[mark49]

Could anyone tell me whether it is possible to work out the probability of this?

The chance of A returning a higher value than B if I know what they return on average.

So if A averages 250
and B averages 200

what percentage of the time will A be greater than B

Sorry if that makes no sense, thanks for any help.

[FourLengthsClear]

Do A and B have a minimum/maximum value? Or is there an expected range?

Does this have anything to do with one-day cricket?

[mark49]

no nothing to do with cricket, I was thinking of passing yards in the nfl and how likely one QB would be to throw for more yards than another.

[YouMama]

take the average, go back each game and calculate how how they did in relation to that average and compare the overlap

[mark49]

would the standard deviation help? I can work those out easily enough.

[YouMama]

its what i would do, its the only way I would think it could be calculated ... I dont have much math/stat knowledge, I just use common sense and problem solving ... but i would think u would need a high/low value to figure out the likely hood of something that averages 250 dropping below 200... cuz u can average 250 and never drop below 200 ... but all this is not useful in my opinion, unless u are using it as part of a whole of something else ... cuz u have to take into consideration other factors like defense faced and the performance of that defense in relation to it average, performance of the WRs as opposed to their average ... weather factors, pass/run play calling balance, etc...

bet on consistency bet against volatility

[samserif]

What you're looking at is the difference of two distributions. This is itself a distribution, with a mean of 50 (since E(A-B) = E(A)-E(B) = 50). It just so happens that Var(A-B) = Var(A) + Var(B). (I'm assuming that these QBs operate independently.) So now you've got a distribution centered at 50 with a variance of Var(A)+Var(B). What you're interested in is the percentage of this distribution that's negative -- in other words, when B > A (and A - B < 0).

To see that this makes sense, imagine two QBs with zero variance: A always throws for 250, and B always throws for 200. The difference of the distributions will never be negative; in fact, it will be the point 50, which makes sense since A will always throw for more than B. If A and B have slight variance, there's a slightly greater chance that B will out-throw A at some point, which again makes sense.

If you can assume that passing yards per game is normally distributed (and this is probably a bad assumption!), you can figure out the percentage of the time that this difference distribution is negative: just use your favorite stat tools to see what percentage of a normal distribution with your given variance is less than -50. (Imagine moving the graph of this difference distribution, centered at 50, to zero, then seeing how much of the curve is to the left of -50.)

And what if passing yards aren't normally distributed? Hell if I know, ha.

[mark49]

Thanks, i will have a go with that. much appreciated.