We are playing a game in my AP Stats class called Greed. To play the game, everyone stands up and the teacher rolls the dice. If any number other than a 5 is rolled in the first round, 100 points are scored. The students have the choice to bank their points by sitting down (they are not allowed to stand back up and continue playing after they bank their points), or they can continue playing to possibly earn more points. The dice is rolled again for round two, and if it is any number other than a 5, they score 200 more points (round three: 300 points, round four: 400 points, etc). However, if a 5 is rolled and you chose to stay standing, you lose all the points you have earned that day. We play one game everyday (One game is when we keep rolling until a 5 is rolled or until everyone is sitting down). My teacher says that there is a strategy to this game, that in the long run the results are predictable. What's the strategy? How can I earn the most points by the end of the semester? After what roll should I sit?
I'll outline some things but avoid explicit solution. The point is for you to try to work it out.
At each step, you will have some amount of points, $M$. If you stand for the next roll you will either win $k$ more points (giving you $M+k$) or you will lose $M$ (leaving you with $0$).
The expected value-maximizing strategy would say that any time a $5/6$ chance of $k$ is worth more than a $1/6$ chance of $M$ -- i.e. when the expected gain from playing another round would be positive -- you benefit (on average) from continuing. A little mental arithmetic lets you see when to sit down for a variety of similar games of this type (as long as its not a game that becomes much, much more profitable later, there's no need to sum series, you can just stay standing until the net gain on the next roll is no longer positive).
Imagine, for example, you played the first round and got $100 $points. Now in the second round you would get $200$ points if you win ($5/6$ chance) and lose $100$ points if you got a $5$ ($1/6$ chance). The expected winnings is $200\times 5/6$ and the expected loss is $100 \times 1/6$, for an expected return of $900/6 = 150$; on average you stand to win much more than lose.
However, if you are trying to be the person with the most points of all at the end of many such games of Greed, maximizing expected value each turn is not the optimum strategy to be the eventual winner!
You may be better taking a smaller chance of a larger return, or banking smaller, surer gains, depending on where you are placed relative to other players.
Imagine you're coming second but the person coming first is a fair way ahead of you. If you sit down at the point that would maximize your expected return for that single round, you may simply make it certain you don't win. Similarly if you're far ahead in the last round, you will be better off sitting down much earlier than would maximize your expected return that round -- you may be unnecessarily risking a near-certain win if you continue.
If you were to be playing many more rounds you would (approximately) seek to maximize expected value per turn, but as you get toward the last few rounds the winning strategy changes: if you're behind, you should take more risk, if you're ahead, less risk. When you're getting close to the end, it's not just the current round of points that matter if you want to maximize the chance of being ahead at the end.
You can work out the exact strategy mathematically, but my guess is that the teacher is trying to get you to undertake the simpler task to maximize your expected winnings (which works quite well in the early to mid stages of the iterated game).
You might like to look at articles on strategy for the dice game Pig (of which this is a variant). There's lots of information on strategy to be found for various versions of this game online, but its more fun to work it out in detail yourself. Here is one document that also makes the point about the expectation-maximizing strategy not being optimal. Your game is a little different (your payoff increases and everyone gets the same rolls) but the basic ideas are the same.
At the n-th roll you have a $1/6$-th chance to lose the arithmetic sum of $1$ to $n-1$ worth of points, or a $5/6$-th chance to gain $n$ points. Therefore there is clearly a turning point when the expected value of the sum is greater than the expected value of $n$ (as the arithmetic sum is quadratic in $n$) so your expected gain for that roll is negative, so stop at this point. As this is a game for a stats class I assume the teacher wants you to actually solve it, so I'll leave the rest to you, but that's the idea.