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  #1  
Old 12-17-2005, 07:35 AM
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Default Question about odds vs. percentages, inspired by Arieh

...on Josh Arieh's blog:

"I'm a very quick math kinda person, but the one thing that confused me was percentages vs. odds on my money. I understand it perfectly now and will now get to play more pots!!!!"

Since Josh has made it this far without understanding the concept, I feel no shame in asking for clarification myself. So, what's the relationship between the price you're being laid and your winning percentage?
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  #2  
Old 12-17-2005, 12:03 PM
binions binions is offline
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Default Re: Question about odds vs. percentages, inspired by Arieh

Percentages and odds are just different ways of expressing the same thing.

If you have a 10% chance to hit your draw, there is a 90% chance you won't. So, the odds are 9:1 against.

In poker, most people use the size of the pot compared to the size of the bet to get pot odds. So, if there is $30 in the pot, and $3 to call, your pot odds are 10:1.

If the pot odds are greater than the odds to hit your hand (i.e. 10:1 pot odds on $3 bet vs. 9:1 odds to hit your 10% draw), then you can call.
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  #3  
Old 12-17-2005, 02:51 PM
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Default Re: Question about odds vs. percentages, inspired by Arieh

[ QUOTE ]
Percentages and odds are just different ways of expressing the same thing.

If you have a 10% chance to hit your draw, there is a 90% chance you won't. So, the odds are 9:1 against.

In poker, most people use the size of the pot compared to the size of the bet to get pot odds. So, if there is $30 in the pot, and $3 to call, your pot odds are 10:1.

If the pot odds are greater than the odds to hit your hand (i.e. 10:1 pot odds on $3 bet vs. 9:1 odds to hit your 10% draw), then you can call.

[/ QUOTE ]

Is there a simple way to compute odds into a percentage? I see the 9:1 and 10:1 examples easily, but what about 3:1. It's not just 1 (divided by) 3, (for 33% to win) is it? Basically I'm just looking for the formula here.

Sorry, I still feel as if I should know this. Something about it just isn't sticking. And why did Arieh's learning this concept mean he can now play more hands?
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  #4  
Old 12-17-2005, 03:08 PM
Leonardo Leonardo is offline
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Default Re: Question about odds vs. percentages, inspired by Arieh

While it is fine for you to ask, for Josh to say he is a 'very quick' math person and that he didn't understand %'s v odds is just stupid. What does he mean by 'very quick'? That he can count to 100 quickly?
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  #5  
Old 12-17-2005, 03:20 PM
Leonardo Leonardo is offline
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Default Re: Question about odds vs. percentages, inspired by Arieh

Add the numbers together and then divide 100 by that number then multiply this number by the number after the word to.

EG. 3 to 1 . 3 + 1 = 4. 100/4 = 25 . 25 * 1 = 25%

EG. 5 to 2 . 5 + 2 = 7 . 100/7 = 14.2 (approx) . 14.2 times 2 = 28.2 .
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  #6  
Old 12-17-2005, 03:30 PM
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Default Re: Question about odds vs. percentages, inspired by Arieh

Easy formula, let's say you're 3:1, you've got 25% chance to hit.

X:Y as percentage is Y / (X+Y)
3:1 as percentage is 1 / (3 + 1) which is 1/4 which is 25%

For stranger ones, you'll just have to guesstimate, e.g.

7:2 is 2/9 which is ~22%

Wow, my first 2+2 posting... you might as well back up the money truck now
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  #7  
Old 12-17-2005, 03:36 PM
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Default Re: Question about odds vs. percentages, inspired by Arieh

[ QUOTE ]
While it is fine for you to ask, for Josh to say he is a 'very quick' math person and that he didn't understand %'s v odds is just stupid. What does he mean by 'very quick'? That he can count to 100 quickly?

[/ QUOTE ]

If he read Josh's blog, it quickly becomes evident that while he is a great poker player, he is not much of an academic.

Thanks everyone, I get it now.
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  #8  
Old 12-17-2005, 04:47 PM
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Default Re: Question about odds vs. percentages, inspired by Arieh

And if you are comfortable with a little algebra, here's the formulae that tie things together:

The probability of an event happening = p (fractional)
The odds (against) = w as in w:1 "doubleyou to one"

w = (1-p)/p

p = 1/(1+w)

If you throw in the number of ways you can lose (not draw the card you want) and the number of ways you can win (draw that nut flush!) then

p = (#of wins)/(#of wins + #of losers)

w = (#of losers)/(#of wins)

Play with the numbers and you'll see, for example, that if p is greater than 1/2, w will be less than 1; this is a positive liklyhood of winning so less than even money (1:1) is good pot odds. (Doesn't happen very often!)
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  #9  
Old 12-17-2005, 08:01 PM
roundest roundest is offline
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Default Re: Question about odds vs. percentages, inspired by Arieh

[ QUOTE ]
[ QUOTE ]
Percentages and odds are just different ways of expressing the same thing.

If you have a 10% chance to hit your draw, there is a 90% chance you won't. So, the odds are 9:1 against.

In poker, most people use the size of the pot compared to the size of the bet to get pot odds. So, if there is $30 in the pot, and $3 to call, your pot odds are 10:1.

If the pot odds are greater than the odds to hit your hand (i.e. 10:1 pot odds on $3 bet vs. 9:1 odds to hit your 10% draw), then you can call.

[/ QUOTE ]

Is there a simple way to compute odds into a percentage? I see the 9:1 and 10:1 examples easily, but what about 3:1. It's not just 1 (divided by) 3, (for 33% to win) is it? Basically I'm just looking for the formula here.

Sorry, I still feel as if I should know this. Something about it just isn't sticking. And why did Arieh's learning this concept mean he can now play more hands?

[/ QUOTE ]

3 to 1 means 1 in 4, so 25%.
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  #10  
Old 12-17-2005, 08:20 PM
ohnonotthat ohnonotthat is offline
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Default Or, to put it another way,

"the only problems I ever had as a shortstop were groundballs and pop-ups". [img]/images/graemlins/confused.gif[/img]

[img]/images/graemlins/grin.gif[/img]
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