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Given a natural number $k$, roll a fair die until the sum of the outcome is more than $k$.

  1. What is the expected number of rolls?
  2. What is the expected number of rolls if we stop after $t$ times even if we did not reach the desired sum?

I've been asked to clarify that this is not homework. Ways I've tried to takle this: I've tried to use the law of total expectation to develop a recurrence formula but I did not reach far enough, and also if I had known that the sum is I could used some combinatorics for the number of rolls that sums to .

Proper Illumination
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    Hi! To avoid down-votes and close-votes, please provide us some context for this question, such as: (a) Is this homework? (b) If so, what course are you taking? (c) What specific topic are you covering at the moment? (d) What do you know that you think might be connected? (e) If you're stuck, what are you stuck on? For example, do you know what to apply, but don't know how to apply it, or do you not know what to apply? Please put these facts in your original post, not as responses to this comment, as comments may be deleted without warning. – Brian Tung Oct 09 '22 at 21:11
  • It is not homework, just an interest. Regarding the topic: I'm not sure, I've tried to use the law of total expectation to develop a recurrence formula but I did not reach far enough, and also if I had known that the sum is $k$ I could use some combinatorics for the number of rolls that sums to $k$. – Proper Illumination Oct 09 '22 at 21:51
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    @ProperIllumination: Good! Can you put that context into your original post? – Brian Tung Oct 10 '22 at 00:13
  • @Masacroso I don't think that Wald's Theorem is relevant, it says that $\mathbb{E}\left(\sum_{i=1}^{N}X_i\right)=\mathbb{E}(N)\mathbb{E}(X_i)$, where $X_i$ are i.i.d. and $N$ is a random variable. I'm asking how to compute $\mathbb{E}(N)$ – Proper Illumination Oct 11 '22 at 11:16

2 Answers2

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Sketch for a solution for the first question: let $\tau :=\min\{n\in \mathbb{N}: S_n\geqslant k\}$ for $S_n:=\sum_{k=1}^n X_k$, then

$$ \operatorname{E}[\tau ]=\sum_{m\geqslant 1}\Pr [\tau \geqslant m]=\sum_{m\geqslant 1}\Pr [S_{m-1}<k] $$

and note that the sum on the right is finite as $\Pr [S_n<k]=0$ when $n\geqslant k$. By last you can use this answer to compute every probability of this sum.

An approximated solution could be found using Wald's theorem and the approximation

$$ \operatorname{E}[S_{\tau }]\approx \frac1{6}\sum_{j=0}^5 (k+j)=k-1+\operatorname{E}[X_1] $$

what is fine for enough large $k$, then you will get

$$ \operatorname{E}[\tau ]\approx \frac{k-1+\operatorname{E}[X_1]}{\operatorname{E}[X_1]} $$

Masacroso
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  • $E[X_1]=3.5$, so I don't get why you left it as $E[X_1]$ – Proper Illumination Oct 17 '22 at 20:17
  • I don't get the first line of the approximation. Here is what I have I don't get the first line of the approximation. Here is what I have $$ \mathbf{E}[S_\tau]=\mathbf{E}\left[\sum_{k=1}^{\tau}X_k\right]=\mathbf{E}[\tau]\mathbf{E}[X_1]=6\sum_{m\geqslant 1}\Pr [S_{m-1}<k]= 6 \sum_{m\ge1}^{ }\sum_{\ell=1}^{k-1}\Pr[\ell,m-1,6), $$where Pr[,,] is from the source you cited, and the second equality follows from Wald's Theorem. – Proper Illumination Oct 17 '22 at 20:20
  • @ProperIllumination observe that $S_{\tau }\in{k,k+1,\ldots ,k+5}$, therefore $\operatorname{E}[S_{\tau }]\in[k,k+5]$ and I'm guessing that for large $k$ we have that $\Pr [S_{\tau }=k+j]$ is almost the same for all $j\in{0,\ldots ,5}$. In any case you have the bound $\operatorname{E}[\tau ]\in [k/\operatorname{E}[X_1], (k+5)/\operatorname{E}[X_1]]$ – Masacroso Oct 17 '22 at 22:42
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Let $X_n$ be the $n^{th}$ roll, and let $S_n$ = $\sum_{j=1}^{n} X_j$ with $T_k = \inf\{j \geq 1 : S_j \geq k\}$.

Also, define $h_k = \mathbb{E}(T_k)$. Then by conditioning on the first roll, we have the simple recurrence:

$$h_1=1, h_k = 1 + \frac{1}{6}\sum_{i = 1 }^ {k-1} h_i \ \text{when}\ 1 < k \leq 6 \\ h_k = 1 + \frac{1}{6}\sum_{i = k - 6 }^ {k-1} h_i \ \text{when}\ k \geq 7 $$

which of course determines $h_k$ completely $\forall k \geq 1$.

The first equation determines $h_{i}$ for $ 1 \leq i \leq 6$, and from there $h_k$ is determined entirely by :$$h_{k+6} = 1 + \frac{1}{6}\sum_{i = k }^ {k+5} h_i \ \text{when}\ k \geq 1 $$

A particular solution is given by $h_n = 2n/7 = n/\mathbb{E}(X_1)$.

The characteristic polynomial is given by $c(t) = 6t^6 - 1 -t -t^2 -t^3 - t^4 - t^ 5$, and has six distinct roots approximately given by the set $\{1,-0.670332,-0.375695 \pm 0.570175 i, 0.294195 \pm 0.668367 i\}$

Hence, the recurrence has general solution : $h_n = 2n/7 + a_1 + a_2(-0.670332)^{n}+a_3(-0.375695)^{n}\cos(0.375695 n) + a_4(-0.375695)^{n+1}\sin(0.375695 n) + a_5 (0.294195)^{n}\cos(0.294195n) + a_6(0.294195)^{n}\sin(0.294195n) $

,where the coefficients are determined by the known values $h_1,h_2, h_3, h_4, h_5$ and $h_6$.

Note that for large $n$ this is linear, and it agrees with Masacroso's answer.

Aditya Dhawan
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