Problem 1

You got this far:

$$S = $0 (1/6) + $1 (1/6) (5/6) + $2 (1/6) (5/6)^2 + ... \\ S = \sum_{i=0}^{\inf} i (1/6) (5/6)^i$$

But I’m not sure you know how to solve that. Let’s start with an easier infinite sum:

What’s the value of:

$$\sum_{i=0}^{\inf} x^i$$

We’re going to call the sum $S$. We don’t know what $S$ is yet, but we can solve for it:

$$S = 1 + x + x^2 + x^3 + x^4 + ...$$

multiply both sides by x

$$xS = x + x^2 + x^3 + x^4 + x^5 + ...$$

$xS$ looks really similar to $S$. Actually, if you add $1$ to $xS$, the right hand side looks like the right hand side of the equation for $S$ again.

$$\begin{align*} 1 + xS &= 1 + x + x^2 + x^3 + x^4 + x^5 + ... = S \\ 1 + xS &= S\\ 1 &= S - xS \\ S &= \frac{1}{1-x} \end{align*}$$

This formula is very important, and useful in solving our actual problem.

Our actual problem was:

$$S = \sum_{i=0}^{\inf} i (1/6) (5/6)^i$$

So, if we didn’t have the $i$ multiplying every element in the sum, this problem would be almost exactly like the one above. Unfortunately, we do. Here’s a tricky way to solve it. I’m going to define new sums $S_1$ to $S_n$.

$$\begin{align*} S = 1 (1/6) (5/6) + 2 (1/6) (5/6)^2 + 3 (1/6) (5/6)^3 + 4 (1/6) (5/6)^4 + ... \\ \\ S_1 = 1 (1/6) (5/6) + 1 (1/6) (5/6)^2 + 1 (1/6) (5/6)^3 + 1 (1/6) (5/6)^4 + ... \\ S_2 = 1 (1/6) (5/6)^2 + 1 (1/6) (5/6)^3 + 1 (1/6) (5/6)^4 + ... \\ S_3 = 1 (1/6) (5/6)^3 + 1 (1/6) (5/6)^4 + ... \\ S_4 = 1 (1/6) (5/6)^4 + ... \\ ... \end{align*}$$

Notice how if you sum $S1 + S2 + S3 + S4 + ...$ (vertically), you get back $S$? Ok, so, that’s kind of cool - but why does this help? Well, you know how to solve for the value of $S1$.

$$S1 = 1/6 [ (5/6) + (5/6)^2 + (5/6)^3 + ... ] \\ S1 = 1/6 [ \frac{1}{1-(5/6)} - 1 ]\\ S1 = 5/6 \\$$

With that you also know how to solve for $S2$.

$$\begin{align*} S2 &= S1 (5/6) \\ S2 &= (5/6) \\ \\ S3 &= S2 (5/6) \\ S3 &= (5/6)^2 \\ \\ S4 &= S3 (5/6) \\ S3 &= (5/6)^3 \\ \end{align*}$$

and so on…

So, to sum it all up, the value of:

$$\begin{align*} S &= \sum_{i=1}^{\inf} S_i \\ S &= S_1 + S_2 + S_3 + S_4 + ... \\ S &= (5/6) + (5/6)^2 + (5/6)^3 + (5/6)^4 + ... \\ S &= \frac{1}{1-(5/6)} - 1\\ S &= 6 - 1 \\ S &= 5 \end{align*}$$

So, if at this point you’re thinking, “really!?…”, that’s understandable. However, there’s a much easier way to solve this problem, which is the way I actually want to teach you. Here goes:

Let’s call the value of the game $S$. If I roll a 6, I lose, and if I don’t I’ve won $1 and I’m in the exact same place as I was one roll ago. Put another way, if I don’t roll a 6, I win $1 and I get to play the same game again - which is worth $S$ (by definition).

$$\begin{align*} S &= (1/6) 0 + (5/6) (1 + S) \\ (1/6) S &= (5/6) \\ S &= 5 \end{align*}$$

QED!