Spartacus

Ok. I have four numbers. We are discussing this from the POV of the defender because the question we are asking is “if I take 2 hits, which die should I take them off”?

chanceOfThisToInv is the probability that in the current situation, the attacker will score zero hits on me.

chanceOfInvToThis is the probability that in the reverse of this situation (I scored no hits so now it’s the attacker’s go) I will score zero hits in return on my attacker.

chanceOfThisToWinGivenNotLoop is the probability that *if* I do not score no hits, then I will go on to win. To calculate it, we work out for each possible number of hits what our best move is, and then for that move take the inverse of the possibility of winning (because that’s the situation our attcker will then be in), and multiply it by the probability of scoring that number of hits.

We then divide the total by the total chance of not scoring no hits.

chanceOfInvToLoseGivenNotLoop is calculated correspondingly.

Now, what’s my win probability?

To figure that out, I will have to sumultaneously figure out winProb and invLoseProb, I think, because they depend on one another.

First, let’s shrink the size of those variable names

ti = this to inv
it = inv to this
rw = raw win pobability, ignoring the loop
ril = raw invers lose probability, ignoring the loop.

wp = win probaibility
ilp = inverse lose probability

wp = (1-ti) rw + ti ilp
ilp = (1-it) ril + it wp

Ahh crapsicks. I hate these. Actually, no I dont, it’s just been a while since I did one.

wp = (1-ti) rw + ti [ (1-it) ril + it wp ] 
wp = (1-ti) rw + ti (1-it) ril + ti it wp
wp -  ti it wp = (1-ti) rw + ti (1-it) ril
wp  = [ (1-ti) rw + ti (1-it) ril ] / (1 - ti it)

ilp = (1-it) ril + it [ (1-ti) rw + ti ilp ]
ilp = (1-it) ril + it (1-ti) rw + it ti ilp 
ilp - it ti ilp  = (1-it) ril + it (1-ti) rw
ilp = [ (1-it) ril + it (1-ti) rw ] / ( 1 - it ti) 

So.

wp  = [ (1-ti) rw + ti (1-it) ril ] / (1 - ti it)
ilp = [ (1-it) ril + it (1-ti) rw ] / ( 1 - it ti) 

Hmm, some common terms there. Rearranging:

wp  = [ ti (1-it) ril +    (1-ti) rw ] / (1 - it ti)
ilp = [    (1-it) ril + it (1-ti) rw ] / ( 1 - it ti) 

Damn! Look at all those multiplications of ril by (1-it)! Recall that we divided by 1-it to get to where we are. Maybe we can just use what we already have? I’ll call them ‘simple win’ and ‘simple inverse lose’

wp  = [ ti sil +    sw ] /  (1 - it ti)
ilp = [    sil + it sw ] / ( 1 - it ti) 

Weird! Can it really be that simple?

On a 1, I lose; on a 2-3, I win; on a 4-0 I pass the dice to you. .1, .2, .7
On a 1-5, you lose. On a 6, you win. On a 7-0, you pass the dice to me. .5, .1, .4

wp  = [ .7 .5 +    .2 ] /  (1 - .28) = .55 / .72 = .763
ilp = [    .5 + .4 .2 ] / ( 1 - .28) = .58 / .72 = .806

The numbers are different because they are in the form of expressing my chance of winning given who is currently holding the dice.

Now, let’s describe the game from the inverse POV.

On a 1-5, I lose. On a 6, I win. On a 7-0, I pass the dice to yoy. .5, .1, .4
On a 1, You lose; on a 2-3, You win; on a 4-0 you pass the dice to me. .1, .2, .7

wp  = [ .4 .1 +    .1 ] /  (1 - .28) = .14 / .72 = 
ilp = [    .1 + .7 .1 ] / ( 1 - .28) = .17 / .72 =

And you can see that .14+.58 is .72 and .17+.55 is also .72.

I feel good. That’s how I’ll calculate it.