FRIDAY'S KILLER QUESTION: You start with no money and play a game where you throw a dice over and over again....

If one of my entry or associate level quants wrote down that answer, I'd give him a right and proper telling-off in private. Its good enough to be able to do sums at primary and secondary school level, but at this level you should be able to explain your answer clearly. Your answer should make sense to a reader familiar with the subject.
I've rarely seen a well presented answer to these Friday quant quizzes or whatever you call them. What is 'S'? 'You can't get to 20' - are we talking about here? Expected winnings or the actual winnings, which is a random variable? Proof read your answer and make sure that it makes sense.
Its great if you have a PhD in Stochastic Filtering of Signals from Outer Space, but you need to make sense to my traders!

Urg, got carried away with expected value formulas. I forgot that each throw assumes you have not rolled a 6 up to that point, thus:

You keep rolling until you have 15 or more, then you stop.

e.g. you roll a 5, then a 5, then a 5. You have 15, the expected value of the next roll is 0, so you stop.

So calculating an n for the number of rolls is not possible, n is not a fixed number. It depends on how lucky you are :)

.n is number of shots
E(Xn)=E(Xn | no 6 after n launches)P(no 6 after n launches) + E(Xn | at least one 6 after n launches)P(at least one 6 after n launches) =
= 3n(5/6)&2+0
Maximized at n =5 and 6 (value of 6.03). 6 gives also 6.03 but has more return with same risk -> I choose 5

An addendum to my previous submission ... this is basically a "russian roulette" problem with the gun replaced with a die. And there are several papers on this online. However, in the original problem there are two variations: in option one the chamber is not spun after each pull of trigger, meaning the probability of getting shot increases with each pull. in option two the chamber is spun each time meaning the prob of getting shot is exactly the same each time. this is option 2 with the added twist of a varying payout.

Personally i would assess my option depending on each turn. I agree with others that the average payout would be 3 a pop so, as said bfor if 5 came up on 1st turn i'd stop. If not i'd continue until i hit 3, (i.e. no more than three turns). Note tho, that of those that specify at least three turns 50% will lose all their money... of those that suggest 5 turns 80% will lose their money (given a large enough sample set!!!).

you should play the game as long as the expected result is positive.
so BC is right about the 15
when you have more than 15, the return of the game is not worth the risk, so you should stop playing
when you have less than 15, you should always play
when you have exactly 15, the expected return is 0, so statistically it does not matter if you play or not giving it is fair dice
but the fair dice is not a given condotion , and again statistically if you reached this far (won 15), the dice is likely to be biased in favour of the player ( less 1/6 chance to hit a 6)
so you can still play

the expected win of this whole game is 6.15.