Are the events mutually exclusive?

I guess I'm not understanding what you're asking. Are you asking for a probability distribution for the number of some predefined events happening per trial?

Ahh, that makes a lot more sense. I don't see how you'd do that without using a computer. Maybe there's a trick, though. I'll look around. You have me interested in this, now.

The chance of winning x number of games out of y total is pretty-straightward combination/permutation math. However, I've never worked a situation where you had different probabilities assigned to each event. So I would probably look up variable probability combination/permutation solutions and see what pops up. If I get free time later I might do that and post back.

This may be useful to you, if you haven't seen it already, specifically methods on applying the Poisson-Binomial distribution: http://www3.stat.sinica.edu.tw/statistica/oldpdf/A7n44.pdf

For my first actuary exams I had to study a bunch of distributions so I looked through my old notes and then did a google search. It's the fifth link.

We could also fairly simply calculate the mean and variance of this thing and then make a normal approximation I think, which should give you a pretty good feel for what the distribution should look like. If P(n)=true probability distribution (n=discrete) and f(x)=normalized approximation then I would think that P(n)~the integral of x=n-.5 to x=n+.5 of f(x), no?

I think by definition it would have to be. I guess which might be somewhat problematic if E[X]~0

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I think by definition it would have to be. I guess which might be somewhat problematic if E[X]~0

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Going off of the general mood of the football board, this may be the case

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Lulz

Just write a small Monte Carlo script. It would probably take less time than attempting to find an analytical solution to your problem.

Sounds like a Random Forest prediction.

Check out bigml.com

If it's the same probability, just use the binomial distribution.

If not, Monte Carlo is the way to go.

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Seriously though, for means that are low numbers and for numbers close to the number of possible events I think that too much of the approximation will be sort of chopped off by the boundary, and I think that'll be the case for a lot of football teams.

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So I got curious about how well the normal distribution would approximate the answer to your question. Here are my results.The answer is for realistic situations I think the normal distribution is an apt approximation. One note: you may want to write your own simulation code just to make sure there are no errors in mine where the data might be skewed, but based on how well the simulation matches with the approximation I think I did everything correctly.

Astro,

Have you had a chance to go through my post? I'm curious to know what you think about using a normal approximation for your problem now.

I might go through and see if I can analytically calculate the maximum proportion of the normal distribution being below zero.

I've finished SOA exam P and FM. Going to start looking for a job soon.

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How difficult were they?

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Difficult, but in a different way than physics problems. If we grade Griffiths' E&M problems on a scale of 1-5 in difficulty with 5 being the most difficult, these problems are all 1s and 2s. However, you have to know a lot more formulas and definitions off the top of your head for the actuarial exams and you only get 3 hours to answer 30 problems.

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Shouldn't the case where all the win probabilities are equal yield the most area behind 0 for any mean?

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I think that's correct. I know that the standard deviation is greatest in that scenario, however the average is also shifted over to the right. I'm not sure how those two facts together play out in terms of what E[X] and sigma[X] should be to get the greatest proportion of the distribution under zero.

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Seriously though, for means that are low numbers and for numbers close to the number of possible events I think that too much of the approximation will be sort of chopped off by the boundary, and I think that'll be the case for a lot of football teams.

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So I got curious about how well the normal distribution would approximate the answer to your question. Here are my results.The answer is for realistic situations I think the normal distribution is an apt approximation. One note: you may want to write your own simulation code just to make sure there are no errors in mine where the data might be skewed, but based on how well the simulation matches with the approximation I think I did everything correctly.

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Glad to know my proposed w/l probability distribution was probably wrong.