?? probability of future results ??
 

Take a hypothetical situation....

Player Able has an established win rate, WR.

Able's WR is expressed as BB/HR or BB/100, whichever
is relevant.

Able's standard deviation (SD) is also expressed in
terms of BB, rather than $$.

Given that, I'd like to be able to predict the probability
of a ceratin result (or greater) over a specific span
of play.

In other words, I'd like a generic function to calculate
the probability of winning (or losing) X BB's over a
specific span Y (HR's or 100's) play, given win (loss)
rate parameters WR and SD.

Help!

Thanks in advance, mosquito.

Re: ?? probability of future results ??
 

**BUMP**

Someone....anyone.....

If a player with a +3.1 BB/100 , SD 4 BB/100 wants to
know the chance that he will go -100 BB in the next
500 hands, as an example. Answer this, and it can be
generalized......

Thanks.....

Re: ?? probability of future results ??
 

You can just a normal approximation, assuming you have enough hands.

That is, after N hands, the amount won has a normal distribution with

mean = win rate (in BB/100)
SD = (sample SD)/sqrt(N/100)

You can find the chance that he's won between x and y by finding the area under the normal curve between x and y.

Cheers,
gm

Re: ?? probability of future results ??
 

[ QUOTE ]
Take a hypothetical situation....

Player Able has an established win rate, WR.

Able's WR is expressed as BB/HR or BB/100, whichever
is relevant.

Able's standard deviation (SD) is also expressed in
terms of BB, rather than $$.

Given that, I'd like to be able to predict the probability
of a ceratin result (or greater) over a specific span
of play.

In other words, I'd like a generic function to calculate
the probability of winning (or losing) X BB's over a
specific span Y (HR's or 100's) play, given win (loss)
rate parameters WR and SD.

Help!

Thanks in advance, mosquito.

[/ QUOTE ]

See my recent post.

Re: ?? probability of future results ??
 

[ QUOTE ]
That is, after N hands, the amount won has a normal distribution with

mean = win rate (in BB/100)
SD = (sample SD)/sqrt(N/100)

[/ QUOTE ]

You mean that this is the mean and SD for the win rate, not the amount won. The amount won has a mean of (win rate in bb/100)*N/100, and an SD of (SD for 100 hands) * sqrt(N/100).

Re: ?? probability of future results ??
 

[ QUOTE ]
[ QUOTE ]
That is, after N hands, the amount won has a normal distribution with

mean = win rate (in BB/100)
SD = (sample SD)/sqrt(N/100)

[/ QUOTE ]

You mean that this is the mean and SD for the win rate, not the amount won. The amount won has a mean of (win rate in bb/100)*N/100, and an SD of (SD for 100 hands) * sqrt(N/100).

[/ QUOTE ]

Yes, I did mean that. thanks.