Here's a question from my probability textbook:
$A$ and $B$ play at draughts and bet $\$1$ even on every game, the odds are $\mu : 1$ in favor of the player who has the first move. If it be agreed that the winner of each game have the first move in the next, show that the advantage of having the first move in the first game is $\$\mu-1$.
Here's the answer in the back of my book:
It is assumed that they go on playing indefinitely.
Let $E$ be the expectation of the first player; then since what one gains the other loses, the second player's expectation is $-E$.
At the first game the first player either wins $\$1$ and remains first player or he loses $\$1$ and becomes second player. And the chances of these two events are as $\mu:1$, so$$E = {{\mu(1 + E) + (-1 - E)}\over{\mu + 1}} = {{\mu - 1}\over{\mu + 1}}(1 + E);$$whence$$E = {{\mu - 1}\over2}, \quad -E = -{{\mu - 1}\over2};$$and the first player's advantage is $\mu - 1$ dollars.
I understand where the ${{\mu}\over{\mu + 1}}(1 + E)$ term comes from. But I don't see where the term ${1\over{\mu + 1}}(-1 - E)$ comes from, specifically the $-1 - E$ part. Can anybody help me see why? How does that term encode the switch to becoming second player?
Player $1$ loses the first game with probability $\frac{1}{\mu+1}$. If this happens, then player $1$ automatically loses the one dollar from the first bet.
Because each of the games are independent of each other, and player $1$ now gets the second move (for game $2$), in effect player $1$ has just swapped roles with player $2$. So starting from game $2$, player $1$'s EV is $-E$.
Together with the one dollar that player $1$ already lost in the first game, this means that player $1$'s total EV in the case that they lose the first game is $-1-E$.
So multiplying this by the probability of losing the first game gives you the $(-1-E)/(\mu+1)$ term.