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Suited connector odds of flopping a str8 or flush draw

  
 
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pokernewb
Old 08-17-2005, 07:05 PM     Post subject: Suited connector odds of flopping a str8 or flush draw #1 (permalink)  
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pokernewb
Hi, does anyone have the odds for the following because I cant seem to find them anywhere:

1. Odds of flopping a flush or straight with open ended suited connectors
2. Odds of flopping a flush draw or outside straight draw with open ended suited connectors
3. Odds of flopping a flush draw or gutshot straight draw open ended suited connectors
4. Odds of hitting a flush or straight by the turn open ended suited connectors

Also, I'm interested in how I'd go about working out those odds. I know how to work out odds for the turn and river but not the flop, especially if i need two or three specific cards to hit.

Cheers

BTW great forum!
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CrunchyNuts
Old 08-17-2005, 07:45 PM     Post subject: Re: Suited connector odds of flopping a str8 or flush draw #2 (permalink)  
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I don't have the time right now to go through the calculations myself, but the technique you'd use here is counting. Figure out how many flops satisfy your conditions then divide by the total number of flops (50 * 49 * 48, if you know only 2 cards) and you've got your chance.

Give it a crack on #1 and I'll come back and see how you fare. Or don't and maybe some day I'll come back and do it for you.
Up my bankroll - buy Saints Row.
 
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pokernewb
Old 08-18-2005, 10:03 AM #3 (permalink)  
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pokernewb
Thanks for the tip but I still don't quite understand. For example, with the straight I need 3 specific cards, but the specific cards I need depend on the other cards.

Say I had a 45 connector, to make the straight I would need either an A23, 236, 367, 678, but I cant add them up and say i have 12 outs because i need all three of either group.

I wish I was better at maths!
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TLR
Old 08-18-2005, 10:53 AM #4 (permalink)  
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TLR is an unknown quantity at this point
 In your example, lets say you have 45h.
Possible flush combinations with 3 cards, order does not matter = 165
(n!/(n-k)!k!
when n=11 and k=2
Total possible combinations on flop
50!/47!3! = 19600
so you have 165/19600 = 0.8% of hitting the flush or str8 flush
Chances of hitting a str8:
You have a few possible str8
A23, 236, 367, 678. There are 4 cards to each suite so a total of 4*4*4 = 64*4=256
There are 4 duplicate combination for str8 and flush
Total chance of hitting str8 or flush is therefore 256+165-4 /19600 = 2.12%

You can do the rest following this logic, I may do them later if I have time


 
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CrunchyNuts
Old 08-18-2005, 02:18 PM #5 (permalink)  
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Quote:
Originally Posted by TLR
In your example, lets say you have 45h.
Possible flush combinations with 3 cards, order does not matter = 165
(n!/(n-k)!k!
when n=11 and k=2
165 is right, but k = 3, it's 11 choose 3. (or is it pick, I can never remember which means order doesn't matter, bleh)
Quote:
Originally Posted by TLR
Total possible combinations on flop
50!/47!3! = 19600
so you have 165/19600 = 0.8% of hitting the flush or str8 flush
Yup yup. I goofed and forgot the /6 on the #flops, oops.
Quote:
Originally Posted by TLR
Chances of hitting a str8:
You have a few possible str8
A23, 236, 367, 678. There are 4 cards to each suite so a total of 4*4*4 = 64*4=256
Think ya just got another typo here. With 4 str8s to make, and 4*4*4 ways to make each (4 suits for each card, each independant), it's 4*(4*4*4) = 4 * 64 = 256
Quote:
Originally Posted by TLR
There are 4 duplicate combination for str8 and flush
Total chance of hitting str8 or flush is therefore 256+165-4 /19600 = 2.12%
w3rd.
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TLR
Old 08-18-2005, 02:24 PM #6 (permalink)  
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TLR is an unknown quantity at this point
 CrunchyNuts Wrote:
[quate]
165 is right, but k = 3, it's 11 choose 3. (or is it pick, I can never remember which means order doesn't matter, bleh)
[/quate]

Correct of course, K=3. my typo

Quote:
TLR wrote:
Chances of hitting a str8:
You have a few possible str8
A23, 236, 367, 678. There are 4 cards to each suite so a total of 4*4*4 = 64*4=256

Think ya just got another typo here. With 4 str8s to make, and 4*4*4 ways to make each (4 suits for each card, each independant), it's 4*(4*4*4) = 4 * 64 = 256
256 was my result as well, what did I miss ?
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pokernewb
Old 08-20-2005, 09:26 AM #7 (permalink)  
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pokernewb
Thanks, I think I understand how to workout the straight now, but I still don't quite get the flush. Would you be able to explain it a bit more? By the way what do the exclamation marks represent in the formula? Like I said I'm pretty thick when it comes to maths!

Cheers
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CrunchyNuts
Old 08-20-2005, 02:00 PM #8 (permalink)  
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CrunchyNuts
Just a typo TLR, 4*4*4 != 64*4=256

! is the mathmatical symbol for "this number multiplied by every number lower than it, down to 1." For instance, 5! = 5*4*3*2*1.

The formula TLR used for figuring out how many flops give you the flush is the generic formula for choosing k things from a set of n things, where order of the things does not matter. So for the flush there's 11 cards of your suit left and you need to choose 3 of them to make a flop of that suit. Plug in for n and k and churn out the answer and that's the number of possible flops of your suit.
Up my bankroll - buy Saints Row.
 
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pokernewb
Old 08-22-2005, 05:43 PM #9 (permalink)  
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Thanks crunchy
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