I got the following quiz asked at the interview: a fair dice is rolled infinitelly; what is the expectation of the rolling iteration I, at which one gets two sixs in a row?

  • $\begingroup$ General methods to answer questions like this (focusing on coin flips but generalizable in obvious ways to dice and other discrete distributions) are discussed in the thread at stats.stackexchange.com/questions/12174/…. $\endgroup$
    – whuber
    Commented Nov 3, 2014 at 16:16

1 Answer 1


Interpretation: Let $K$ denote the iteration at which the second six is obtained. Determine $E[K]$.

Answer: Let $K_0$ denote the time when we reach the first six. Then $E[K_0]$ is $\frac{1}{1/6}=6$ since it follows a geometric distribution.

When we arrive at the first six, there is a 1 in six probability that we will arrive at the second in the next roll. If we fail to do so, the game starts over. That is,

$E[K] = E[K_0] + 1 + \frac{5}{6}E[K]$

Simplifying gives $E[K] = 6 E[K_0] + 6 = 42$.

Other versions:

Standard probability theory:

First by the tower-property, $E[K] = E[E[K\mid K_0]]$. Computing the inner expectation gives

\begin{align} E[K\mid K_0] &= \sum_{k=K_0 + 1}^\infty k P(K = k) = \sum_{k=K_0 + 1}^\infty k\big( P(K = k\mid K = K_0+1)P(K = K_0 + 1)\\ &+ P(K = k\mid K \neq K_0+1)P(K \neq K_0 + 1)\big) \\ &=\frac{K_0 + 1}{6} + \frac{5}{6}\sum_{k=K_0 + 2}^\infty k P(K = k\mid K \neq K_0+1) \\ &=\frac{K_0 + 1}{6} + \frac{5}{6}E[K]. \end{align} The last equality follows from that we can put $s=K_0+2$ to obtain \begin{align} E[K] &= \sum_{s=1}^\infty k P(K = s) = \sum_{s=1}^\infty k P(K = s\mid K \neq s-1) \\&=\sum_{k=K_0 + 2}^\infty k P(K = k\mid K \neq K_0+1), \end{align}

and the one before from $P(K = K_0 + 1) = \frac{1}{6}$.

Markov theory:

This is a Markov chain with three states, $K$, $K_6$ and the absorbing state $K^*$. The states are for the last two dice-throws. Let $x$ be any integer $1\ldots 5$ and $y$ any integer $1\ldots 6$. Then the three states represent $(y,x)$, and $(x,6)$ and $(6,6)$ respectively.

There is a theorem which states that the time to absorption is given as the solution to the equation system:

\begin{align} K &= 1 + \frac{1}{6}K_6 + \frac{5}{6}K\\ K_6 &= 1 + \frac{5}{6}K. \end{align}

Solving for $K$ gives the solution $K_6 + 6 = 42$. That is, once we are in state $(x, 6)$ we expect $36$ additional rolls. To get the total we have to add six rolls to reach state $(x, 6)$.

  • $\begingroup$ +1 This is correct. That the answer differs from $6^2=36$ begs for an explanation, though: can you supply some useful intuition to account for the difference? $\endgroup$
    – whuber
    Commented Nov 3, 2014 at 15:50
  • $\begingroup$ seems correct, but could you explain why do you have 5/6 E(K) as the second summation term? Could you think of a strict prove with probabilities? $\endgroup$
    – coffee
    Commented Nov 3, 2014 at 23:21
  • $\begingroup$ Added content according to comments. $\endgroup$
    – Hunaphu
    Commented Nov 5, 2014 at 17:27
  • $\begingroup$ ... but actually I did not. So the definition of $K$ is the iteration on which the second six was obtained. However, is should be completely defined by $K_0$, since we're looking for two sixs in a row. So $E(K|K_0) = K_0+1$ (it is constant given $K_0$), is not it true? $\endgroup$
    – coffee
    Commented Dec 1, 2014 at 2:24
  • $\begingroup$ You say that a second six always follow the first which is not true. $\endgroup$
    – Hunaphu
    Commented Dec 1, 2014 at 12:33

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