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

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 $5-4p$.

• How do you get $E(X)$? How can we say if you approach is correct if you don´t tell us your thoughts and/or show your calculation. – callculus Jul 29 '18 at 15:40
• After your edit I changed my downvote into an upvote. – callculus Jul 29 '18 at 15:53
• In your second edit, the probability distribution you've defined has probabilities that, in general, do not add up to one. You may wish to consult the Wikipedia article I link to in my answer below. – Theoretical Economist Jul 29 '18 at 16:24

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

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

• Ok, so since after the fourth game, the expected number of losses is 1, I get that now. But how is the number of losses in first four games $4(1-p)$? – barnaby-b Jul 29 '18 at 16:04
• @barnaby-b This is a property of the binomial distribution, which I linked to in my edit. Alternatively, the expected number of losses from a single game is just the probability of losing, which is $1-p$. Since expectation is linear, the expected number of losses in four games is $4(1-p)$. – Theoretical Economist Jul 29 '18 at 16:16

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.