Net gain on dice rolling

Question

Suppose you have to pay $15$ to roll a dice repeatedly until the first $6$ comes up. Then you will get paid as many dollars as the square of the no of rolls $X^2$.

I have a equation as follows: $$G = \text{Gain} = X^2 -15 \\ E(G) = E(X^2 - 15) = E(X^2) -15 $$

Since this is a geometric distribution I found the mean to be $E(X) = 1/p = 1/(1/6) = 6$ and variance $= (1-1/6)/(1/6)^2 = 30$

Therefore I found $E(X^2)$ to be $66$ using $V(X) = E(X^2) − (EX)^2$

The overall net gain was $66-15 = 51$

I am just wondering whether I am on the right track. I thought the net gain would be very low for this $Q$ as you need to roll at least 4+ times to get a gain.

Also can someone direct me how to find the probability that the users net gain will be positive. I was thinking of calculating the probability of rolling at least 4 times.

Answer 1

Your calculation of $E(X^2)$ is absolutely correct, it indeed equals $66$. You indeed need to be able to roll the die at least four times to make profit, which means that you need to not throw a $6$ three times in a row. The probability of not throwing a $6$ on a given roll equals $\frac{5}{6}$, so the probability $P(X > 3)$ of making a profit equals:

$$P(X > 3) = \left(\frac{5}{6}\right)^3 = \frac{125}{216} \approx 0.579$$

Answer 2

You are on the right track. Intuitively a large reward is plausible because of the squared increase in rewards. Very unlikely events have a huge reward which in this case outweigh the small losses.

Regarding your last question: Calculating the probability of rolling at least 4 times is correct, however, rather difficult. It is easier to calculate the probability of a loss and then using $P(\text{gain}) = 1 - P(\text{loss})$ since we know the two probabilities must sum up to 1.

(I get $P(\text{gain}) \approx 0.58$)