Probability flip 2 successive heads will take longer than 3 successive heads Suppose person A tosses a fair coin until he gets 2 successive heads. Let $X$ denote the number of flips A makes. Suppose person B tosses a fair coin until he gets 3 successive heads. Let $Y$ denote the number of flips B makes. What is $Pr(X>Y)$?
The brute force computation is $\sum_{n=4}^\infty Pr(X=n)P(Y<n)$, but I'm unsure how to evaluate this. As $n$ gets larger, the computation seems to be a pain.
Ways to generalize this to any number of successive heads?
Thanks!
 A: Let $p[x,y]$ be the probability of the event $X > Y$, given that player $A$ already has $x$ consecutive heads, and player $B$ already has $y$ consecutive heads.

Our goal is to find $p[0,0]$.

From any given state $[x,y]$, the next roll has $4$ equally likely outcomes:
$$(T,H),\;\;(T,T),\;\;(H,H),\;\;(H,T)$$
where the ordered pair indicates the flip results for $A,B$, respectively.

Using the $4$ outcomes in the order shown above,  we get the equations
\begin{align*}
p[0,0]&={\small{\frac{1}{4}}}\bigl(p[0,1]+p[0,0]+p[1,1]+p[1,0]\bigr)\\[4pt]
p[0,1]&={\small{\frac{1}{4}}}\bigl(p[0,2]+p[0,0]+p[1,2]+p[1,0]\bigr)\\[4pt]
p[0,2]&={\small{\frac{1}{4}}}\bigl(1+p[0,0]+1+p[1,0]\bigr)\\[4pt]
p[1,0]&={\small{\frac{1}{4}}}\bigl(p[0,1]+p[0,0]+0 + 0\bigr)\\[4pt]
p[1,1]&={\small{\frac{1}{4}}}\bigl(p[0,2]+p[0,0]+0 + 0\bigr)\\[4pt]
p[1,2]&={\small{\frac{1}{4}}}\bigl(1+p[0,0]+0 + 0\bigr)\\[4pt]
\end{align*}
Thus, we have a system of $6$ linear equations in $6$ unknowns.

Solving the system, we get
$$p[0,0] = \frac{361}{1699} \approx .212478$$
