Calculating Odds Question

[ZManODS]

How would you calculate pulling a backdoor flush? I would think it would be...

Probablity = 1 - p[38/47 * 37/46)

But that ends up over 30% and i know thats not right. Can someone explain how it is done. Thank you.

[morgan]

What you calculated was the odds of completing a flopped 4-flush by the river. What you want is

10/47 * 9/46 = 4.16%

(say you have 3 clubs after the flop, then you must hit one of the 10 remaining clubs out of the 47 unseen cards, and then one of the 9 out of the last 46 cards)

Take care,

Morgan

[CrackerZack]

You have to take the chance of missing on the first card which is 37/47 + making the first 10/47 * missing the 2nd 37/46 and do 1 - [37/47 + (10/47 * 37/46)]... which = about 4.2% which is about 22.8 - 1.

In other words, not good.

[Dynasty]

The odds of you making a backdoor flush are 23:1.
Because the turn bet is larger, you need 27:1 to profitably call a flop bet with nothing but a backdoor flush draw.

[Flashy]

Are the odds really that bad?
With 4 different suits there are 16 combinations of suits.
Four of these combinations are of the same suit.
One is the combination you want.
So aren't the odds 1/16?
Now I know that you are suppose to account for more or your suit already in play. That's what drives the figure to 1/23.
But that also assumes there hasn't been an even distribution of suits to all the hands dealt. For all you know the deck may be richer in your particular suit.
So I believe the odds are somewhere between the 1/16 to 1/23. That's my belief although I am probably missing something.
Of course, the odds seem even better when someone else is going runner runner to crack your set.
By the way, I greatly enjoy your posts and hunt for them. I am sure alot of players do.
Let me know if I am totally off base.

[Ed Miller]

Now I know that you are suppose to account for more or your suit already in play. That's what drives the figure to 1/23.

That's correct.

But that also assumes there hasn't been an even distribution of suits to all the hands dealt. For all you know the deck may be richer in your particular suit. So I believe the odds are somewhere between the 1/16 to 1/23.

For all you know, there may be none of your suit in the deck. Therefore, I believe the odds are infinite. You can't make any assumptions about the distribution of cards that you haven't seen (unless you've done some hand-reading). Think about your logic... your point is that some cards are in other people's hands... so they can't possibly come on the river. My point is... most of the cards (all but one) are on the bottom of the deck... and they can't possibly come on the river either. Maybe all the cards in your suit are all in order on the bottom of the deck. There are three burn cards... maybe they are all your suit. The bottom line is that the river card will be one card chosen from the 46 cards you haven't seen... and you have no reason at all to think that any one card is any more likely than any other card.

[Dynasty]

There is a good essay in David Sklansky's Poker, Gaming, and Life about backdoor flush draws. If you ask Mason, he may put it up in the Essays section of this website.

[tewall]

There are 5 cards you know about, the 3 on the board and 2 in your hand. Of those cards, 3 are the same suit, the suit of the backdoor flush. That means there are 10 cards that can help you for the turn, and assuming that comes to pass, 9 on the river. You need both of those things to happen, so you multiply the probabilities together. 10/47 * 9/46 = a little over 4%.

There's an easy way to estimate these probabilities that is simple enough to do at the table. You take your number of outs and multiply by 2. So the estimate would be 10 * 2 = 20% times 9 * 2 = 18%, or 3.6% total. This is off by about .5%. Usually the estimate is closer, but here it's off by this much because you're multiplying two estimates together. If you only are dealing with one estimate, it's usually quite close.

For example, say you have 6 outs. The odds of catching on the turn by estimate would be 12%. The actual answer is 12.7%. Then if you want to see what odds that corresponds to take the estimate and divide it into a hundred and subtract 1. For example, 12 goes into 100 a little over 8 times, so 6 outs would correspond to a little over 7:1.

[Flashy]

I appreciate your post. My only point was that if you knew all suits had been distributed evenly, the odds of runner-runner flush are closer 1/16. As you point out, since you don't know how the suits were distributed, but you know the suits of five cards, then that's all you can go on. So you must calculate the odds at 1/23. I get the point.
I just wonder if randomn simulations would show runner runner flush occurring more than 1/23 times with nine players at the table.
My quess is that it would.

[Dynasty]

I just wonder if randomn simulations would show runner runner flush occurring more than 1/23 times with nine players at the table. My quess is that it would.

Of course, it won't. The 1/23 figure is basic mathematics.

[Ed Miller]

I just wonder if randomn simulations would show runner runner flush occurring more than 1/23 times with nine players at the table. My quess is that it would.

It absolutely won't. Reread my post... I don't think you fully understand the point. Forget about what cards are where. Given that you have two cards and see three on the board, tell me which card is most likely to come on the turn. Is it the A /forums/images/icons/spade.gif ? Maybe it's the 6 /forums/images/icons/diamond.gif . I know.. it's the 2 /forums/images/icons/club.gif . The obvious fact is that any card that you haven't seen is equally likely to be the turn card. Given that all cards are equally likely to come, the chance that your suit will come is just a fraction... in this case, 10/47. Saying that the odds are actually something different is saying that some individual cards are more likely than others to come.

[Bob T.]

When the question is will THIS runner runner flush come in, the answer is clearly 1 out of 23. However lets say the board is rainbow, and there are a lot of loose players in the pot. The chance that any runner runner flush will appear is more like 3 out of 23. So in one sense flashy's perception is correct. Runner runner flushes do probably occur more often than one hand out of 23 because there are three possible suits on a rainbow board that someone might hit a running flush on.

Good luck,
play well,

Bob T.

[Flashy]

Thanks for the defense, I need all the help I can get.

My only point is that a 1/23 assumes the worst distribution of suits. It is designed to estimate the odds in heads up play having gone only 7 cards into the deck.

However, when I am playing in a nine player game, 18 cards go to the players, 2 are burned, and 3 are flopped. I would assume that between 5 to 7 of each suit was distributed to the players. The fact that I can account for three of one suit doesn't mean the deck is depleted in a suit.

To my detractors, I offer this bet. I will pick diamonds as a suit. We get a deck. We deal cards up until 3 diamonds emerge. Now we have accounted for 3 diamonds. Then we go to the 24th card and deal 2 more. I will pay you a dollar for every time 2 more diamonds don't come next. You pay me $23 every time they do. Heck I will even make it $22 so you can have the best of it. I will also pay a dollar if 3 diamonds don't emerge in the first 18 and the 21st-23rd cards.

How fair would that bet be?