Expectation for a dice throw A single die is rolled as many times as you want. For each throw, you receive $n$ dollars if dice shows $n$, if $n<6$. If dice shows $6$, you lose all the money accumulated. What is the expectation and the best strategy for this game?
I have understood the best strategy: we roll until we have an amount $\geq 15$.
EDIT:If at some point we reach zero before reaching 15, we would again start tossing, and continue this until we have >=15 , imo in this way probability of reaching a value >=15 would be 1
But I was stuck at the expected value using  this strategy, with this strategy we could end up with 15,16,17,18,19,20. What is the expectation? Manually calculating the probability for reaching first to 15, 16 etc would be quite burdensome, is there any simple way to this.
Thanks!!
 A: 
With this strategy we can end up with 15,16,17,18,19,20

You can also end up with 0 by hitting a $6$ before you reach $15$, which is the main incentive for not playing forever (at which point does the risk of hitting a $6$ and losing it all outweigh the potential benefit of gaining more?)
Let's say you already have $n$ dollars. Then the expected value of you continuing to play is $\sum_{k=1}^5 \frac{1}{6}(n+k) = \frac{1}{6} (5n + 15) $. Setting $n \leq  \frac{1}{6} (5n + 15)$ gives $ n \leq 15$. So $15$ is the point at which the expected value of further playing is equal to the current amount.
Given that there's a high chance we roll a $6$ before we reach $15$, what's the actual expected value of playing until 15?
Let $S_{k}$ denote the probability of a play succeeding in getting at least $k$ dollars. Let $E_{k}$ be the expected value of a play that doesn't stop until it either fails or has at least $k$ dollars. We have:
$$E_{n} = \sum_{k=1}^5 \frac{1}{6} (k S_{n-k} + E_{n-k})\\
(n\leq 0) \Rightarrow (E_{n}=0)
$$
What this states is the expected value of attempting to gain at least $n$ dollars depends on rolling a non-$6$ value $k$, then adding it to the expected value $E_{n-k}$ (attempt at gaining an additional $n-k$ dollars),  provided it succeeds. We may deduce:
$$S_{n} = \sum_{k=1}^5 \frac{1}{6} S_{n-k}\\
(n\leq 0) \Rightarrow (S_{n}=1)
$$
This is a recurrent formula for the probability of a play for at least $n$ dollars to succeed.
Plugging these into a program as recursive functions (with memoization), we obtain:
$$E_{15} = E_{16} \approx 6.15374 $$
Here is a plot of $E_{k}$, $\;k=\{1\dots 30\}$ .

Edit: I was considering that  dice showing $6$ is a losing condition, and the game stops. If you allow the player to just restart, then the probability of reaching $15$ eventually goes up (in limit) to $1$. But then so does the probability of reaching any other number, say $200$. Let a sub-game be defined as playing until you throw a $6$. With no stopping condition, the game consists in playing as many consecutive sub-games as you want. For any number $n$, the probability of winning at least $n$ before ending a sub-game is greater than $0$ . The probability of getting $n$ on a sub-game corresponds to our initial definition of $S_{n}$, and we have $S_{n} \geq (\frac{5}{6})^{n} > 0$. That is, you can win any amount as long as you avoid getting $6$ for long enough. The probability of not winning at least $n$ before finishing a sub-game is $1-S_{n}$. The probability of not winning at least $n$ in $p$ consecutive sub-games is $(1-S_{n})^{p}$. So the probability of winning $n$ dollars in at least one of $p$ sub-games is $1 -(1-S_{n})^{p}$, which goes to $1$ as $p$ goes to infinity. In short, if you keep playing for long enough, you'll get lucky eventually and hit $n$ dollars, regardless  of what $n$ is. I believe this is an instance of an infinite monkey theorem.
