Playing 4 games, then playing until a loss - expected number of losses

Question

Consider a game where you win with probability $p$. We play 5 such games, and if we win game number 5, we keep playing until we lose.

a) Find the expected number of games played

b) Find the expected number of games lost

Considering the a) part of problem, I defined X to be a random variable that measures number of games, starting at 5 (since we are sure the game is played at least 5 times) and came up with $$E(X) = \frac{5-4p}{1-p}$$ which appears to be correct.

I have no idea how to approach the second part of the problem.

The official solution states that it is the expected number of games played times the probability of loss, i.e. $E(X)\cdot(1-p) = 5-4p$ with no explanation whatsoever. So any sort of intuition or explanation (or an alternative solution) would be greatly appreciated.

$\bf{edit}$ I have arrived at E(X) by defining a discreet random variable with the following distribution $$X\sim \left( \begin{array}{ c c c } 5 & 6 & 7 & ... & k \\ 1 - p & p(1-p) & p^2(1-p) & ... & p^{k-5}(1-p) \end{array} \right) $$ and applying the definition of expectation.

$\bf{edit 2}$ I tried defining a random variable $Y$ which measures the number of losses in first four games. So for 0 losses we must win every game, so the probability is p^4, etc. I arrived at the following distribution: $$X\sim \left( \begin{array}{ c c c } 0 & 1 & 2 & 3 & 4 \\ p^4 & p^3(1-p) & p^2(1-p)^2 & p(1-p)^3 & (1-p)^4 \end{array} \right) $$

However the expectation of this variable doesn't match the solution $5-4p$.

Answer 1

Your expected number of losses is the expected number of losses in the first four games, plus the expected number of losses in all games played after the first four.

The number of losses in the first four games is binomially distributed with "success" probability $1-p$ and four trials. Thus, the expected number of losses in the first four games is $4(1-p)$. The expected number of losses in all games played after the first four is $1$. (Since you stop immediately after a loss.)

Hence, the expected number of losses is

$$ 4(1-p) + 1 = 5 - 4p. $$

Answer 2

Another answer has already derived the expected number of losses in an elementary manner. To derive the same result in the manner in which the official solution derives it, apply the general version of Wald's identity. You have an infinite sequence of (i.i.d.) binary random variables $X_n$ that take the value $1$ for a loss and the value $0$ for a win, as well as a random number $N$ of them that you sum, and they satisfy the premises of the identity. In particular, even though $N$ is not independent of the $X_n$ (as required for the basic version of the identity), we have

$$ \mathsf E\left[X_n1_{\{N\ge n\}}\right]=\mathsf E\left[X_n\right]\mathsf P(N\ge n)\;. $$

Thus, the expected number of losses is the expected number of games played times the probability of a loss.