Expectation of number of equal points in tennis game Two people play a tennis game up to $7$ points (first to $7$ wins), with equal probability $\frac12$ of winning each point. What is the expected number of times in the game where the score of the two players are equal (not including $(0,0)$? 
I am having trouble enumerating the paths honestly. Should I focus on trying to enumerate the paths in $\mathbb{Z}^2$ starting from $(0,0)$ up to the top and right walls of a side of a $7\times7$ square with its bottom left corner at the origin (excluding the paths where the endpoint is $(7,7)$ since that's not a possible score) to see how many times they hit $(i,i)$? This seems extremely cumbersome, and I want to know if there's an organized way of looking at it. Anything will help, thank you. 
 A: I am really not sure if this is right, but see if it helps anyhow. Just to have an ideia of what is going on write the first cases. $n$ represent the number of points played, and we are interested in $n$ going from $1$ to $13$
\begin{align*}
n=1:\quad &(1,0)(0,1)\\
n=2:\quad &(2,0)(1,1)(0,2)\\
n=3:\quad &(3,0)(2,1)(1,2)(0,3)\\
n=4:\quad &(4,0)(3,1)(2,2)(1,3)(0,4)\\
&\vdots\\
n=13 \quad &(13,0)(12,1)\cdots (1,12)(0,13)
\end{align*}
We are only interested in the even cases. Suppose $A$ represents a point to the first player and $B$ a point to the second player, then the probability of $(1,1)$ happening will be $\frac{1}{2^2}$ times all the cases that lead to this result, that is, all permutations with repetition of $AB$, which gives $\frac{1}{2^2}2!\,$. The probability of $(2,2)$ will be $\frac{1}{2^4}$ times all permutations with repetition of $AABB$ which gives $\frac{1}{2^4}\frac{4!}{2!2!}$. In general the probability of $(\frac{n}{2},\frac{n}{2})$ happening when you make $n$ plays is $\frac{n!}{\frac{n}{2}! \, \frac{n}{2}! \, 2^n}$
So you can take the sum of all of them 
$$
I = \{2,4,6,8,10,12\}, \qquad \sum\limits_{n\in I} \frac{n!}{\frac{n}{2}! \, \frac{n}{2}! \, 2^n}
$$
Which is approximately $1.93$
A: Here's one method (not sure if it's best). Each game can be a path from $(0,0)$ to $(7,7)$, where the game itself stops after hitting the $7$ wall (once you hit the wall, the rest of the path to $(7,7)$ is determined anyway, so adding on these path segments doesn't hurt). Counting paths from $(0,0)$ to $(7,7)$ is a well-known combinatorial problem. Then the probability of hitting $(i,i)$ ($0<i<7$) is gotten by counting paths from $(0,0)$ to $(i,i)$ to $(7,7)$. Finally, as rikhavshah pointed out in the comments, linearity of expectation tells us we can add these six probabilities to get the answer.
Edit: I'll expound a bit on how linearity of expectation applies here. Say the random variable $Y$ counts number of times the score is equal -- i.e., the number of spaces $(i,i)$ the path passes through ($1 \leq i \leq 6$). (So $Y$ can take any integer value between $0$ and $6$ inclusive.) The question asks for $\mathbb E[Y]$. Now we can express $Y$ as a sum of random variables $$ Y = \sum_{i=1}^6 I_i$$ where $I_i$ is the indicator random variable for the path passing through $(i,i)$ (i.e., it takes on the value $1$ if the path passes through $(i,i)$ and $0$ if it does not). Now linearity of expectation tells us 
$$\mathbb E[Y] = \sum_{i=1}^6 \mathbb E[I_i]$$ and this result holds regardless of any dependence relations between the $I_i$s. The expectation of an indicator random variable is a probability; i.e., $\mathbb E[I_i] = \mathbb P[A_i]$ where $A_i$ is the event that the path passes through $I_i$. Thus $$\mathbb E[Y] = \sum_{i=1}^6 \mathbb P[A_i]$$
