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eric.wikman@gmail.com science forum beginner
Joined: 05 Oct 2005
Posts: 3
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Posted: Wed Oct 05, 2005 2:15 am Post subject:
Figuring % chance A will be larger than B
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I have two completely independent numbers and the mean and standard
deviation for each. I'd like to figure out what percent chance one has
of being larger than the other. I cannot say for sure if it follows the
normal distribution or some other distribution, because I have a small
sample size. For example:
Team A has a mean of 80.375 and a StDev of 13.02
Team B has a mean of 70.875 and a StDev of 25.20
I did a Monte Carlo simulation (and I assumed it was a normal
distribution) and found that Team A has a 63.26% chance of winning
after 100,000 simulations.
Each team can only effect the number of points they score, they cannot
effect the number of points the opposing team can score, that's why I
say the numbers are independent. Team A and B are both fantasy
football teams.
What is a better way to come up with 63.26% (or some more accurate
number) using math instead of brute force?
Eric |
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iandjmsmith@aol.com science forum beginner
Joined: 13 Sep 2005
Posts: 15
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Posted: Wed Oct 05, 2005 11:54 am Post subject:
Re: Figuring % chance A will be larger than B
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eric.wikman@gmail.com wrote:
| Quote: | I have two completely independent numbers and the mean and standard
deviation for each. I'd like to figure out what percent chance one has
of being larger than the other. I cannot say for sure if it follows the
normal distribution or some other distribution, because I have a small
sample size. For example:
Team A has a mean of 80.375 and a StDev of 13.02
Team B has a mean of 70.875 and a StDev of 25.20
I did a Monte Carlo simulation (and I assumed it was a normal
distribution) and found that Team A has a 63.26% chance of winning
after 100,000 simulations.
Each team can only effect the number of points they score, they cannot
effect the number of points the opposing team can score, that's why I
say the numbers are independent. Team A and B are both fantasy
football teams.
What is a better way to come up with 63.26% (or some more accurate
number) using math instead of brute force?
Eric
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If the random variables are independent then the mean of the difference
is the difference of the means and the variance of the difference is
the sum of the variances. This is regardless of the distributions of
the random variables.
The value you want if they are both normally distributed is
cdf_normal(9.5/sqrt(804.5604)) which is approx 63.16% and is close to
your simulated value 63.26%.
Ian Smith |
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eric.wikman@gmail.com science forum beginner
Joined: 05 Oct 2005
Posts: 3
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Posted: Wed Oct 05, 2005 5:18 pm Post subject:
Re: Figuring % chance A will be larger than B
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Thank you Ian. That works great, I tried it with the other 4 matchups,
and it was always within 1% of the number I got from simulation.
Do you know of any web references or books that could help me learn
more about why this works, and hopefully dive into a tougher discussion
on how to handle it when there are dependencies (therefor not
independent variables)? |
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iandjmsmith@aol.com science forum beginner
Joined: 13 Sep 2005
Posts: 15
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Posted: Thu Oct 06, 2005 7:57 am Post subject:
Re: Figuring % chance A will be larger than B
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eric.wikman@gmail.com wrote:
| Quote: | Thank you Ian. That works great, I tried it with the other 4 matchups,
and it was always within 1% of the number I got from simulation.
Do you know of any web references or books that could help me learn
more about why this works, and hopefully dive into a tougher discussion
on how to handle it when there are dependencies (therefor not
independent variables)?
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You will never really be able to deal with complex problems without a
grounding in the basics. However, I am afraid I would not know what to
recommend. I've never looked at web references for learning about
probability and statistics and books I used were probably too
mathematical and are probably no longer available.
In the short term using Google groups or something similar, where you
can search to see if similar problems have been answered, might be as
good as any approach. Then check the answers, as you have done in this
case, to make sure the answers make some sense.
If you can't find something similar then post your own question.
Ian Smith |
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Google
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