Hi, I think you guys are right about needing to condition on the first couple tosses then applying recursion. I looked at a simpler case, let X be the number of flips for 1 head and Y be the number of flips for 2 heads. (I'll apologize for not knowing how to use TeX.)
We condition the results of the first toss of each sequence, so we have four cases for the set of first tosses: {H,H}, {H,T}, {T,H}, {T,T}.
Note: {T,H} means the sequence where we try to get 1 head got a tails, and the sequence where we try to get 2 heads got a head).
P(X > Y) = P(X > Y | {H, H}) P({H, H}) + P(X > Y | {H, T}) P({H, T}) + P(X > Y | {T, H}) P{T, H}) + P(X > Y | {T, T}) P({T, T})
P(X > Y) = 0 * P({H, H}) + 0 * P ({H, T}) + P (X > Y | {T, H}) P({T, H}) + P(X > Y) P({T, T})
P(X > Y) = P (X > Y) | {T, H})(1/4) + P(X > Y)(1/4)
Now the next step is to get another equation relating P(X > Y) and P (X > Y) | {T, H}), then we can solve them in a system of equations. Let's now condition on the 2nd toss.
Note: P (X > Y) | {T, H}) != 1/2
P (X > Y) | {T, H}) = P (X > Y) | {TH, HH}) (1/4) + P (X > Y) | {TH, HT}) (1/4) + P (X > Y) | {TT, HH}) (1/4) + P (X > Y) | {TT, HT}) (1/4)
P (X > Y) | {T, H}) = 0 * (1/4) + 0 * (1/4) + 1 * (1/4) + P(X>Y) (1/4)
P (X > Y) | {T, H}) = 1/4 + P(X>Y) (1/4)
Now we have two equations and two unknowns, and if we solve for P (X > Y) we get .0909.