I roll 6-sided dice until the sum exceeds 50. What is the expected value of the final roll?
I am not sure how to set this one up. This one is not homework, by the way, but a question I am making up that is inspired by one. I'm hoping this will help me understand what's going on better.
Best Answer
Let $u(n)$ be the expected value of the first roll that makes the total $\ge n$. Thus $u(n) = 7/2$ for $n \le 1$. But for $2 \le n \le 6$, conditioning on the first roll we have $$u(n) = \left( \sum_{j=1}^{n-1} u(n-j) + \sum_{j=n}^{6} j \right)/6$$
That makes $$ u_{{2}}={\frac {47}{12}},u_{{3}}={\frac {305}{72}},u_{{4}}={\frac {1919}{432}},u_{{5}}={\frac {11705}{2592}},u_{{6}}={\frac {68975}{15552 }}$$ And then for $n > 6$, again conditioning on the first roll, $$u(n) = \frac{1}{6} \sum_{j=1}^6 u(n-j)$$ The result is $$u(51) = \frac {7005104219281602775658473799867927981609}{1616562554929528121286279200913072586752} \approx 4.333333219$$ It turns out that as $n \to \infty$, $u(n) \to 13/3$.
EDIT: Note that the general solution to the recurrence $\displaystyle u(n) = \frac{1}{6} \sum_{j=1}^6 u(n-j)$ is $$ u(n) = c_0 + \sum_{j=1}^5 c_j r_j^n$$ where $r_j$ are the roots of $$\dfrac{6 r^6 - (1 + r + \ldots + r^5)}{r - 1} = 6 r^5 + 5 r^4 + 4 r^3 + 3 r^2 + 2 r + 1 = 0$$ Those all have absolute value $< 1$, so $\lim_{n \to \infty} u(n) = c_0$. Now $6 u(n+5) + 5 u(n+4) + \ldots + u(n) = (6 + 5 + \ldots + 1) c_0 = 21 c_0$ because the terms in each $r_j$ vanish. Taking $n = 1$ with the values of $u_1$ to $u_6$ above gives us $c_0 = 13/3$.