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A die is rolled repeatily. Let $X$ be the random variable that denotes the number of rolls to get a 4 and $Y$ be the random variable that denotes the number of rolls to get a 1. What is $E[X|Y=7]?

My thoughts were $\dfrac{1}{\dfrac{1}{6}} + 7$ since the expected value for rolling a 4 is 6 and we are given that we rolled 7 times (but we know on the 7th roll we did not get a 4)) but I know the answer is not right. Since we must factor in the probabilites of rolling a 4 in the first 6 rolls. How do I do this?

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    $\begingroup$ Is $X$ the number of rolls needed to obtain a 4? similarly, for $Y$? $\endgroup$ – Nana Dec 18 '13 at 6:20
  • $\begingroup$ ahhh yes I misworded $\endgroup$ – adam Dec 18 '13 at 6:23
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Hint: Note that $X$ is a geometric random variable. $Y=7$ implies that rolls one through to 6 was not a $1$. So we can consider two cases: $X \le 6$ and $X\gt 7$

By definition $$\begin{align*} E(X \, | \, Y = 7) & = \sum_{k=1}^{\infty} \, k \,P(X = k \, | \, Y = 7)\\ & = E(X \, | \, Y = 7, X\lt 7) \cdot P(X \le 6 \, | \, Y = 7) \\ &\,\,\,\,\,\,\,\,\,\,\,+ E(X \, | \, Y = 7, X\gt 7) \cdot P(X \gt 7 \, | \, Y = 7) \end{align*}$$

Can you take it from here?

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  • $\begingroup$ is this bayes theorem or what is this $\endgroup$ – adam Dec 18 '13 at 7:31
  • $\begingroup$ I just want to know what property we used to come to this conclusion need a little more explanation as to how you come to this equation. I get that you need to consider the two cases. But the means on how you consider them is confusing $\endgroup$ – adam Dec 18 '13 at 7:44
  • $\begingroup$ @adam: I used the so called partition theorem which says that if $A_n$ is a partition of a sample space the $E(Z) = \sum_{n}E(Z|A_n)P(A_n) $ for a random variable $Z$. $\endgroup$ – Nana Dec 18 '13 at 7:57
  • $\begingroup$ ahhh ok thanks. And the commas in the equation? what do they represent? the constraints on $X$ how would I read this in plain english $\endgroup$ – adam Dec 18 '13 at 8:09
  • $\begingroup$ the stuff after the comma, represent another conditioning term. you'd read it in the usual way; e.g. expectation of x given that y=7 and x is less the 7...etc $\endgroup$ – Nana Dec 18 '13 at 8:53
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Your figure of $6+7$ is the expected number given that there is no $1$ in the first seven rolls. The probability of this given the seventh roll is a $4$ and the previous rolls are not $4$ is $\left(\frac45\right)^6$. You also need to consider the possibility that earlier rolls are $1$.

So I suspect the answer is $$13 \times \left(\frac45\right)^6+1\times \frac{1}{5}+2\times \frac{4^1}{5^2}+3\times \frac{4^2}{5^3}+4\times \frac{4^3}{5^4}+5\times \frac{4^4}{5^5}+6\times \frac{4^5}{5^6}$$

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  • $\begingroup$ does this include for the case that $X>7$? So I need to consider the case in which one of these rolls is a 1 and the other case that none of these rolls are a 1. But why 13 though? $\endgroup$ – adam Dec 18 '13 at 7:54
  • $\begingroup$ Your $\dfrac{1}{\tfrac{1}{6}} + 7 =13 = E[X|X\gt 7]$ $\endgroup$ – Henry Dec 18 '13 at 7:56
  • $\begingroup$ ahhhh I see because If we didn't roll a 1 in 7 rolls than those rolls are "wasted" thus we just add $E[X]$ to however many times we rolled. I find nana's answer a little complicated im not sure what intution she used though I would like to know the theoretical way of seeing this problem so I may generalize it to $k$ rolls $\endgroup$ – adam Dec 18 '13 at 8:05
  • $\begingroup$ looking at your geometric sum is there a better way to express it? The series is arithmetic but at the same time geometric. Or do I have to use brute force $\endgroup$ – adam Dec 18 '13 at 8:30
  • $\begingroup$ @adam: here it is easier to do the calculation. In general consider the derivative of a geometric series $a+ax+ax^2+ax^3+\cdots$ $\endgroup$ – Henry Dec 18 '13 at 8:33
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Rolling a die is a statistically independent event. The odds of getting any number from $1$ to $6$ is always $1/6$. And the expected value is always $3.5$.

But I think that is not what you are asking. I don't understand your question.

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  • $\begingroup$ I thought the same way but we rolled 7 times but we do not know anything about the other 6 rolls. They have probabilites that have not been factored. $\endgroup$ – adam Dec 18 '13 at 6:05

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