The "Successive Wins" problem in Mosteller's book 50 challenging problems in probability has a second answer that is often overlooked. The problem is as follows. To encourage her daughter, Sara, her mom offers her a grant money if she wins (at least) two tennis sets in a row in a three set series to be played against her mom and the champ of the tennis club alternately. Mom-Champ-Mom or Champ-Mom-Champ, according Sara's choice. The cham is a better player than her mom. Which series should Sara choose?
We denote the probability of winning a set against mom, is $m$, whereas the probability of winning a set against the Champ is $c.$
The second answer is as follows.
We have a tendency to suppose that the higher the expected number of sucesses, the higher the probability of winning a prize, and often this supposition is useful. In our problem the expected number of wins under Champ-Mom-Champ is 2c+m, which is less than the expected number of wins for Mom-Champ-Mom, 2m +c.
My question is why is the expected number of wins 2c+m for Champ-Mom-Champ ... how does Mosteller come up with this, and why is the expected number of wins for Mom-Champ-Mom is 2m+c?
It can't be as simple as Champ-Mom-Champ = c+m+c = 2c + m (?! How does this even follow as expected number of wins), and analogously for Mom-Champ-Mom = m+c+m = 2m+c. (This can't be right.)
So, how do we proceed on this?
Explanation of expected number of wins
The solution claims that the expected number of wins for the MCM pattern is $2m+c$, and the expected number of wins for the CMC pattern is $2c+m$. This is a consequence of linearity of expectation.
Let's focus on the MCM pattern for now. Let $X_1$ be a random variable which is equal to $1$ if Sara wins her first game, and $0$ if she loses. Similarly, define $X_2$ and $X_3$ to be $0$ or $1$ based on whether Sara wins her second and third games. This means the total number of wins is $X_1+X_2+X_3$. Therefore, $$ \text{expected # of wins}=E[X_1+X_2+X_3]=E[X_1]+E[X_2]+E[X_3] $$ We can distribute the $E$ over addition, because this is one of the most basic properties of expected value. All that remains is to compute $E[X_i]$, for $i\in \{1,2,3\}$. This is easy because each random variable can only take the values $0$ or $1$. For example, $$ E[X_1]=0\cdot P(X_1=0)+1\cdot P(X_1=1)=0\cdot (1-m)+1\cdot m=m $$ Similarly, $E[X_2]=c$ and $E[X_3]=m$. Putting this altogether, the expected number wins is $2m+c$ for the MCM pattern, as they claimed.
Why expected number of wins is not important
In order to win two games in a row, Sara needs to win the middle game, and must win one of the other two games as well. Since the middle game is so crucial, it is better to have the middle game be as easy to win as possible. Therefore, the preferred schedule is Champ, Mom, Champ.
It may be surprising that Sara would prefer playing more matches against the more skilled player. And it is true that Sara will win more matches, on average, if Sara chooses Mom-Champ-Mom. However, with the MCM setup, it is likely that Sara will get the pattern Win-Lose-Win, which is a lot of wins, but which is useless towards Sara's goal. Basically, expected value is useless because this only tells us the average number of wins, but the number of wins does not necessarily correlate with getting two wins in a row.
Rigorous solution
Let $x$ be the probability of winning against the first opponent (so $x$ is the odds of winning aginst either mom or champ), and let $y$ be the probability of winning against the second opponent ($y$ is whatever $x$ is not). In order to win two games in a row, Sara must win the middle match, and then must not lose both first and last matches. Therefore, $$ P(\text{Sara wins two in a row})=y\cdot (1-(1-x)^2)=yx(2-x) $$ Now, let $m$ be the probability Sara wins against Mom, and let $c$ be the probability Sara wins against the Champ. In the case where Sara plays Mom-Champ-Mom, the probability is $c\cdot m(2-m)$ and in the case of Champ-Mom-Champ, the probabilty is $m\cdot c(2-c)$. Since $c>m$, it is clear that $cm(2-m)>cm(2-c)$, which proves that Champ-Mom-Champ is preferable.