"Expected rolls to get a 6": recursion vs infinite sum

Question

For the question of

How many throws on average to get a $6$ on a dice?

The recursive answer is quoted as $\mathbb{E}[X] = p + q(1 + \mathbb{E}[X])$, where $p = \frac{1}{6}$ and $q = 1 - p$.

I know that this has been explained in other answers but I still do not understand the intuition behind the second term. Would someone be able to explain how one would get to this? Are there any good resources on recursive probabilities that anyone would recommend? I have not come across this in Feller's Intro version 1 book yet.

Also, how would this be expanded to equal the original infinite equation of $$\mathbb{E}[X] = p + 2qp + 3q^2p + 4q^3p + \dots?$$

Because when I expand it just by one recursive step I get $$\mathbb{E}[X] = p + q + q(p + q(1 + \mathbb{E}[X])) = p + q + qp + q^2(1 + \mathbb{E}[X])$$ which seems like it would not equal the original infinite sum. Have I expanded incorrectly?

Thanks.

Answer 1

$P(\neg 6_1) = \frac{5}{6}$

$P(\neg 6_2) = \left(\frac{5}{6}\right)^2$

$P(6_2) = 1 - \left(\frac{5}{6}\right)^2$

$P(6_n) = 1 - \left(\frac{5}{6}\right)^n$

From this point on, the decision might be arbitrary. What does highly probable mean to you, and how much of a risk-taker are you?

The generally acceptable probability that 9 out of 10 would consider as worthy of an attempt = $99$%.

$1 - \left(\frac{5}{6}\right)^n \geq 99\%$

Answer 2

There's nothing wrong with your expansion, which you can simplify to $$ \mathbb{E}(X)=1+q+q^2\mathbb{E}(X)\ . $$ Taking this one step further gives \begin{align} \mathbb{E}(X)&=1+q+q^2(1+q\mathbb{E}(X))\\ &=1+q+q^2+q^3\mathbb{E}(X)\ \end{align} and you can obviously continue by induction to get $$ \mathbb{E}(X)=1+q+q^2+\dots+q^n+q^{n+1}\mathbb{E}(X) $$ for any $\ n\ .$

However you can also rewrite your expansion as $$ \mathbb{E}(X)=p+2pq+2q^2+q^2\mathbb{E}(X)\ $$ and taking this one step further gives \begin{align} \mathbb{E}(X)&=p+2pq+2q^2+q^2(1+q\mathbb{E}(X))\\ &=p+2pq+3q^2+q^3\mathbb{E}(X)\\ &=p+2pq+3pq^2+3q^3+q^3\mathbb{E}(X)\ . \end{align} Again you can obviously continue by induction to get $$ \mathbb{E}(X)=p+2pq+\dots+(n+1)pq^n+(n+1)q^{n+1}+q^{n+1}\mathbb{E}(X)\ . $$ Since both series $\ \sum_\limits{n=0}^\infty q^n\ $ and $\ \sum_\limits{n=0}^\infty (n+1)pq^n $ converge to the same limit, $\ \frac{1}{1-q}=\frac{1}{p}\ ,$ which is also the value of $\ \mathbb{E}(X)\ ,$ there's no inconsistency here.