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  #1  
Old 10-22-2007, 07:34 PM
ICMoney ICMoney is offline
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Default Simple Odds Question

I found an odds calculator but I am getting different percentages.

For example, it says if you have 11 outs you should hit 42% on either street.


I am getting 47.3% by:

11/47 + 11/46.


A 5% difference in equity seems huge.


Am I right or how should I be calculating it?

Does it have something to do with redraws for the opponent or what?

Thanks.
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  #2  
Old 10-22-2007, 07:42 PM
CruS CruS is offline
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Default Re: Simple Odds Question

11 out, 36 "anti outs"
(36/47)*(35/46) = ~0.583
Chance of not hitting = 58%, thus
chance of hitting = 42%
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  #3  
Old 10-22-2007, 07:48 PM
ICMoney ICMoney is offline
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Default Re: Simple Odds Question

Looks good to me.

Thanks man
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  #4  
Old 10-22-2007, 07:48 PM
CruS CruS is offline
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Default Re: Simple Odds Question

just to clarify why your math is flawed:
if I were to flip a coin two times, I have "two streets" with 50% to hit.
50% + 50% = 100%! Which is obv wrong.
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  #5  
Old 10-22-2007, 07:52 PM
ICMoney ICMoney is offline
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Default Re: Simple Odds Question

[ QUOTE ]
just to clarify why your math is flawed:
if I were to flip a coin two times, I have "two streets" with 50% to hit.
50% + 50% = 100%! Which is obv wrong.

[/ QUOTE ]

I would just call it rigged.

But yeah, I see your point.
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  #6  
Old 10-22-2007, 09:13 PM
members_only members_only is offline
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Default Re: Simple Odds Question

I think what you're missing is that chance of hitting one of your outs on either street = chance you hit turn + chance you DON'T hit turn AND hit river. You were just adding chance you hit turn and chance you hit river, which leads you to overestimate the probability.
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  #7  
Old 10-22-2007, 09:32 PM
CruS CruS is offline
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Default Re: Simple Odds Question

[ QUOTE ]
I think what you're missing is that chance of hitting one of your outs on either street = chance you hit turn + chance you DON'T hit turn AND hit river. You were just adding chance you hit turn and chance you hit river, which leads you to overestimate the probability.

[/ QUOTE ]

Would seem logical, but in fact it's not. This has nothing to do with it at all, read my posts again and you will understand. Working with probabilities like this one requires that you use multiplication rather than addition to get the right results.
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  #8  
Old 10-22-2007, 09:39 PM
WhiteWolf WhiteWolf is offline
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Default Re: Simple Odds Question

[ QUOTE ]
[ QUOTE ]
I think what you're missing is that chance of hitting one of your outs on either street = chance you hit turn + chance you DON'T hit turn AND hit river. You were just adding chance you hit turn and chance you hit river, which leads you to overestimate the probability.

[/ QUOTE ]

Would seem logical, but in fact it's not. This has nothing to do with it at all, read my posts again and you will understand. Working with probabilities like this one requires that you use multiplication rather than addition to get the right results.

[/ QUOTE ]
members_only's method is correct, and yields the same answer as your method. You can add the probabilities of events so long as they are mutually exclusive.

11/47 + 36/47*11/46 = 41.7%
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  #9  
Old 10-22-2007, 09:42 PM
CruS CruS is offline
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Default Re: Simple Odds Question

huh, I thought he was right at first but I drew it up and it was not correct, guess I missed something.
Sorry about that, I will make my trials better before making a statement.
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  #10  
Old 10-22-2007, 09:45 PM
members_only members_only is offline
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Default Re: Simple Odds Question

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
I think what you're missing is that chance of hitting one of your outs on either street = chance you hit turn + chance you DON'T hit turn AND hit river. You were just adding chance you hit turn and chance you hit river, which leads you to overestimate the probability.

[/ QUOTE ]

Would seem logical, but in fact it's not. This has nothing to do with it at all, read my posts again and you will understand. Working with probabilities like this one requires that you use multiplication rather than addition to get the right results.

[/ QUOTE ]
members_only's method is correct, and yields the same answer as your method. You can add the probabilities of events so long as they are mutually exclusive.

11/47 + 36/47*11/46 = 41.7%

[/ QUOTE ]

Phew, thanks! I was beginning to wonder...
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