Reply: Android: Netrunner:: General:: Re: Netrunner OCTGN Tournament: 1

by kingjames01

hollis wrote:

I'm pretty sure what you've said so far hasn't been in dispute (neither by my probably-incorrect formula, nor FileAccess's).

But I was interested in how to calculate a player's chance of winning when playing a specific side if:

A) You know Player 1's skill relative to player 2's
B) You know Player 1's overall win rate against Player 2 (assume equal number of games played on side A and side B)
C) You know Side A's chance of winning when two equally skilled opponents face off (even if P1 and P2 are not equally skilled themselves)

You're addressing what a player's overall win rate will be, but not what their win rate will be when only playing a specific side of the game. As far as I can tell, your math looks right for the case of overall win rate.

It could be that my question is ill-defined. However, I have not thought of a way to prove that (yet). My guess is it's not ill-defined.

I've actually addressed this but only for the case where:

A) Player 1's skill is equal to player 2's
B) This is irrelevant if both players are equally skilled as the math shows
C) The chances of Faction C winning when two equally skilled players compete is defined as I.

If you only want 1 game using a specific side, then make that number equal to 1 and the other number equal to 0.

For example, suppose you and I are equally matched. You play as Faction C which gives you an intrinsic advantage 70% of the time. Further suppose that we only play 1 match under these conditions.

Then

Win Percentage = E / N
               = E / (N_C + N_R)
               = (N_C * I + N_R * (1 - I)) / (N_C + N_R)
               = (1 * 0.70 + 0 * (1 - 0.70)) / (1 + 0)
               = 0.70
               = 70%

Once I find some more time to continue my proof, I'll express the general relationship.

It's just as simple but the individual terms are modified.