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A game where you roll a fair die, repeatedly, adding up the faces that show up, until the face 6 appears. What is the expected sum (including the 6)?

All I can think of is that the expected value of each roll is $7/2$.

I'd appreciate your help!

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Solution 1: We roll the die and add the value. With $5/6$ probability we start again. Hence,

$$E(X) = \frac{7}{2} + \frac{5}{6}E(X)$$

and this gives $E(X) = \boxed{21}$.

$E(X) = \infty$ is also a solution, so we should show that the expectation is finite. This is equivalent to showing that the expected number of rolls is finite, since the expected value is bounded between the number of rolls and six times the number of rolls. By standard methods, the expected number of rolls is

$$\sum_{n=0}^{\infty} (n+1) \frac{1}{6} \left(\frac{5}{6}\right)^n$$

which converges to $6$. We can use this to motivate a secondary method:

Solution 2: On average we roll $6$ times. The average on a roll is $\frac{7}{2}$. Hence the answer is $21$.

You might ask, but doesn't it matter that we can't roll a six or else we stop? Each roll still has the same expectation of $\frac{7}{2}$, it is just that we stop arbitrarily when we happen to roll a six.

(This perhaps unintuitive logic is similar to that in the problem involving a town where couples birth babies until their first boy and then stop. The proportion of boys to girls in this town is still $1:1$ since each birth is still just an independent event; they just choose to stop arbitrarily when the first boy comes)

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  • $\begingroup$ +1 for multiplying averages in a logical way. Can you explain where the 3 comes from in your first formula? You said we add the value that we roll to E(x), so why are you adding three? $\endgroup$ – theREALyumdub Jun 22 '16 at 0:30
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    $\begingroup$ @theREALyumdub Expanded, the RHS is $\frac{1}{6} (E(X) + 1) + \frac{1}{6} (E(X) + 2) + \dots \frac{1}{6} (E(X) + 5) + \frac{1}{6}(6)$. I group the first five together. In other words, $3$ is the average roll of $\{1, 2, 3, 4, 5\}$. $\endgroup$ – MCT Jun 22 '16 at 0:32
  • $\begingroup$ Ah of course, not the average of them all and 6. That make sense. $\endgroup$ – theREALyumdub Jun 22 '16 at 0:34
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    $\begingroup$ $E(X) = \infty$ is another solution to that equation, so to make this a rigorous proof we need to find a reason why the expectation should be finite. $\endgroup$ – user14972 Jun 22 '16 at 4:13
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    $\begingroup$ @Hurkyl You're right. I've added. $\endgroup$ – MCT Jun 22 '16 at 4:41
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Let $N\sim\mathcal{Geo}_1(1/6)$ be the count of rolls until you roll a six.   This is a geometric random variable with support $\{1,2,\ldots\}$.   Then you know $\mathsf E(N)$.

Let $X_i\sim\mathcal U\{1,2,3,4,5,6\}$ be the value of the $i$-th roll.   Noting that roll $N$ is the first time a $6$ appears, you can calculate $\mathsf E(X_i\mid i<N)$ and $\mathsf E(X_i\mid i=N)$.

Then the expectation you want is given by the Law of Iterated Expectation:

$$\begin{align}\mathsf E\left(\sum_{i=1}^{N} X_i\right) ~=&~ \mathsf E\big((N-1)\,\mathsf E(X_i\mid i<N)+\mathsf E(X_i\mid i=N)\big)\\ =&~ 5\cdot 3+ 6\\ =&~ 21\end{align}$$

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