Exercise :
A casino slot machine has been programmed so as to offer total winnings $X$ euros per hour, regardless of other hours, with $μ_x = E[X] = 180$ and $σ_x = V[X] = 10.125$ . From long term data we know that the value $Y$ in euros that players bet per hour (independent of the values that are played at other hours and the winnings values) follows the uniform distribution over the interval $[100,300]$.
(a) Calculate approximately the probability that in the duration of one month the slot machine has yielded total winnings of over $135.000$ euros.
(b) Calculate approximatel the probability that in the duration of one month the clear profit of the casino is over $10.000$ euros.
(c) Let it be that we want to change the parameter $\sigma$ in the program of the slot machine. What is the biggest value of $\sigma$ for which the clear profit that the slot machine yields for the casino is over $10.000$ with probability at least $95%$ ?
(Suppose that the casino operates the 24hrs per day and that every month has 30 days).
Attempt – Question :
(a) Since we are asked to calculate the probability of winnings over the course of one month, we should multiply the variable $X$ by $24 \cdot 30 = 720$. Thus, what we need to find, is :
$$P[720X > 135.000] = 1 – P[720X \leq 135.000] = 1 – P[X \leq 187.5] = 1 – P[X-180 \leq 7.5] = 1 – P\bigg[\frac{X-180}{45\sqrt{5}} \leq \frac{7.5}{45\sqrt{5}}\bigg] = 1 – P\bigg[Z \leq \frac{7.5}{45\sqrt{5}}\bigg]$$
which can be calculated approximately via the $\Phi$ probability function boards.
(b) The clear profit of the casino in the duration of one month is the total value of the bets by the players minus the total winnings, thus $720(Y-X)$. Now, the probability of the clear profit of the casino being over $10.000$ per month is :
$$P[720(Y-X) > 10.000] = 1 – P[720(Y-X) \leq 10.000] = 1 – P[Y-X \leq 125/9$$
Since $Y \sim U(100,300)$, it will be $E[Y] = 200$ and $V[Y] = 10000/3$.
How should I proceed now to calculate the probability above ? How would I form a normal distribution simulation for $2$ variables this time, which follow different distributions ?
(c) For the probability of the casino earnings (total profit) being more than $10.000$ with at least $95%$ probability, I assume that something like this is needed :
$$P[720(Y-X) > 10.000] = 0.95$$
but proceeding in this case I have the same issue as before, thus leading to the initial question.
How would I proceed with implementing the parameters of the distributions both the variables $X$ and $Y$ into a normal distribution simulation fraction in order to produce an approximate probability ?
Also, is my approach for part (a) correct ?
Best Answer
For (b)
The Central Limit Theorem does not "care" for the distributions of $X, Y, Z = 720(Y - X)$. They just have to be independent and identically distributed.
By properties of expected value and variance you have:
$$E[Z] = 720 (E[Y] - E[Χ])$$
and since $Cov(X, Y) = 0$
$$V[Z] = 720^2 \times (V[X] + V[Y])$$
From these, you can use the CLT since $\{Z_i\}$ are iid to calculate your result.
Same goes for (c). You will make an equation (in the same manner) with $\sigma$ and solve it (using $\Phi$ tables to use the 1-1 property and solve for $\sigma$).