| 0 |
Subject: Excel / Stats Question
Posted by: TacoJohn
- [529141115] Tue, Oct 11, 2005, 17:16
Ok, I'm betting there's a relatively easy way to do this.
If I have two fantasy teams' mean scores and std deviation from a sample of performances, assuming a normal distribution, how can I calculate the percent chance that one team would score higher than the other in a given trial? |
| 1 | Sludge
ID: 27751510
Tue, Oct 11, 2005, 17:25
|
Assuming you know how to find normal probabilities of the form P(Z > a):
If X is the score of team 1 with mean m1 and standard deviation s1, and Y is the score of team 2 with mean m2 and standard deviation s2, we have a neat theorem that says that if X and Y are normal, then X-Y is also normal with mean m1-m2 and standard deviation sqrt(s1^2+s2^2). So P(X-Y>0) = P(Team 1 Wins) is equal to the following:
P(X-Y > 0) = P(Z > (m2-m1)/sqrt(s1^2+s2^2)), where Z~N(0,1).
P(X-Y<0) = P(Team 2 Wins) = P(Z < (m2-m1)/sqrt(s1^2+s2^2)), where Z~N(0,1).
|
| 2 | TacoJohn
ID: 529141115
Tue, Oct 11, 2005, 17:28
|
Aha, I knew that at one point in my life. Thank you.
|
|
|
|
Post a reply to this message: (But first, how about checking out this sponsor?)
|
|