Going from amount bet as % of pot to pot odds......

[Shroomy]

I was thinking about pot odds and figured it should be pretty easy to compute pot odds if you just had the percent of pot that was bet that you had to call.

After dusting off my algerbra and a few false starts I simplified a couple equations and came up with


P + 100
________
P


Will give you your pot odds to one if you use the whole number for P ...

for example

Villain bets half the pot. (50%)
p=50 (not .5)

so 50 + 100 = 150

150 / 50 = 3

so you are getting 3 to 1 odds.


It seems obviously correct, but I wanted to make sure I didnt make a mistake that happened to give me the correct answer for the few common numbers I tested to see if the equation was correct.


btw .. I was just trying to come up with a simple way for going from the amount bet to the number of outs you needed to call (with the typical assumptions .. no more betting, youre sure your losing at moment, and you will win if you hit etc)

I know Im unlikely to stumble accross a simpler method than the typical
get your pot odds
1 / your pot odds + one (your break even percent)
divide by two (or .2) to get number of outs needed for one card to go.

but ... its interesting and gets me more comfortable converting back and forth between odds, outs, percentages etc .. I just hope Im doing it right [img]/images/graemlins/smile.gif[/img]

Anything glarinly wrong? Thx [img]/images/graemlins/smile.gif[/img]

[jay_shark]

Maybe this may help .

If you're getting pot odds of p:q , then you need to win at least q/(p+q) to make the call . So if you're getting pot odds of 3:1 , then you need to win at least 1/4 or 25 % of the time . Anything more , is icing on the cake .

If you have x outs and you're drawing only one turn card , then you'll win x/47 . This is the same as 2x/94 which is very close to 2x/100 . This means that if you have x outs and you're drawing only one card , then apply the 2x rule to give you the percentage .

If you have x outs with two card to draw (turn and river) , then the probability you hit one of your outs is :
x/47 + x/47 -c(x,2)/c(47,2)

We can disregard the negative term as it will be close to 0 and approximate it as x/47+x/47 which is very close to x/50 +x/50 =2x/50 =4x/100 . This is how the 4x rule was derived .

This means that if you have 9 outs on a flush draw and you wish to draw two times , then the probability you hit is very close to 9*4=36 % of the time . This is a pretty neat formula and very useful to know .

[jay_shark]

One other thing . If you estimate the probability you hit one of your outs to be x%, then you need pot odds of (100-x%):x% . I prefer to think one street at a time and so I'll usually apply the 2x rule and work from there .

I think that covers everything , if not, let me know .