View Full Version : how to calculate odds I need to be ahead?

09-05-2003, 10:20 AM

this is a follow-up question from a thread that is pretty much dead. so I'll summarize here or you can follow the link at the end.

the example is: I have pocket aces on a rainbow board and am raised on the turn getting effective odds of about 6.5-1 on calling down through river. the question was what percentage of the time do I have to be ahead to make the call correct. (i.e. two outs to a win)

I was given the following equation- So let X be the % of time you are ahead. We want X(.95) + (1-X)(.05) = .13;

what does (.95) represent?
the amount I am ahead now but outdrawn?

thanks for the help,


link: http://forumserver.twoplustwo.com/showflat.php?Cat=&Board=smallholdem&Number=319703& fpart=&PHPSESSID=

09-05-2003, 11:14 AM
no, from what I saw in the last thread, the 95% is the proability that IF you are ahead now you do not get outdrawn, and the 5% is the probability that IF you are behind now you do outdraw him. The X% is the probability you are ahead, so combined, X*.95 is the probability you are ahead and stay ahead, and (1-X)*.05 is the combined probability that you are behind but pull ahead. (The .95 and .05 assume that you can only lose to a 2 outer or pull ahead with a 2 outer, so all that is being considered here is sets and trips).

Note that ahead and stay ahead or behind and pull ahead are the only scenarios you win in. Those are equated to the 13% of the time you need to win to break even. With 6.5/1 odds you are nearly pot bound and should call with any draw to trips or better if hitting is virtually certain to win.

09-05-2003, 09:21 PM

I think that is what I meant to say. thanks for the explanation, makes things clearer.