hey you 2 who are viewing!
 

trying for 2 hours lol, like an idiot...

if i'm btn and sb plays 40% of hands and BB plays 40% of hands, what equation shows the % that at least one will play this hand?

Thanks!
M

Re: hey you 2 who are viewing!
 

4/10 + 4/10 - (4/10)^2

= 8/10 - 16/100
= 80/100 -16/100
= 64/100

Re: hey you 2 who are viewing!
 

1-(1-.4)(1-.4) = .64.

Re: hey you 2 who are viewing!
 

I love you Jason_t.

Re: hey you 2 who are viewing!
 

[ QUOTE ]
trying for 2 hours lol, like an idiot...

if I'm btn and sb plays 40% of hands and BB plays 40% of hands, what equation shows the % that at least one will play this hand?

Thanks!
M

[/ QUOTE ]

There is an easier way to calculate this yourself if you don't know the formulas. This will work when you want to know if at least one, and not exactly one, will play.

Odds that you won't play =60% Odds the opp. won't play =60%
Odds of both happenning are 60% X 60% = 36%

Which you know becomes 64% that either of you will play.

Re: hey you 2 who are viewing!
 

The trouble with these calculation is they assume SB and BB decide independently of each other and of prior betting.

Let's take the simplest situation that you care about, everyone folds to you and you raise (if you fold, you don't care, and if you only call, it's guaranteed that at least one of SB or BB will stay in the pot). I doubt that SB is calling 40% of your raises, but if she is, BB is probably more likely to call or raise if SB folds. Also if the player in the BB spot calls 40% of the time overall, he's much more likely to call in the BB because he's already posted some of the bet and he knows what everyone else has done. The same applies, to a lesser extent, to the SB.

Therefore, the odds of at least one of them calling you is probably considerably greater than the 64% computed assuming independence.

Re: hey you 2 who are viewing!
 

[ QUOTE ]
The trouble with these calculation is they assume SB and BB decide independently of each other and of prior betting.

Let's take the simplest situation that you care about, everyone folds to you and you raise (if you fold, you don't care, and if you only call, it's guaranteed that at least one of SB or BB will stay in the pot). I doubt that SB is calling 40% of your raises, but if she is, BB is probably more likely to call or raise if SB folds. Also if the player in the BB spot calls 40% of the time overall, he's much more likely to call in the BB because he's already posted some of the bet and he knows what everyone else has done. The same applies, to a lesser extent, to the SB.

Therefore, the odds of at least one of them calling you is probably considerably greater than the 64% computed assuming independence.

[/ QUOTE ]

I understand the other issues but I just wanted to know the simple calculation. Thanks guys.