gallabru, on 01 July 2020 - 07:16 PM, said:

Jylpah could you explain us what the 0-1... results are ?

A loss, the reds had one player left.

gallabru, on 01 July 2020 - 07:16 PM, said:

If we discard draws, supremacy (strong approximation) and last second suicide (7-7), battle results should be N-7 (N=0:6). Then 'steamrolls' should account for 1/7 (counting red crushing just double all numbers) of the results, i.e. 14%. If we assume that supremacy accounts for half the battles and that steamrolls never happen in supremacy (strong approximation again) then 7-0 could account for 7% of the battle results.

At the other end of the statistical spectrum, we could assume that a there are more 4-7 than 14% of the games because there are many ways to get 4 deads. Then one N-7 result can happen 7!/N!(7-N)! times. For N=0, this is 1 again but the number of possible battle result amounts to 114. So, 0.8 % of steamrolls.

Simple(ton) maths for purely random results. But, obviously, we are hoping to have an impact on the result (good or bad) otherwise why play ?

You are interpreting the numbers other way than intended (the numbers denote the number of remaining players per team).

If I understood you correctly you are assuming that each (atomic) outcome (0-5, 0-3) are equally probable and then you are using combinatorics to calculate the probabilities, right (like calculating the probabilities for a sum of several dices)? Your steamroller estimate 0.8% is far from the real data so maybe the assumptions do not hold?

**Edited by jylpah, 01 July 2020 - 05:42 PM.**