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View Full Version : Percentage of getting a call???


dabluebery
05-24-2004, 11:07 AM
I'm doing some analysis of my home NL-ring game. It's low stakes, but taken very seriously. We play .25 / .50 blinds, and then it's NL from there. The buy in is usually around $20.

Let's say I hold AQs. If I make a $5 raise, (which is huge for our game). With what hands will people call with? I figured the percentage chance of someone holding a calling hand to that bet, for hands like AK, AKs, big pockets, some suited connectors, and suited Aces.

The chances of one other player holding a "calling" hand by my definition is about 9%. If there are 9 other players, though, what are the chances that one of the 9 will hold one of those calling hands???

I have this math, but I think it's wrong;

=(1-(.09))^9
which is (100%-9%) for chances of them not having a hand, multiplied by itself however number of players there are. Where am I going wrong?

Dubious
05-24-2004, 02:05 PM
Yeah, you've got about the right theory. If you think you'll get called by 9% of all possible hands and you consider the hands to all be independent (which I think is fair in this case since you're considering a lot of calling hands and the 9% is just an estimate anyway), the chances nobody has a calling hand is (1-(0.9))^9 = 0.428. So, the chances that at least one person will call your bet is 1-0.428=0.572 ==> 57.2%