Percentile and standard deviation

[Grendel]

Hey, I've got a question for you guys. Assuming you have an accurately measured winrate (in BB/100) and standard deviation:

If you sit down and play 100 hands, with a result of X BB, I have a good idea how to calculate what percentile ranking that is. For instance, if X=winrate, that's the 50th percentile. If X=winrate+1SD, that's the 84th percentile. My statistics background is quite rusty, but I remember there being a lookup table for this sort of calculation.

My question is if you sit down and play a number of hands other than 100, with a result of X BB, how do you calculate that percentile?

Purely hypothetically, say my winrate is 2.0 BB/100 and my SD is 15.0. Now say I play a 500-hand session and lose 80 BB. This is an unusual result, but well within the realm of possibility. My question is basically how unusual this is. Is it the bottom 2% of 500-hand sessions, or the bottom 10%?

And to think, I was a math major once... [img]/images/graemlins/blush.gif[/img]

Thanks in advance,
-Grendel

[pzhon]

[ QUOTE ]
Purely hypothetically, say my winrate is 2.0 BB/100 and my SD is 15.0. Now say I play a 500-hand session and lose 80 BB. This is an unusual result, but well within the realm of possibility. My question is basically how unusual this is. Is it the bottom 2% of 500-hand sessions, or the bottom 10%?

[/ QUOTE ]
To express a result in standard deviations, compute the standard deviation for the session. A session of length n*100 will have a standard deviation squareroot(n) times the standard deviation for 100 hands.

In your example, after 500 hands, your expected win is 5*2.0 = 10 BB, and the standard deviation is sqrt(5)*15 ~ 33.5. Losing 80 BB is 90 BB below average, or 90/33.5 ~2.68 standard deviations below average.

For a normal distribution, the probability of an outcome at least 2.68 standard deviations below average is about 0.368%, or 1/272.

[Grendel]

Very nice. Thank you for the info.

[ QUOTE ]
For a normal distribution, the probability of an outcome at least 2.68 standard deviations below average is about 0.368%, or 1/272.

[/ QUOTE ]

Is this the sort of conversion I need to look up in a table? Are there any tools in, say, MS Excel that can do this part?

-Grendel

[gaming_mouse]

[ QUOTE ]
Is this the sort of conversion I need to look up in a table? Are there any tools in, say, MS Excel that can do this part?

[/ QUOTE ]

Yes. And yes, there is an excel function for it, though i forget the name. There are also online calculators. Search for "normal distribution calculator" and you will find many.

gm

[BruceZ]

[ QUOTE ]
Very nice. Thank you for the info.

[ QUOTE ]
For a normal distribution, the probability of an outcome at least 2.68 standard deviations below average is about 0.368%, or 1/272.

[/ QUOTE ]

Is this the sort of conversion I need to look up in a table? Are there any tools in, say, MS Excel that can do this part?

-Grendel

[/ QUOTE ]

Excel function =NORMSDIST(-2.68) returns 0.368%
Excel function =NORMSINV(0.368%) returns -2.68

Note that NORMSDIST gives you the area from minus infinity to -2.68, so it always gives you the probability of being between minus infinity and the entered number of standard deviations from the mean. If you put in +2.68 for 2.68 above average, it would give you the area from minus infinity to +2.68 or 1-0.368% = 99.632%, indicating a probability of 99.632% of being between minus infinity and 2.68 standard deviations above the mean. NORMSINV is the inverse of NORMSDIST.