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During a session, most of the flops you play, you miss. Most of the times that you connect with the flop it's a weak connection, often with "backdoor" outs, i.e. Having to hit multiple cards, over multiple streets, in succession. How can I convert backdoor outs in terms of equity?

For example: My hole cards are 5:diamonds:6:diamonds:. The Flop is T:hearts:4:diamonds:J:spades:.

  • How do I measure the probability that my straight backdoor draw will be realized at the river?

and also,

  • The probability of making three of a kind ("trips") of 5 or 6?
  • What about two pairs?

I guess that I can convert the chance of making straight, trips and two pairs at the turn and river in terms of outs, and the convert it in terms of probability?

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You need to expand on this question. At the moment it is not very clear what you are asking. –  Gaz Winter Mar 4 '13 at 20:00
3  
@emanuele I've made a few edits to your question. If this wasn't your intention for the question to be structured this way, please don't hesitate to re-edit. After all, it's your question. I thought it was worth editing, and believed this is what you are trying to ask. –  Toby Booth Mar 5 '13 at 19:45
    
Thanx, I am sorry for my poor english. –  emanuele Mar 6 '13 at 0:25

2 Answers 2

up vote 5 down vote accepted

To convert backdoor draws into probabilities, you need to multiply the probability of hitting the first card by the probability of hitting the second card given that you hit the first card.

How do I measure the probability that my straight backdoor draw will be realized at the river?

It depends on the number of gaps you have in your draw;

Suppose we modifiy your example so that your straight draw has 2 gaps:

Hole cards: 2:diamonds:3:diamonds:. Flop: T:hearts:6:diamonds:J:spades:.

The only combination of ranks that will give you a straight is 4-5. Then,

  • the first card must be one of the 8 cards of these ranks (4c, 4d, 4h, 4s, 5c, 5d, 5h, 5s).

  • the second card must be one of the 4 cards of the other rank (if, for example, a 5 was dealt on the turn, the 4 cards are 4c, 4d, 4h, 4s).

Probability of hitting the first card: 8/47; Probability of hitting the second card: 4/46;

By multiplying the two (because you need the first AND the second card), you get (8/47 * 4/46) =~ 1.48%

With one gap in your straight draw. For example:

Hole cards: 2:diamonds:3:diamonds:. Flop: T:hearts:5:diamonds:J:spades:.

Now, there are 2 types of combinations that will give you a straight: A-4 and 4-6. Each of them has 1.48% probability, so you have 2.96% chance of hitting.

And with no gap, like in the example you gave, there are 3 types of combinations that will give you a straight: 2-3, 3-7 and 7-8. Each of them has 1.48% so you have 4.44% chance of hitting your straight.

The probability of making three of a kind ("trips") of 5 or 6?

What about two pairs?

The easiest way is to combine these two outcomes.

Not considering the turns and rivers with ranks matching the cards of the board, you will hit a trips or two pair if:

  • the first card is one of the 6 cards matching the rank of your hole cards (5c, 5h, 5s, 6c, 6d, 6s).

  • idem for the second card (but there are only 5 of them left in the deck).

Probability of hitting the first card: 6/47; Probability of hitting the second card: 5/46;

By multiplying the two, you get (6/47 * 5/46) =~ 1.4%

By the way, the same method will give you the probability of hitting a flush:

  • the first card must be one of the 10 diamond cards left in the deck

  • idem for the second card (with 9 diamond cards left):

Probability of hitting the first card: 10/47; Probability of hitting the second card: 9/46;

By multiplying the two, you get (10/47 * 9/46) =~ 4.16%

Randomly dealing a turn and a river on the example hole cards and flop from your example 10 000 times got me these results:

Two Pair: 99

Trips: 56

Straight: 432

Flush: 392

Straight Flush: 21

Two Pair with board cards: 691

Trips with board cards: 81

Other: (No Pair, One Pair): 8228

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Thanx. Very good theoritical answer, but I am looking for a quick and dirty method of converting backdoor draws in outs counting. –  emanuele Mar 6 '13 at 9:18
    
Nicely explained Chris. @emanuele using the "Rule of 4-2" will be valid for this if you modify it to use it more like a "Rule of 2-2" where you calculate the odds of ONLY the next combination of cards that you want. Look here to get some info on estimating the probability of drawing outs. –  Toby Booth Mar 6 '13 at 11:02

I am not sure if this is the answer you are looking for, but I will give it a go.

You would calculate your outs the same as you always do. So if you need 1 card to hit your set you have 2 outs. On the turn you multiply that by 4, so there is an 8% chance of you hitting that card. On the river there is only a 4% chance as you half the previous percentage.

So i guess for your example of hitting the straight. I think you would have 8 outs of hitting the turn. Either the 3's or the 7's. So thats a 32% chance of hitting them both by the river. Or 16% chance of hitting one of them on the turn. You would then have another 8 outs assuming you hit 1 of these. Only this time you would have an 8% chance of hitting.

As I say I am not sure that this is what you want, but I thought I'd give it a go at least.

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