# "Fair Game" problem

#### testyou

##### New Member
Hello, I'm new here. I joined in the hope that someone would be able to help me with this problem:

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Players A and B play a fair game for $20: the winner takes$20 from the loser.
They play until one of them has lost all of their money.
Player A starts with $100 and Player B starts with$60.
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1.
What is the probability that A wins all of B's money?

This is simple:
A/(A+B) = 100/160 = 62.5%

What's proving difficult is the second part.

2.
It was described as a "fair game" because A and B had the same probability of winning each single game.
However, A has an advantage over B by having more money with which to play.
To make this really fair, so that A and B would have the same probability of winning all of the other player's money:
what should A's probability of winning for each single game be?

I really have no idea how to do this without building some sort of infinite probability tree.

My first thought was:

100(x) = 60(1-x)
x = .375

But this is wrong.

I have no idea how to do this, and I hope that someone here will be able to shed some light on it.

#### testyou

##### New Member
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.