In a casino a player can win 1 euro with a probability of 18/28 and loses 1 euro with a probability of 20/38.

Question

In a casino, for each bet on the wheel, a player wins 1 euro with the probability of $\frac{18}{38}$ and loses 1 euro with a probability of $\frac{20}{38}$.

a) What is the average value won per game?

b) What is the probability the player loses money if he plays 6 times?

Not sure how to solve either. Not sure how I can calculate the average without knowing how may games he player, and for b) I tried to use the binomial distribution with $x = 0$ and $p = \frac{18}{38}$ but didn't work. Help?

The right answers are $\frac{1}{19}$ for a) and $0.394296$ for b).

Answer 1

For the first part, you are asked for the expected winnings per game; that is to say, after one play. If $X$ is a random variable representing the winnings after one game, then either $X = 1$ with probability $18/38$, or $X = -1$ with probability $20/38$, where we use $-1$ to represent a loss of $1$ euro. We can express this as $$\Pr[X = 1] = \frac{18}{38} \\ \Pr[X = -1] = \frac{20}{38}.$$ The expected value of $X$ after one play is therefore $$\operatorname{E}[X] = (1) \Pr[X = 1] + (-1)\Pr[X = -1].$$ All we did is take each possible outcome ($1$ or $-1$), multiply it by the probability of observing that outcome, and then taking the sum.

For the second part, the player loses money in $n = 6$ plays if they have at least 4 losses, because if they win $3$ times and lose $3$ times, their net earnings is $0$. So they have to lose more times than they win. If $Y$ is a binomial random variable that represents the number of wins in $6$ tries, where the probability of any try resulting in a win is $p = 18/38$, then $$\Pr[Y = y] = \binom{n}{y} p^y (1-p)^{n-y}.$$ You then must find $$\Pr[Y \le 2] = \Pr[Y = 0] + \Pr[Y = 1] + \Pr[Y = 2],$$ representing the sum of probabilities for all outcomes in which there are fewer wins than losses.

Answer 2

For question (a), it's asking for the expected average value per round, which is the same thing as the expected value or the mean of a random variable. In case of a discrete random variable $X$ with finitely many possible values, you find the mean or the expected value as $$\mu_X=\mathbf{E}(X)=\sum xp(x),$$ where the summation runs over all possible values $x$ of this random variable. In your case, the only possible values are $x=1$ (winning) and $x=-1$ (losing).

For question (b), note that to lose money the player needs to lose more games than he won. Out of six games played, this would happen if $4$ or more are lost, or equivalently if $2$ or fewer are won. You're absolutely correct in deciding to use the binomial distribution here, but what you calculated is not the answer to the question. Say, if you focus on the games won, then you need to use the binomial distribution to find the probability $\mathbf{P}\{Y\leqslant2\}$, where $Y$ represents the number of games won.