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A fair die is rolled repeatedly. Let $X$ be the number of rolls needed to obtain a 6 and $Y$ the number of rolls needed to obtain a 5. I need help in computing $E(X |Y = 1)$ and $E(X|Y=5)$.

I know that $X$ is a geometric random variable, so $E(X)=6.$ My guess for the first one is 7, but I'm not sure. Given $Y=1$, it means that the first roll can't be a 6, so it will take at least 2 rolls to obtain a 6. Is my reasoning right?

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Yes, the first one is $7$, the wasted toss plus the expected waiting time after the first toss. –  André Nicolas Dec 19 '11 at 5:35

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up vote 1 down vote accepted

EDIT : Thanks to Andre, he gave me a slap in the face right when I needed it, i.e. before my applied analysis exam tomorrow. XD I'll make this answer right before I bring anymore shame on me.

My ex-approach for the case $Y=5$ is indeed making everything more complicated. The more easy and non-complicated way to do this is just use the definition of the expectation : $$ \mathbb E(X \, | \, Y = 5) = \sum_{n=1}^{\infty} \, n \, \mathbb P(X = n \, | \, Y = 5). $$ Now we just need to compute those probabilities. We know that $P(X = 5 \, | \, Y = 5) = 0$, so that removes this term. For the first four terms, note that $$ \mathbb P(X = n \, | \, Y = 5) = \left( \frac 45 \right)^{n-1} \left( \frac 15 \right) = \frac{4^{n-1}}{5^n}, n=1, 2, 3, 4. $$ because the condition $Y=5$ only means for $X$ that the first four rolls cannot take the value $5$, hence the value of those rolls become independent and uniformly distributed over $\{1,2,3,4,6\}$.

For the next rolls, $Y=5$ gives information on the first five rolls but no information on the ones after. Thus $$ \mathbb P(X = n \, | \, Y = 5) = \left( \frac 45 \right)^4 \left( \frac 56 \right)^{n-6} \left( \frac 16 \right). $$ The $n-6$ stands for the number of rolls after the first five rolls which are not a 6 before you actually get your first $6$ (which gives the $1/6$ term). Therefore, the series we look at in the expectation can be computed, after the first 5 terms, as the derivative of a geometric series in $\left( \frac 56 \right)$. I don't wanna compute it right now because I am going to mess it up, I am definitively too tired for this.

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Thanks Patrick: I got $\mathbb P(X < 5 \, | \, Y = 5) = \frac{369}{625}$. So $P(X > 5 \, | \, Y = 5)=\frac{265}{625}$. Are these right? I'm also finding difficulty in computing the expectations. Are those 4 and 10 respectively? –  jojo Dec 19 '11 at 5:54
    
Sorry, I meant $P(X > 5 \, | \, Y = 5)=\frac{256}{625}$. If you can give a detailed solution too that'll be nice. Thanks. –  jojo Dec 19 '11 at 6:07
    
I am a little busy right now, don't have time to explicitly compute. I trust Andre Nicolas, he's very usually right. =) –  Patrick Da Silva Dec 19 '11 at 6:17
    
@AndréNicolas: Thanks. I figured my 10 should be 11, using the memoryless property. How do I get $\mathbb E(X \, | \, Y=5, X < 5)$? –  jojo Dec 19 '11 at 6:29
    
@jojo : You can easily determine the distribution of $X \, | \, Y = 5, X < 5$ as a random variable. The condition $Y=5$ does not influence probabilities so only the condition $X<5$ is relevant. How do you determine the distribution of $X \, | \, X < 5$? Hint : $\mathbb P(X = i \, | \, X = 5) = \mathbb P(X=i \cap X < 5) / \mathbb P(X < 5)$. –  Patrick Da Silva Dec 19 '11 at 6:39

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