Better complete this now seeing it's Census night.

I was hoping more would have tried this.

If Mad divides one more of his rows by 4 and sums, then he should get 18.75%. Which is what ATC and I get.

Anyone wanting to figure out how to try such problems should look up Probability Trees. Mad seemed to be using them without knowing about them, but he forgot about the initial branches.

But there seems to be a more powerful way of solving the problem (in one spreadsheet cell!) for any number of remaining Rounds and Handicaps!

If you look at how many more wins B gets than A then that seems to be a Binomial Distribution for 4 games.

n B-A
1 2
4 1
6 0
4 -1
1 -2

16 Total

Apply the 1 game handicap to B and

Probability = 1/16 + (4/2)/16 = 3/16 = 18.75%

Anyone prepared to try something akin to the current situation with Freo and Geelong?

4 Rounds
2 Handicap for Geelong


I think for now ATC should disqualify himself pending a testosterone test.