Game of draughts, expected value of first move advantage

Solution 1:

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.