Expected value of a dice Game with 18 rolls at most - land on a 6 to lose it all, anything else wins you the dice value

Question

I've been given a question with a game: '' You accumulate points. you start with a point value of 0, and you roll the die at most 18 times. every time you roll the die you win as much points the number shown, unless you rolled a 6 in which case you lose it all and the game ends. what's the expected number of points at the end of the game? '' I'm honestly quite stumped - I tried reasoning as follows: X = dice i's result, W = total wins. my attempt: $$E( \text{pts in turn i} | X_i \neq 6) = \frac{1+2+3+4+5}{6} = \frac{15}{6}$$ We can at most play 18 turns, and the dice rolls are independent. We can compute E to be: $$E[W| U_{i=1}^{18} X_i \neq 6] = 18 \cdot \frac{15}{6} $$ $$E[W] = E[W| U_{i=1}^{18} X_i \neq 6]P( U_{i=1}^{18} X_i \neq 6) + E[W| (U_{i=1}^{18} X_i \neq 6)^C]P(( U_{i=1}^18 X_i \neq 6)^C)$$ so that should equal $$E[W] = 18 \cdot \frac{15}{6} \left(\frac{5}{6}\right)^{18}$$ which isn't the case - I'm unsure as to whether my formula is correct (couldn't find it) or if I'm thinking about this problem properly.

Answer 1

You roll 18 times (just continue after a six, even though you lost). The probability of no six is $(5/6)^{18}$. If there is no six the expected number of eyes is three per roll, or 54 in all 18 rounds. Your expected winnings are

$54 \cdot (5/6)^{18}$ ≈ 2.028

Answer 2

Actually there is a small mistake in your answer! This problem can be couched within the probability space $\{(1,...,6)^{18}\}$ equipped with uniform measure, so that each outcome has probability $1/6^{18}$. We aim to calculate the expected value of the random variable $$ f(x_1,...,x_{18}) = \left(\sum_{j=1}^{18}x_j\right) {1}\{{x_1\ne 6,...,x_{18}\ne 6}\} $$

The expected value of $f$ is $$ \sum_{i_1=1}^5 \cdots \sum_{i_{18}=1}^5 \frac{i_1+\cdots + i_{18}}{6^{18}}= \frac{5^{17}\times 15\times 18}{6^{18}} = (5/6)^{17} \times 18 \times \frac{15}{6}. $$ I have confirmed via simulation that this appears correct.

I believe the problem with your calculation is that you need to calculate the conditional mean of the die roll given that the roll is not a 6. This is not equal to 15/6 as in your calculation.