I should know how to do this . . .
 

Assume my PFR is R. How likely is it that in a stretch of N hands, I will actually raise a fraction of hands r?

The question is really about confidence in PFRs after a small number of hands. For example, if my actual PFR is 10%, how likely is it that in any given stretch of 140 hands, my observed PFR is only 3%?

The question came up in regards to not 3-betting AKo from an UTG pre-flop raiser with a PFR of 3%, but only 144 hands. I advised that 144 hands was nowhere near enough of a raise to simply cold-call (thus letting in 5 players behind Hero when he has an unpaired offsuit hand that only makes 1 straight), and that a raise was in order. I'd like to back it up mathematically but I seem to be coming up stupid tonight.

Like I said, I should know how to do this, but I'm really sort of at a loss.

Edit: In case it was unclear, R != r.

Re: I should know how to do this . . .
 

[ QUOTE ]
Assume my PFR is R. How likely is it that in a stretch of N hands, I will actually raise a fraction of hands r?

[/ QUOTE ]

The probability of raising a fraction r or less is:

=BINOMDIST(r*N,N,R,true)

where BINOMDIST is the Excel function for the cumulative binomial distribution

sum{k = 0 to int(r*N)} C(N,k)*R^k*(1-R)^(N-k).


[ QUOTE ]
The question is really about confidence in PFRs after a small number of hands. For example, if my actual PFR is 10%, how likely is it that in any given stretch of 140 hands, my observed PFR is only 3%?

[/ QUOTE ]

The probability of 3% or less is

=BINOMDIST(3%*140,140,10%,TRUE) =~ 0.12%.

You can also use a normal approximation to the binomial distribution, where the mean is R*N, and the standard deviation is sqrt[N*R*(1-R)]. This works better when N is large.

Re: I should know how to do this . . .
 

That's perfect, Bruce. Thanks ten to the sixth.