Expected length of a sequence The following problem has kept me pondering for a while now and since I can't get through, I'm posting it here.
Say that you can draw a number $x_i$ uniformly from the set $\mathcal{X} \sim \{0,0,1,2,3,4,5,6\}$. The sum of those numbers we call $X$. We stop drawing numbers when $X$ is at least $5$, the amount of numbers drawn we will call $N$. I would write the stopping condition as:
$$X = \sum_{i=0}^{N}x_i \geq 5 $$
What is the expected value of numbers we need to draw, $E[N]$?
I'll attempt some visualisations based on empirical outcomes.
 A: Define the function $h(x)=\mathbb{E}_x(N)$ to be the expected number of steps needed, if we start with a sum of $x$. First step analysis gives
\begin{eqnarray*}
h(4)&=&1+h(4)/4 \\
h(3)&=&1+h(3)/4+h(4)/8\\
h(2)&=&1+h(2)/4+h(3)/8+h(4)/8\\
h(1)&=&1+h(1)/4+h(2)/8+h(3)/8+h(4)/8\\
h(0)&=&1+h(0)/4+h(1)/8+h(2)/8+h(3)/8+h(4)/8.
\end{eqnarray*}
Solving this system we get  $E(N)=h(0)={2401\over 972}\approx 2.47.$
A: It is possible to solve a more general problem. We are looking for $$N(t)=\min \left\{ n : \sum_{i=1}^n x_i >t\right\}.$$
Now set $$m(t)=\mathbb{E}[N(t)]$$
We can condition on $x_1$ to obtain
$$m(t)=\cases{ 1 & if $\quad x_1\ge t$\cr
        m(t-x_1) & if $\quad x_1<t$\cr}$$
We know also that $$\begin{align}m(t)=\mathbb{E}[\mathbb{E}[N(t)|x_1]]&=\sum_{i=0}^{t-1} m(t-i)\mathbb{P}(x_1=i)+\sum_{i=t}^6\mathbb{P}(x_1=i)\\ &=p\sum_{j=1}^{t} m(j)+(6-t)p\end{align}$$
Where, for brevity and because $x_i$ is uniform, we have written $p$ for $\mathbb{P}(x_1=i)$
Assume we know the expressions for all $m(j)$ with $j=1,...,k$, then
$$\begin{align}m(k+1)&=p\sum_{j=1}^{k+1} m(j)+(6-k-1)p\\ &= pm(k+1)+p\sum_{j=1}^{k} m(j)+(6-k-1)p \\ &\Rightarrow m(k+1)=\frac{p}{1-p}\sum_{j=1}^{k} m(j)+(6-k-1)\frac{p}{1-p}\end{align}$$
We can calculate $m(1)$ from the recursion, namely
$$m(1)=p m(1)+5p\Rightarrow m(1)=\frac{5p}{1-p}$$
Now 
$$m(2)=\frac{p}{1-p}\frac{5p}{1-p}+(6-1-1)\frac{p}{1-p}=5\left(\frac{p}{1-p}\right)^2+4\frac{p}{1-p}$$
$$m(3)=5\left(\frac{p}{1-p}\right)^3+9\left(\frac{p}{1-p}\right)^2+3\frac{p}{1-p}$$
$$m(4)=5\left(\frac{p}{1-p}\right)^4+14\left(\frac{p}{1-p}\right)^3+12\left(\frac{p}{1-p}\right)^2+2\left(\frac{p}{1-p}\right)$$
And finally 
$$m(5)=5\left(\frac{p}{1-p}\right)^5+19\left(\frac{p}{1-p}\right)^4+26\left(\frac{p}{1-p}\right)^3+14\left(\frac{p}{1-p}\right)^2+\frac{p}{1-p}$$
