Not to be confused with the number of rolls needed to get two of the same number, suppose you keep rolling a dice until you get a 3 and then 4 consecutively. Calculate the expected number of rolls required.

My logic was that the probability of rolling a three is 1/6 and the probability of rolling a four is 1/6, so we can multiply the two to get 1/36. Then since the distribution is geometric, the expected value of the geometric distribution is $1/p = 1/(1/36) = 36$.

However, multiple people have gotten an answer of either 9 or 12. I get how one would arrive at 12, but how do you get 9 rolls out of this? Just what does this problem have to do with Harmonic Numbers? Any help would be appreciated.

  • 1
    $\begingroup$ You can take a look here. I don't think $9$ or $12$ is correct. $\endgroup$
    – VHarisop
    Nov 14, 2018 at 21:40

1 Answer 1


Just to have a different method, let's do it with states. There are only three states of interest. Either you are just starting or the last throw was not a three (call it $\mathscr S_{\emptyset}$) or the last throw was a $3$, call it $\mathscr S_{3}$, or you are done.

We also denote by $E_{\emptyset}$ ansd $E_3$ the expected number of moves it should take to finish from there. Of course $E_{\emptyset}$ is the answer we seek.

Considering the first toss we see that $$E_{\emptyset}=\frac 16\times (E_3+1)+\frac 56\times (E_{\emptyset}+1)$$

And if you are in $\mathscr S_{3}$ we see that $$E_3=\frac 16\times 1+\frac 16\times (E_3+1)+\frac 46\times (E_{\emptyset}+1)$$

This system is easy to solve and we get $$E_{\emptyset}=36\quad E_3=30$$

Which, of course, confirms your result.


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