Hey Guys,

Let's say I'm playing a game, and on my turn I'm allowed two plays. Only the better result of my plays count, and I'm not allowed to make the same play twice on the same turn. I have these two plays:

Play A averages 25 points with a standard deviation of 10.
Play B averages 15 points with a standard deviation of 8.

If on my turn I make Play A & Play B, what's my average score for the turn?

What if the game allowed 3 plays on the same turn, under the same rules (only the best play counts, can't repeat the same play) and we had a Play C with an average of 12 points with a standard deviation of 7?

I know the formula to compute how often Play B is superior to Play A, but unfortunately, I don't know how to take that any further. If anyone could even just let me know terms to Google, that would probably be enough to get me going in the right direction here. Thanks a lot.

Not enough info. What is the distribution of points? Normal? Uniform? Binomial? Something else? Is it continuous or discrete? Can you have negative points?

If a(x) and b(x) are the density functions for A and B (or distributions if discrete), then you want the double integral (or double sum if discrete) x*a(x)*b(y) as x ranges from y to the maximum value and y ranges over all values, plus y*a(x)*b(y) as y ranges from x to the maximum value and x ranges over all values.

Hey Bruce, good to see you still kicking around.

The distribution of the points is a normal continuous distribution. Zero is the lowest score possible, so no negative points.

I've done some Googling on density functions, so ty for that term. What I've learned so far has been...daunting. My days of doing integrals are 15 years in the past, and I remember roughly 0 of it. I'm going to keep studying and Googling on the subject, but if you (or anyone else) happens to know of an Excel formula for this, or some sort of web-based solver, I think that would be pretty awesome.

Thanks for the help Bruce.

Those statements are incompatible with each other. A normal distribution allows scores to -infinity or it isn't a normal distribution.

If we use normal distributions but throw away negative points, the average is 26.7 for the first case, and 27.1 for the second case by R simulation of 10 million games. If we were to replace negative points with zero, it would only decrease the average scores by about 0.1 in this case, but that would cause the probability density to increase discontinuously at zero. R is a free download that you can get running very quickly. I think it's possible to do the integral if you needed a formula.



Output:

> m = pmax(a,b)[!(a<0 | b<0)]
> sum(m)/length(m)
[1] 26.71047
>
> m = pmax(a,b,c)[!(a<0 | b<0 |c<0)]
> sum(m)/length(m)
[1] 27.1327

Sorry about the terminology screw up, you're of course correct that if the values can't be negative it's no longer a normal distribution. I obviously wasn't thinking about how the second part of my answer affected the first.

Thank you very much for both turning me onto R and providing me with the code to run the sims. That will serve my purposes pretty well.

Being able to solve the integrals would be good, but really, if I didn't have a way to get it into Excel to do some the calculations in batches, it wouldn't do me too much good over and above what I can do with R.

Thanks again for your help Bruce.