Hang on, I think I figured it out myself.

I drew a big tree, and made these formulas as result:

p = A win = A + 20, B - 20

q = B win = A - 20, B + 20

100 / 20 = 5

60 / 20 = 3

...

p^3 + (p^4 x q) + (p^5 x q^2) + (p^6 x q^3) ...infinite series... = .5

q^5 + (q^6 x p) + (q^7 x p^2) + (q^8 x p^3) ...infinite series... = .5

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p^3 (1 + pq + (pq)^2 + (pq)^3 + (pq)^4 ...infinite series...) = .5

q^5 (1 + pq + (pq)^2 + (pq)^3 + (pq)^4 ...infinite series...) = .5

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a/(1-r)

...

p^3 x (1 / (1 - pq) = .5

q^5 x (1 / (1 - pq) = .5

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p^3 = .5 - .5pq

q^5 = .5 - .5pq

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p^3 - q^5 = 0

p^3 = q^5

p + q = 1

q = 1 - p

p^3 = (1 - p)^5

solve for p = .4123201971

solve for q = .5876798029

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So, for it to be a fair bet between player A ($100) and player B ($60):

P(player A wins) = .41 (2dp)

P(player B wins) = .59 (2dp)

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It looks done to me, but I'm not sure.