We have a fair, six-sided dice. The questions are

  1. What's the expected number of rolls to get 1 followed by 1?
  2. What's the expected number of rolls to get 1 followed by 2?

Let $E$ be the expected time to get '11'. From the geometric distribution, we know that it takes on average 6 rolls to get a 1. Let us roll until we have a 1. Then there is a $1/6$ chance of being done in the next roll $$E = 6 + \frac{1}{6} \cdot 1 + \frac{5}{6}(1 + E) \qquad \Rightarrow \qquad E = 42.$$ I am fairly certain that my solution is correct for the first one, but I am a bit confused about the logic for the second one. I know its very similar to the first one but with a twist. Let $E$ be the expected time to get '12'. Then we first have to roll on average 6 times, after which we have a $1/6$ probability of being done (2), a $4/6$ chance of starting all over (3,4,5,6), and a $1/6$ probability of making no progress (1). Then $$E = 6 + \frac16\cdot 1 + \frac46 \cdot (1 + E) + \frac16 \cdot (1+6) \qquad\Rightarrow\qquad E=48.$$ Is the $(1+6)$ part right? For some reason I don't think it is.

  • 4
    $\begingroup$ Indeed the $(1+6)$ term is incorrect and should be replaced by $(1+F)$ where $F$ denotes the mean number of throws to get 12 starting from 1. Thus, by the reasoning you used to get the equation on $E$, one gets $F=\frac16+\frac16(1+F)+\frac46(1+E)$. Solving the $(E,F)$-system yields $E=36$. $\endgroup$
    – Did
    Sep 3, 2016 at 20:18
  • $\begingroup$ Aha this makes sense. Is there a way to solve the problem without solving a system (as in, make it all one easy equation like the '11' case)? In particular, this was an interview question where I did not have pen and paper and a time limit. $\endgroup$
    – user369210
    Sep 3, 2016 at 20:52
  • 1
    $\begingroup$ Yes: by the law of large numbers, after $n\gg1$ throws, there are approximately $n/36$ occurrences of the motive 12. Since after one such occurrence, one starts afresh to produce another motive 12, the mean time to produce one is $36$. The argument breaks for the motive 11 because once some 11 is realized, we already have one 1 at our disposal, possibly, to produce the next 11. Hence, after the first 11, the next ones are closer from each other, which explains why the time needed to produce the first motive 11 is $>36$. $\endgroup$
    – Did
    Sep 3, 2016 at 21:02


You must log in to answer this question.

Browse other questions tagged .