E(Peace, A) = 1 * P(A>B) * P (Peace, B) + 2 * P(A > B) * P(War, B) + 1 * P(A<B) * P(Peace, B)
= (1-A) * Pp + 2(1-A)*(1-Pp) + A*(Pp) = Pp - APp +2 -2A - 2Pp + 2APp +APp = (2 -2A - Pp +2 APp)
E(War, A) = 2 P(A>B)(Pp) + 2 P(A>B) (Pw) = 2P(A>B) = 2 - 2A
A should optimally declare war only when the expected value of war is greater than the expected value of peace
2 - 2A > 2 - 2 A - P + 2 A P
0 > 2AP - P
2A - 1 < 0 (probabilities can't be negative)
A < 0.5
if the value of winning is 5
Ep = (1-A)P + 5(1-A)(1-P) + A(P) = P - AP + 5 -5A - 5P + 5AP + AP = 5 - 5A - 4P + 5AP
Ew = 5 - 5A
Ew > Ep
0 > -4P + 5AP
5a < 4
A < 0.8
if B has already declared war, eliminate the P(p,B) elements
E(Peace, A) = 2 * P(A > B) * P(War, B) = 2(1-A) = 2 -2A
E(War, A) = 2 * P(A>B) = 2 - 2A
Ew is equal to Ep
A is going to war, it's irrelevant whether or not it declares war.
if B has already declared peace, elminate P(w,B)
E(Peace, A) = 1 * P(A>B) * P (Peace, B) + 1 * P(A<B) * P(Peace, B)
= (1-A)P + (A)P = P = 1
E(War, A) = 2 P(A>B)(Pp) + 2 P(A>B) (Pw) = 2P(A>B) = 2 - 2A
so Ew > Ep only when 2 - 2A > 1
2A < 1
A < 0.5
edit: last one's weird, I'd double check it but I'm bored now.
double edit: just noticed a bunch of flipped >. apparently they're all weird, so either I'm completely wrong or the likely weaker one should attack.
triple edit, just to verify
expected value of A declaring war given A has 0.25 army would be
if A wins, 2 trillion, if A loses, 0
A has a 0.25 chance of winning, so 0.5
expected value of A declaring peace given A has 0.25 army would be
if B is also at peace, 1 trillion, if B declares war, and A wins, 2 trillion, if B declares war, and A loses, 0
1 trillion * Probability B declares peace + (2 (0.25) + 0 (0.75)) * Probability B declares war
P + 0.5(1-P) = 0.5P + 0.5
Ew = 0.5
Ep = 0.5(1+P)
sooo Ep > Ew...
yep, I was completely wrong there, something went wrong somewhere because keyboard typing a bunch of functions is the most annoying thing in the world
quardruple edit: real solution on next page