Calculate Expected Value in dice game.

Question

We play a game in 2 stages:

In stage one, we throw a dice until we get number 6, Let N represent the number of times played until we got 6 for the first time.

In stage two, we throw N dices (each one only once).

Question: Let $X$ represent the sum of results we got in stage 2, calculate $E(X|N=n)$:

What I know? I know that $N$ is $\operatorname{Geo}(1/6)$ and this $E(N)=1/(1/6)=6$ to continue I need to know the distribution of $X|N=n$, can I get help?

Was asked already Expected value of the sum of n dice rolls., you only need linearity of expectation. — Vepir, Dec 25 '20 at 18:48
That is different question, I am talking about conditional expected value which is making the whole issue of formality here — MrCalc, Dec 25 '20 at 19:12
If you don't want to use linearity, you can look at Conditional expectation: Computation, and use $$E(X|N=n)=\sum_{j=n}^{6n}j\cdot P(X=j,N=n),$$ where $P(X=j,N=n)=\frac{1}{6^n} \sum\limits_{k=0}^{\lfloor (j-n)/6 \rfloor} (-1)^k \binom{n}{k} \binom{j-6k-1}{n-1}$ according to Probabilies of rolling n dice to add up to a specific sum. When I plug this into Mathematica, I get the simplification $E(X|N=n)=3.5n$ as expected. — Vepir, Dec 25 '20 at 19:57

Joe · Accepted Answer · 2020-12-25T20:01:05.693

3

If we throw $n$ dice, then the expected value of their sum is $3.5n$. This follows directly from the fact that the average score on one die is $3.5$ (and expectation is linear).

Let $A_i$ equal the outcome of the $i$th roll of the die. $E(A_i)$ can be calculated in the following way:$$\frac{1+2+3+4+5+6}{6}=3.5 \, .$$ Let $B$ equal the sum of $n$ rolls. \begin{align} E(B)&=E(A_1)+E(A_2)+\ldots+E(A_n) \\ &= \underbrace{3.5 + 3.5 + \ldots + 3.5}_{\text{$n$ times}} \\ &= 3.5n \, . \end{align}

edited Dec 25 '20 at 20:01

answered Dec 25 '20 at 18:32

Joe

22,603

Sorry, but this isn't a formal answer – MrCalc Dec 25 '20 at 18:33
@MrCalc Is everything ok now? – Joe Dec 25 '20 at 18:39
Your point is clear but I need to calculate E(Y|N=n) not E(Y) – MrCalc Dec 25 '20 at 18:41
@MrCalc That's what I just did, no? I assumed that you were rolling it $n$ times, and calculated the expectation. – Joe Dec 25 '20 at 18:42
@MrCalc Sorry, I used a different notation compared to you. Maybe that's what's confusing. – Joe Dec 25 '20 at 18:43
then you should change E(Y) accordingly, Plus I learnt that in order to calculate E(|=) I need to know what kind of distribution |= has (which you didn't mention) – MrCalc Dec 25 '20 at 18:44
@MrCalc I've changed my notation because it conflicted with yours. The expectation of $B$ does change according to the value of $n$. If $n=1$, the expectation is $3.5$, if $n=2$, the expectation is $7$, etc. The underlying distribution is a discrete uniform one, but I think you are getting bogged down by formality and might be overcomplicating things... – Joe Dec 25 '20 at 18:46
|= isn't uniform for sure. I can prove this – MrCalc Dec 25 '20 at 18:50
@MrCalc No, $B|N=n$ is not uniform, but $A_i$ is, and that's all you need. Just try to think about things intuitively. If I roll $100$ dice, what do you think the average of the rolls is going to be? – Joe Dec 25 '20 at 18:52
I understand you, but again how may I know what kind of distribution |= has? – MrCalc Dec 25 '20 at 18:53
Let us continue this discussion in chat. – MrCalc Dec 25 '20 at 18:57
@MrCalc I don't think there's a name for this distribution. See here. But you don't need to know what the distribution is. Since you know the distribution for a single roll, you can work out the expected value for $n$ rolls. Linearity of expectation is key here. – Joe Dec 25 '20 at 18:58

Calculate Expected Value in dice game.

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