Probability: Three-flush to Flush

[Warik]

I know that if I have 1 card to make my flush by the turn, then the probability of hitting it on the river would be 37/9 ~ 4.11:1 (right?)

But if I flop 4 cards to a flush and need to make my flush by the river... how would that calculation work? Like this?

P(getting it on the turn) + P(getting it on the river assuming I didn't get it on the turn) + P(next two cards being of my suit)

P(turn) = 1 / (9/38 + 9/37 + 9/37 + 8/38) = 1.07083:1 (right? that seems ridiculously good to me)

If I do 1 - P(not getting any more of my suit) instead, I come up with 1.537:1 which seems right, but then I wonder how the heck I got 1.07083 before.

I'm curious about the previous calculation, but my main question is this: what if I flop only THREE cards to the flush and I need to hit the next two cards have to help my flush, how would I do that calculation?

help? /images/graemlins/smile.gif

[Robk]

The probability of making your flush with two cards to come is 1 - P(none of your suit come) = 1 - (38/47)*(37/46) = 1-.65 = .35, or about 1.86:1.

The probability of making a runner-runner flush is just the product of the probabilities of one of your suit coming on each of the next two cards = (10/47)*(9/46) = .0416.

P.S. You seem to have a fundamental misunderstanding about odds and probability. You wrote that the odds against hitting a flush card on the river are 37:9, which is correct. But the probability you make your hand is 9/(37 + 9), NOT 9/37.

[Bozeman]

Let me try to sort through the errors made here:

"P(getting it on the turn) + P(getting it on the river assuming I didn't get it on the turn) + P(next two cards being of my suit)

P(turn) = 1 / (9/38 + 9/37 + 9/37 + 8/38) = 1.07083:1 (right? that seems ridiculously good to me)"

First, as has been pointed out, odds are different than probabilities. Probs. can be added, but odds can not.

The number you have used for probs are the odds, they should be 9/47 instead of 9/38, etc..

While you can express it as a sum of probs., you need to make sure that all events are mutually exclusive to do this, so P(flush)=P(turn)+P(river given not turn) (P(turn) includes P(turn and river). P(flush)=9/47+9/46*38/47=35.0%

Or

P(flush)=P(turn not river)+P(river not turn)+P(river and turn)=9/47*38/46+38/47*9/46+9/47*8/46=35.0%

(or of course the easy way: P(flush)=1-P(not turn, not river))

In your equation, you found the prob. of turn and river hitting by adding, instead of multiplying.

(The probability of one of two exclusive events happening is the sum of their individual probabilities, while the probability of two independent events both happening is the product of their individual probabilities.)

Craig

[Warik]

Ok I think I understand where I screwed up, and I understand how you calculated the probability of making the runner-runner flush. If I wanted to express that as odds, how would I do so? I can't just take the reciprocal of 0.0416 to get 24:1 right?

You mentioned 9/(37 + 9) probability is equivalent to 37:9 odds... so the reciprocal wouldn't work. How do I get from one to the other? (Or better yet, how do I calculate the runner runner in terms of odds as opposed to probability?)

[Robk]

Try using this rule. To convert a:b to a probability, just take b/(a+b). EG 9:1 against is 1/(1 + 9) = 1/10 = .1. To go the other way, .0416 = 1/(a + 1) <=> .0416 a + .0416 = 1 <=> a = (1-.0416)/.0416 = 23. So a runner-runner flush is 23:1 against. (Note that 1/(23 + 1) = .0416)

[Robk]

On a poker note, you can't call a bet on the flop getting 23:1 if you have only a runner runner flush draw. That's (mostly) because you'll also have to pay a double size bet on 4th street to draw at your hand on the river. Sklansky went through this calculation in depth in one of his lesser read books (maybe PGL??) and determined that you need something like 28:1, I don't really remember. But backdoor draws can add value to a hand that has some to begin with, sometimes swinging a fold to a call. See TOP for a great discussion of this. A classic example in HE is overcards on the flop that also have a 3 flush and 3 straight. (Note that drawing at a flush/straight on 4th street will also allow you to often catch an winning pair you wouldn't have otherwise.)

[Bozeman]

Take the reciprocal (of probability) and subtract 1 from it (to get odds to 1)

[Warik]

[ QUOTE ]

Try using this rule. To convert a:b to a probability, just take b/(a+b). EG 9:1 against is 1/(1 + 9) = 1/10 = .1. To go the other way, .0416 = 1/(a + 1) <=> .0416 a + .0416 = 1 <=> a = (1-.0416)/.0416 = 23. So a runner-runner flush is 23:1 against. (Note that 1/(23 + 1) = .0416)

On a poker note, you can't call a bet on the flop getting 23:1 if you have only a runner runner flush draw. That's (mostly) because you'll also have to pay a double size bet on 4th street to draw at your hand on the river. Sklansky went through this calculation in depth in one of his lesser read books (maybe PGL??) and determined that you need something like 28:1, I don't really remember. But backdoor draws can add value to a hand that has some to begin with, sometimes swinging a fold to a call. See TOP for a great discussion of this. A classic example in HE is overcards on the flop that also have a 3 flush and 3 straight. (Note that drawing at a flush/straight on 4th street will also allow you to often catch an winning pair you wouldn't have otherwise.)

[/ QUOTE ]

Thanks. I understand the conversions perfectly.

Makes sense about the runner-runner draws. To call a 50 cent bet on the flop I'd need $11.50 (23 x $0.50) or actually $14 (28 x $0.50 like you mentioned) in the pot. The highest I've ever seen post flop is about $5 in the loosest of games. We'd need quite a few maniacs in there to get the pot that high, and I don't think I'd like to need 2 cards to make my hand vs. multiple maniacs. /images/graemlins/smile.gif

Now, does this conversion and odds determination hold true after the river card is seen and the board is analyzed?

Let's say, for example, I have KJo and flop is KJ6 rainbow. We go through the standard bet/raise/call procedure and see that the turn is an ace. I get to the river and see another ace.

So, I have two pair, aces and kings, but any ace beats me, so do two jacks and two kings... but pokercalc.com says I have an 80% chance of winning.

So...

.80 = 1/(a+1)
.80a + .8 = 1

.80 a = .2

a = .25

That's .25:1... so as long as there's a quarter in the pot it's OK to call a raise? That doesn't make sense given the board.


or worse... if I have Q3 clubs and the board has 4 hearts and a Q, pokercalc says I'm 29.15% to win.

.2915 = 1 / (a + 1)

.2915a + .2915 = 1
a = .7085 / .2915
a = 2.43

There will always be at least $2.43 in the pot on the river... but obviously it's a mistake to call here.

I take it these calculations are only good when determining if I should wait out my draw but are of no significant meaning when it comes down to the end?

[webiggy]

when determining whether or not to play this hand through. Clearly a -EV unless you're getting tremendous odds as to make this hand pay, you will likely need to make at least one BB if your first card DOESN'T hit and at least 2 BB's if it does. Since the odds of hitting your miracle hand is alomost 25:1, the pot should be offering close to 30:1 on the flop in order to play this hand if your hand has no other value. On the otherhand, if you have top pair or if you have 6 other outs from holding two suited overcards, you can go with a slightly lesser pot.