John keeps rolling a standard six-sided dice until it comes up as a 6. What is the expected number of dice rolls?

Question

For this problem, I found a very complicated solution. Are there any easier methods? I am asking mainly because the final answer was what I intuitively guessed before solving the problem. It is also surprising that after so much work I get such an intuitive result.

Problem: John keeps rolling a standard six-sided dice until it comes up as a 6. What is the expected number of dice rolls?

My Solution: The probability that John rolls only once is $\frac{1}{6}$. The probability that John rolls twice is $\frac{5}{6}\cdot\frac{1}{6}$. The probability that John rolls three times is $\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{1}{6}$. Seeing the pattern, the probability that John rolls $k$ times is $\frac{1}{6} \cdot \left(\frac{5}{6}\right)^{k-1}$. We verify this by adding all $k$ from $1$ onwards to $\infty$ and checking if that is $1$. This works because positive integers are the only number of times that John can roll. $$\sum_{n=1}^{\infty} \frac{1}{6} \cdot \left(\frac{5}{6}\right)^{i-1} = 1$$What we want is the sum of the arithmetico-geometric sequence $$\sum_{n=1}^{\infty} \frac{i}{6} \cdot \left(\frac{5}{6}\right)^{i-1} = \hspace{1mm}?$$To get a better idea of what this sequence is, we expand out a few terms. $1 \cdot \frac{1}{6} + 2 \cdot \frac{5}{6}\cdot\frac{1}{6} + 3 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^{2} + \cdots$. Now we factor the initial term, $\frac{1}{6} \left(1 + 2 \cdot \frac{5}{6} + 3 \cdot \left(\frac{5}{6}\right)^{2} + \cdots \right)$. We let $S = 1 + 2 \cdot \frac{5}{6} + 3\left(\frac{5}{6}\right)^{2} + \cdots$. We can now use the trick for any geometric sequence on this sequence. $\frac{5}{6}S = \frac{5}{6} + 2\left( \frac{5}{6}\right)^2 + 3\left(\frac{5}{6}\right)^{3} + \cdots$. Subtracting the second equation from the first gives us $\frac{1}{6}S = 1 + \frac{5}{6} + \left(\frac{5}{6}\right)^2 + \left(\frac{5}{6}\right)^3 + \cdots$. This leads to $S = 36$. The expression we want is $\frac{S}{6} = \boxed{6}$

Edit: My question is different than the other one because I am asking for an explanation on why the answer is the intuitive one and if there are any better methods to solve while that question is only asking for the answer.

Answer 1

Imagine rolling the die a huge number of times, say $6N$, where $N$ is extremely large. Of these rolls, approximately $N$ of them will be a six. For each of these, record the number of rolls since the previous six (or since the beginning, for the first six). The total of these numbers will be somewhat less than $6N$ since some number of rolls at the very end are likely to be non-sixes. But if $N$ is extremely large, the average of the numbers will be approximately $6N/N=6$.

Answer 2

It can be shown using summation by parts with the result memorized for convenience that when X is non-negative (almost surely) then

$$E[X] = \sum_i P[X\ge i]$$

In the case of a geometric random variable

$$E[X] = \sum_i (1-p)^i = \frac{1}{1-(1-p)} = \frac{1}{p}$$

So in your case when $p = \frac{1}{6}$, the expected value is 6.

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For intuition about the special case of the geometric distribution, recall that it has the memoryless property

$$P[X \ge (i+j) | X \ge j] = P[X \ge i]$$

This means that at any point in time, the waiting time for the next 6 to occur is the same. So if $W$ is the expected waiting time then we have a $\frac{1}{6}$ chance of rolling a 6 immediately making our waiting time 1. If not, our excess waiting time is the same as our waiting time $W$.

$$W = \frac{1}{6}1 + \frac{5}{6} (1+W)$$

Collecting $W$'s and solving we get $W = 6$.

Answer 3

You can also use the fact that:

$$\sum_{i=1}^{\infty}ik^{i-1}=\left(\sum_{i=0}^{\infty}k^{i}\right)'=\frac{1}{(1-k)^2} $$

So:

$$\sum_{n=1}^{\infty} \frac{i}{6} \cdot \left(\frac{5}{6}\right)^{i-1} =$$

$$\frac{1}{6}\sum_{n=1}^{\infty} {i} \cdot \left(\frac{5}{6}\right)^{i-1} =$$

$$\frac{1}{6}\frac{1}{(1-\frac{5}{6})^2} = \frac{36}{6} = 6 $$

For a far more informal (indeed, in my eyes not worthy of the name 'proof') way to think about this: as you throw a dice a bunch of times, the expected number of 1's, 2's, 3's, etc. that you will have thrown should be the same. So, when you get one 6 (or: if you want: when the expected number of 6's thrown reaches 1), you can expect to have thrown one 1, one 2, etc. as well, i.e. you can expect to have thrown the dice 6 times. Again, I myself am actually not fully convinced by this line of reasoning, and prefer the mathematical solution, but I have obtained what turned out to be the correct answer with this line of thinking at other places ... So maybe it can be considered a way to get a hypothesis going, to be proven by a proper mathematical proof.