Probability of Backdoor Flush?

[joker122]

What is the probability of a backdoor flush? That is to say, if there is one spade on the flop, and you hold two spades, what is the probabilty of two more spades falling on the turn and river?

A related question, what if you hold 3 to a straight flush? Example:
You hold 8 /images/graemlins/spade.gif9 /images/graemlins/spade.gif

and the flop is: A /images/graemlins/diamond.gif7 /images/graemlins/spade.gif2 /images/graemlins/heart.gif

What is the probability that you will make either a flush or a straight?

[uuDevil]

I'm sure this has been answered before, but:

P(Backdoor Flush)= (10/47)(9/46)= 0.0416 or 4.16%
On the turn, 10 of the unseen 47 cards will make a 4-flush. If one of these hits, 9 of the last 46 cards will complete your flush. This probability is the same as odds of about 23:1 or roughly the equivalent of 2 outs.

P(Backdoor straight)= (16/47)*(4/46)= .0296 or 2.96%
On the turn, there are 16 of the 47 unseen cards that will contribute to your straight chances. If one of these hits the board, 4 of the remaining 46 cards will make your straight. This is the same as odds of about 33:1, or the equivalent of 1.4 outs.

I think this is right. /images/graemlins/grin.gif

[Bozeman]

For just a bd str8 (open end), I get:

8/47*8/46+8/47*4/46=8/47*12/46=4.44%

For the bd oe-str8 and flush,

5/47*9/46+6/47*8/46+6/47*4/46+2/47*15/46+2/47*12/46=171/2162=7.9% or 11.6:1

[uuDevil]

Bozeman,

On the backdoor straight odds, you are correct. Thanks for catching that /images/graemlins/blush.gif. I guess I'll have to chase a little more....

Elaborating, there are 8 cards that will make an open-ended st on the turn, and 8 that will complete it on the end. In addition, there are 8 cards that will make a gutshot on the turn and 4 to complete it: (8/47)*(8/46)+(8/47)*(4/46)=.0444, which is about 21.5:1 and worth about 2 outs.

For either a st or flush (or st fl), I don't quite get the same answer. It seems to me you can just add the flush and st probabilties, except you have to subtract out the probability of the st flush (so it isn't counted twice):
.0416+.0444-((2/47)*(2/46)+(2/47)*(1/46))= .0832, or about 11:1. Or did I go wrong again? /images/graemlins/crazy.gif

[bigpooch]

Best to look at the combinations out of C(47,2) = 1081.
Didn't I answer a similar post awhile back? In any case,
there are 45 ways to make a flush and in addition, either
15, 30 or 45 additional ways to make a straight. With three
suited cards in a sequence with conceivably three different
straights you can make, there are altogether 90 ways to make
a pat hand out of 1081 or about a probability of 0.08326.

If you just have a three-straight, you have 48 combinations
of straight chances if you can make three different ones,
32 if you can make two of them and just 16 if you can make
only one kind of straight. With the possibility of making
both a straight and a flush (i.e., a straight flush), you
have to subtract one from each kind of straight to avoid
double counting.

[Bozeman]

You (and bigpooch) are right, though my method works, I made one error (I subtracted 4 from 9 in the first term instead of ten to get the bare flush draws, too much thinking about 4flushes): (will use 78s ah2d9s)
<font class="small">Code:</font><hr /><pre>
Turn River cards that make your hand
6s,ts 2 15 (9 flush or str8fl, 6 str8)
5s,js 2 12 (9 flush or str8fl, 3 str8)
6cdh,tcdh 6 8
5cdh,jcdh 6 4
234qkas 6 9</pre><hr />

for 180/(47*46) (or 90 combinations out of 1081) = .0833

Craig