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If I repeatedly roll a fair, $X$-sided die, on average how many rolls should I expect to make before I happen to roll the same value $N$ times in a row?

I've found questions on here with answers when $X=2$, or where $N=2$, or where you're looking for a specific result $N$ times in a row; I'm interested in the situation where I don't care what the specific value is, I just want it to be $N$ times in a row. The closest I can find is Suppose we roll a fair $6$ sided die repeatedly. Find the expected number of rolls required to see $3$ of the same number in succession., but I can't quite figure out how to generalize it beyond $N=3$.

philomory
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3 Answers3

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Let $E_k$ denote the expected number of rolls that are still needed if $k$ rolls have passed with equal result. We write recurrence relations for $E_k,0\leq k\leq N-1$ and calculate the wanted expectation $E=E_0$ from them.

We obtain \begin{align*} E_0&=E_1+1\tag{1}\\ E_1&=\frac{1}{X}(E_2+1)+\frac{X-1}{X}(E_1+1)\\ E_2&=\frac{1}{X}(E_3+1)+\frac{X-1}{X}(E_1+1)\\ E_3&=\frac{1}{X}(E_4+1)+\frac{X-1}{X}(E_1+1)\tag{2}\\ &\ \ \ \vdots\\ E_{N-3}&=\frac{1}{X}(E_{N-2}+1)+\frac{X-1}{X}(E_1+1)\\ E_{N-2}&=\frac{1}{X}(E_{N-1}+1)+\frac{X-1}{X}(E_1+1)\\ E_{N-1}&=\frac{1}{X}+\frac{X-1}{X}(E_1+1)\tag{3}\\ \end{align*}

Comment:

  • In (1) we note that any roll of the $X$-sided die will do.

  • In (2) we are in the situation that a run of length $3$ is given. With probability $\frac{1}{X}$ we get a run of length $4$ and with probability $\frac{X}{X-1}$ we get a different result and start with a run of length $1$.

  • In (3) we get a run of length $n$ with probability $\frac{1}{X}$ and can stop or we are back at the beginning and start again with a run of length $1$.

This system can be easily solved. We iteratively obtain be respecting all equations from above besides the last one (tagged with (3)). \begin{align*} E_1&=E_0-1\\ E_2&=E_0-(1+X)\\ E_3&=E_0-(1+X+X^2)\\ E_4&=E_0-(1+X+X^2+X^3)\\ &\ \ \ \vdots\\ E_{N-1}&=E_0-(1+X+X^2+\cdots+X^{N-2})\\ &=E_0-\frac{X^{N-1}-1}{X-1}\tag{4}\\ \end{align*}

From equation (4) and (3) we finally conclude \begin{align*} E_{N-1}&=E_0-\frac{X^{N-1}-1}{X-1}\\ (1-X)E_0+XE_{N-1}&=1 \end{align*} from which \begin{align*} \color{blue}{E=E_0=\frac{X^N-1}{X-1}} \end{align*} follows.

Note: This approach is a generalisation of @lulu's answer of this MSE question. In fact we get as small plausibility check with $N=3$ and $X=6$ the result \begin{align*} E=\frac{6^3-1}{6-1}=\frac{215}{5}=43. \end{align*} in accordance with @lulu's result.

[Add-on] The strongly related problem: Rolling a fair $X$-sided die until we get a run of length $N$ of a specific value can be solved by the system above with the first equation (1) replaced with \begin{align*} E_0&=\frac{1}{X}(E_1+1)+\frac{X-1}{X}(E_0+1) \end{align*} which gives the result \begin{align*} \color{blue}{E=E_0=\frac{X\left(X^N-1\right)}{X-1}} \end{align*} This is also plausible compared with the result above, since the probability to roll a specific value is $\frac{1}{X}$.

Markus Scheuer
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Suppose $N\ge2$, otherwise the problem is trivial. We need at least two rolls. Let $Y_1,Y_2,\ldots$ be the values of the rolls, that is independent random variables each uniformly distributed in $\{1,2,\ldots,X\}$. Define $Z_j\in\{0,1,\ldots,X\}$ by $Z_j\equiv Y_j-Y_{j+1}\pmod X$. Then the $Z_j$ are also independent and uniformly distributed. In effect we are looking for the expectation of $T$, which is the least $t$ with $Y_t=Y_{t-1}=\cdots= Y_{t-N+1}$, equivalently the least $t$ such that $Z_{t-1}=Z_{t-2}=\cdots= Z_{t-N+1}=0$. Thus $E(T)=1+E(U)$ where $U$ is the number of rolls needed to get $N-1$ consecutive occurrences of a specific value. You can use the Markov chain method in the linked solution to find that.

Angina Seng
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$N_1$ - No of tosses for 1st Heads
$N_2$ - No of tosses for 2 consecutive Heads
$N_3$ - No of tosses for 3 consecutive Heads
$\mathbb E[N_1] = 2$
$\mathbb E[N_2] = (1+\mathbb E[N_1])2 = 6$
$\mathbb E[N_3] = (1+\mathbb E[N_2])2 = 14$

A general formula can be created $\mathbb E(1) = X$
$\mathbb E(2) = (X+1)X$
$\mathbb E(3) = \big((X+1)X+1\big)X$
and so on.

$$\mathbb E(N) = \sum_{i=1}^N X^i = \frac{X(X^N-1)}{X-1} $$

I have made a video explanation and made a simple formula at the end

https://youtu.be/d72rcsBblRE

Too irRational
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  • Please don't answer old questions if your answers have nothing more to add to the ones already existing. – Alex M. Oct 15 '20 at 13:41