How fast do different PFR% converge?

[Mike Kelley]

I ran into a hand tonight from a player that I had 750 hands on

His PFR% was 3.5% The hand that he played would make his PFR% about 6% so this got me to thinking how fast do these percentages converge?

It seems like the higher the PFR% the more accurate it would be in less hands.

Has anyone ever worked on this?

[Mike Kelley]

Like a 2.5% pfr%? How many hands before you have 95% confidence it is accurate to within .5%

5%

7.5%

10%

12.5%

15%

Does the number of hands decrease as the pfr% rises? That seems correct to me intuitively

[Mike Kelley]

Can anyone point me in the right direction? I can probably figure the math out if someone just gets me started.

[DrVanNostrin]

The SD of someone's PFR% = sqrt((1-p)*p/n)

So it converges pretty fast, especially when a player's PFR is low.

But this assumes that the events are independant when they're not. If someone is raising that little they can easily notice other players' PFR tendancies and adjust. Or maybe he decided to mix things up or recognized a special situation, etc.

[Mike Kelley]

[ QUOTE ]
The SD of someone's PFR% = sqrt((1-p)*p/n)

So it converges pretty fast, especially when a player's PFR is low.

But this assumes that the events are independant when they're not. If someone is raising that little they can easily notice other players' PFR tendancies and adjust. Or maybe he decided to mix things up or recognized a special situation, etc.

[/ QUOTE ]

does p = the pfr%
n = number of hands?

I'm not sure what the special situation was. He was UTG and Raised AJs, I quizzed the strat forum I post in and no one put that in his range.

[DrVanNostrin]

[ QUOTE ]
p = the pfr%
n = number of hands?

[/ QUOTE ]
Correct.

I think it's safe to say during the 750 hands his raising standards changed. If he's raising AJs UTG, there's almost no way his PFR is that low for 750 hands.

edit: He was probably reading a poker book while playing. It took him a while to make it to the preflop chart.

[Mike Kelley]

Intuitively this doesn't make sense to me. Logic would seem to dictate that if we had 1326 ping pong balls and 47 of them were red (but we didn't know they were red). It seems like we would have to take many more samples from the group to get an accurate representation of those red balls, especially if in some light the red ball appeared to be white...


Now suppose that 663 of the balls are red. If we were to take samples from that group it would converge toward actual much faster. Atleast that is how I picture it working in my mind.

It seems like the bigger portion of the population the test group you are trying to get information about is, the less trials it would take to get an accurate representation of that group.

[Mike Kelley]

[ QUOTE ]


I think it's safe to say during the 750 hands his raising standards changed. If he's raising AJs UTG, there's almost no way his PFR is that low for 750 hands.

edit: He was probably reading a poker book while playing. It took him a while to make it to the preflop chart.

[/ QUOTE ]

It could be he's a donk and plays AKs, AQs, AJs,TT+ for a raise and limps AKo also.

[Mike Kelley]

[ QUOTE ]
The SD of someone's PFR% = sqrt((1-p)*p/n)

Is this right?

3.5% of 1326 = 47 hands so


PFR% = sqrt((1326-47)*47/750)=9
1 standard deviation is + or - 9 hands
2 SD's would be + or - 18 Hands
3 SD's would be + or - 27 more hands

which if I understand correctly from a little reading should cover 97% of his range.

3 SD's should = 97% of all results right?

[DrVanNostrin]

[ QUOTE ]
Intuitively this doesn't make sense to me. Logic would seem to dictate that if we had 1326 ping pong balls and 47 of them were red (but we didn't know they were red). It seems like we would have to take many more samples from the group to get an accurate representation of those red balls, especially if in some light the red ball appeared to be white...


Now suppose that 663 of the balls are red. If we were to take samples from that group it would converge toward actual much faster. Atleast that is how I picture it working in my mind.

It seems like the bigger portion of the population the test group you are trying to get information about is, the less trials it would take to get an accurate representation of that group.

[/ QUOTE ]
The number you're looking at is the coefficient of variation; that is, the ratio of SD to mean. The coefficient of variation relates to the % uncertainty.

But we're dealing with the absolute uncertainty (which happens to have units of %), not the % uncertainty.

Instead of looking at how many balls are red you can look at how many are not red. Knowing P(not red) is the same as knowing P(red). Having a confidence interval for P(not red) is the same a having a confidnece interval for P(red). The size of your confidence interval for P(not red) must equal the size of your confidence interval for P(red).

Var(X) = E(X^2) -[E(X)]^2

If the probability X = 1 is p and the probability X = 0 is 1-p (as it is in this case)...

E(X^2) = p

E(X) = p; [E(X)]^2 = p^2

var(X) = p - p^2 = p*(1-p)

sd(X) = sqrt(var(X)) = sqrt(p*(1-p))

If you define your random variable to be the average, not the total, sd is proportional to 1/sqrt(n)