Up date of what I think I learned regarding pot odds and breaking even

[ThomasDayton]

I looked in Phil Gordon's book and I came up with this:
Pot odds = pot/what I'm asked to call. Example pot 7 a bet of 2 so i would get 9/2 or 4.5 to 1 on my money. To break even I am told = 1/pot odds+1 or in this case 1/4.5+1 or 1/5.5 18% I need one win for every 5.5 loses. 1 win 11 5.5 loses 11. I was trying to say this in my posts before on pot odds and stuff (or I meant to say it) just got some things mixed up. I hope this one is correct

[GetPhysical]

I`m just gonna offer a friendly word of advice. what is it that you want to achieve, are you really going to sit there and crunch numbers constantly whilst playin? Are you playing at such high levels where this is even important?
Not that i`m saying it isn`t but generally all you need to know are your odds of completeing your hand against the odds you are getting from the pot to call, alright the i need one win for every five losses ...........teah great but do you really care about that. Seriusly you just seem worried about posting somethin now mate, theres plenty of people on here posting complex mathematical solutions to the wierd and wonderful world of odds, but just concentrate on playing cards and enjoying yourself...

but seriously though if you haven`t seen this http://www.pr0crast.com/2+2.NL.Anthology.v1.htm

Then its definitely worth a look

[Poker Clif]

[ QUOTE ]
I`m just gonna offer a friendly word of advice. what is it that you want to achieve, are you really going to sit there and crunch numbers constantly whilst playin? Are you playing at such high levels where this is even important?
Not that i`m saying it isn`t but generally all you need to know are your odds of completeing your hand against the odds you are getting from the pot to call, alright the i need one win for every five losses ...........teah great but do you really care about that. Seriusly you just seem worried about posting somethin now mate, theres plenty of people on here posting complex mathematical solutions to the wierd and wonderful world of odds, but just concentrate on playing cards and enjoying yourself...

but seriously though if you haven`t seen this http://www.pr0crast.com/2+2.NL.Anthology.v1.htm

Then its definitely worth a look

[/ QUOTE ]

Very nice consolidation of information in one spot. Thanks!

[metsandfinsfan]

[ QUOTE ]
To break even I am told = 1/pot odds+1 or in this case 1/4.5+1 or 1/5.5 18% I need one win for every 5.5 loses. 1 win 11 5.5 loses 11. I was trying to say this in my posts before on pot odds and stuff (or I meant to say it) just got some things mixed up. I hope this one is correct

[/ QUOTE ]

this part is wrong. it is 1 win 11 for every 5.5 losses of 2 .. which makes it an EV of zero. It costs you 2 to call not 11. SO if you have better than a 1 in 5.5 chance of winning a call is profitable.
Example - on the turn you know you have 12 outs. You have a 12/46 chance in hitting your card, so a call is profitable.

12 times you win $11 = +$132
34 times you lose $2 = -$68
so in 46 times you win +$64
$64/48 = $1.33 so you win $1.33 every time you make this call

[ThomasDayton]

My point is 1 to 5.5 is a break even ration, if in the example above I win 1 time and loses 5.5 times I break even (1x11=11 5.5x2=11). So true any draw that offers better then 1 to 5.5 would be a profitable call, or @18% and better.

[Small Fry]

On 4.5 :1 odds break even is 4.5:1. Meaning you lose 4.5 times for every 1 win. How many TOTAL events is this? 4.5 losses + 1 win = 5.5 total events.

Pot is 9 you put in 2 and win. How much does the dealer shove your way? 11. He shoves 11 chips your way.

So how many times must we put in 2 before we equal 11. 5.5 TOTAL times. How many must we then lose to hit our break even. 5.5 total minus the 1 win. or 4.5 times. We lose 4.5, win 1. 4.5:1.

If you want to look at this from an amount won perspective then in the above example how much do you win if the pot is 9 and you put in 2 and win? 9. How many times must you put in 2 before losing 9 ? 4.5 times. Lose 4.5 times win 1 for break even. 4.5:1.

How often do you win with odds of 4.5:1. In both scenarios above you lost 4.5 times and won once. So out of a total of 5.5 tries you acheived 1 win. or 18%. 4.5:1 odds is equal to an 18% chance of winning. As I've previously explained we can also expand the 4.5:1 odds so that when adding both sides together they equal 100. In this form the actual numbers also represent their corresponding percentages. Since we said 4.5:1 equals 18% lets multply each side by 18 and see what happens. 4.5*18=81. 1*18=18. So we get 81:18. Thanks to rounding it only equals 99.....lol. Looks like 4.5:1 is pretty close to 18% to me

Bottom line. If you have 4.5:1 odds to hit your hand and the pot offers you better than 4.5:1 then you have whats known as positve expected value(+EV)as the pot is offering you more money than the opportunity to hit your hand offers

[ThomasDayton]

You took up a lot of room to say you need 18% chance or better to profit, which my fomular gave you in a shorter time. However 4.5:1 as your bottom line states is 22% not 18%

[Nightlight87]

The lot of room is taken up with helpful explanations. I'd recommend you re-read his first big post because you are still converting odds to percentages incorrecctly.

4.5:1 is not the same as 1/4.5 = 22%

It is winning 1 time, losing 4.5. So you won 1 out of 5.5
So 1/5.5 = 18%

EDIT: Also glad you stuck around. It can be confusing to begin with, but if you stay around and read and absorb what people here say you can learn a lot and almost definately become a winning player. Good luck [img]/images/graemlins/smile.gif[/img]

[ThomasDayton]

[ QUOTE ]
4.5:1 odds is equal to an 18% chance of winning.

[/ QUOTE ] It has been a while since I've been in school is this a "newer math"

[jmitchell42]

no, with their explanation it seems pretty right on.
4.5:1 odds=18% win once in 5.5 trials
3.5:1 odds=22% win once in 4.5 trials.