Very sorry about the end of my last comment: I used a “less than” sign and a “greater than” sign, and it took it as formatting. I can’t seem to edit my comment. Here’s what I meant to say:

…But then I believe it turns out that:

R(3600) = 1 (for the moment, let’s forget about overtime games), and
R(t) = min( W(t), R(t+1) ), for all t less than 3600.

So you can start at t=3600, and proceed backward: at every time t, the value of R(t) is the smaller of W(t) and R(t+1). You can also do your “integration” from right to left, since addition is commutative! So as you find each value of R(t), add it to your running total. What follows is some pseudo code:

total = 0
t = 3600
r = 1
while t is greater than 0:
{
total = total + r
t = t-1
r = min(w[t], r)
}
total = total / 3600.0

Again, sorry I screwed up the comment. Thanks for an interesting discussion!

Hi,

Hope I’m not too late to the discussion. Dan, I like the idea of your RDWP function (in my head I’m pronouncing it R-Dip).

Doug, I don’t think it would be hard to implement this idea. First, when Dan (earlier Dan, not me) says “integrate”, he really just means “average”. I think the thing to do is just break the game into 3600 one-second intervals, and then measure the RDWP function at the end of each interval. That is, find RDWP after the first second, the second second, etc….or in other words, find RDWP at the 14:59 mark of the first quarter, the 14:58 mark of the first quarter, and so on, up to the 0:00 mark of the fourth quarter.

Now just add all 3600 of these values together, and divide by 3600 to get an average RDWP over the entire game. A computer could easily add 3600 values together and divide by 3600, so that’s not a problem…but how do you calculate the RDWP function at each of these points?

Well, let’s let W(t) be the win probability at time t, and let R(t) be the RDWP function at time t. Then R(t) is the minimum of W(u), over all u greater than or equal to t. But then I believe it turns out that:

R(3600) = 1 (for the moment, let’s forget about overtime games), and
R(t) = min( W(t), R(t+1) ), for all t 0:
total = total + r
t = t-1
r = min(w[t], r)
total = total / 3600.0

It seems to me that overtime games could be handled similarly, but you’d probably want to start at a higher t-value (i.e., whenever the game actually ended), and divide by that same number at the end, rather than 3600. Of course, I suspect that most overtime games will probably fit into the “had any number of things gone differently at the end, the winning team would have lost” category.