FreeProbabilityEasy

At Least Six I

Asked atCitadel

You roll a fair six-sided die again and again, keeping a running total. Let $T$ be the number of rolls it takes for the total to reach $6$ or more. Find $E[T]$ .

Solution

Try it yourself first. Open a hint if you get stuck.

Hint 1

Work backwards. Let $E_i$ be the expected number of further rolls when the total is $i$ . What is $E_5$ ?

Hint 2

From total $i < 6$ , one roll takes you to $i + k$ with $k = 1, \dots, 6$ equally likely, so $E_i = 1 + \tfrac16\big(E_{i+1} + \dots + E_{i+6}\big)$ , where any $E_j$ with $j \ge 6$ is $0$ .

Show solution

Set up. Let $E_i$ be the expected number of further rolls when the running total is $i$ . Once the total is $6$ or more we stop, so $E_i = 0$ for $i \ge 6$ .

From a total $i < 6$ we always spend one roll, and then land on $i + 1, \dots, i + 6$ with chance $\tfrac16$ each:

E_i = 1 + \tfrac16\big(E_{i+1} + E_{i+2} + \dots + E_{i+6}\big).

Only the terms that land below $6$ are nonzero.

Work backwards from $5$ .

$E_5 = 1$ , since any roll finishes.
$E_4 = 1 + \tfrac16 E_5 = \tfrac76$ .
$E_3 = 1 + \tfrac16(E_4 + E_5) = 1 + \tfrac16 \cdot \tfrac{13}{6} = \tfrac{49}{36}$ .
$E_2 = 1 + \tfrac16(E_3 + E_4 + E_5) = \tfrac{343}{216}$ .
$E_1 = 1 + \tfrac16(E_2 + E_3 + E_4 + E_5) = \tfrac{2401}{1296}$ .
$E_0 = 1 + \tfrac16(E_1 + E_2 + E_3 + E_4 + E_5) = \tfrac{16807}{7776}$ .

Answer:

E[T] = E_0 = \frac{16807}{7776} \approx 2.16139.

Notice the pattern: $E_4 = \tfrac76$ , $E_3 = \left(\tfrac76\right)^2$ , …, $E_0 = \left(\tfrac76\right)^5$ . At Least Six II explains why.

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