try, you should have pot odds of 23-1.4 = 32-1 to call on the flop.

32-1 is right, but I didn't mean to say 23-1.4. We're saying it costs 1.4 and the odds of making it are 23-1, so x/1.4 = 23/1, or x = 32, so we need 32-1. Really if we are taking into account the extra bet we have to pay on the turn, we ought to take into account the extra bets we win on the turn too. We don't know how much that will be in general, but if we are putting in an extra 0.4 bets on the turn on average, then we must win at least this extra amount when we win. So the above would become (x+.4)/1.4 = 23/1, x = 31.8, so we would still need 32-1, and this is still 1.4 outs, it doesn't make much difference unless we win even more.

Taking into account extra bets we win or lose on the next round is normally outside the scope of simple pot odds, and comes under implied odds and effective odds. However, since the backdoor draw requires two cards to make, it is hard ignore future bets. The pot odds we need and hence the outs which the backdoor draw is worth ultimately depend on the future bets we win and future bets we must put in. A first order approximation is to completely ignore future bets on the turn and say we need to be getting 46-1 pot odds to call on the flop, so it is worth 1 out. A better approximation would be to say it costs 1.4 bets on average, and it is 23-1 to make the hand, so we need 32-1 to call on the flop, and it is worth 1.4 outs.