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:
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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?
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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