# Probability of Finishing in a Certain Place

1. Dec 12, 2011

### dward1

I have a projection issue. We have a weekly contest where scores are accumulated over 14 weeks so I have mean, STDEV, all that good stuff. Now I want to predict the likelihood of each of these 6 people finishing in 1st, 2nd or 3rd. We have 3 more weeks remaining and you get a score each week. Each of the 6 have different totals right now and are averaging different weekly scores with different standard deviations.

For example
Team A currently has 1010 has been averaging 80 with a STDEV of 8
Team B has 1000, averaging 70 with a STDEV of 11
Team C has 1100, averaging 90 with a STDEV of 5.
Team D has 1050, averaging 85 with a STDEV of 15

those are just made up numbers but I just want to get how to do it? what are the odds of Teams A, B, C, and D finishing 1st, 2nd, or 3rd? thanks

2. Dec 14, 2011

### Stephen Tashi

This is an interesting problem and, as far as I can see, it can't be solved by simple arithmetical operations. Assuming the teams scores are independent random variables, the probability that they finish in a given order can be expressed as a multiple integral.

Let $\mu_a =$ the mean of team A's score
Let $\sigma_a =$ the standard deviation of team A's score
Let $$\phi_a(x) = \frac{1}{\sqrt{2\pi}\sigma_a} e^{- \frac{(x-\mu_a)^2}{2 \sigma^2_a}}$$
and use similar notation for the other teams.

The probability $p_{abcd}$ that the teams finish in the the order A,B,C,D is:

$$p_{abcd} = \int_{-\infty}^{\infty} \int_{-\infty}^{x_a} \int_{-\infty}^{x_b} \int_{-\infty}^{x_c} \phi_a(x_a) \phi_b(x_b) \phi_c(x_c) \phi_d(x_d) dx_d dx_c dx_b dx_a$$

Since this is your first post, so I don't know if you are familiar with integral calculus. I don't know if you are interested in trying to simplify this expression or compute it numerically.

If you want to know the probability that team A finishes first without specifying the order in which the other teams finish you must add together the probabilities of all the possible orders that have team A first. $p_{abcd} + p_{abdc} + p_{acbd} + ....$ etc.

3. Dec 14, 2011

### PatrickPowers

I think that the answer is a messy convolution that can only be solved by banging on it with a computer via Monte Carlo integration.

4. Dec 14, 2011

### Stephen Tashi

My answer in the previous post doesn't capture the fact that scores are accumulated. It is only correct for the ranking at the end of 1 week when all start with a score of 0.

The answer to problem in your example is more complicated. It can be worked out if you're interested, but it will also involve multiple integrals.

5. Dec 19, 2011

### dward1

thanks, this is what I found a day or so after making this post and ran it. did not know how to do it before this and it was very interesting and enjoyable to do. the powers of excel!