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Roll a standard 4-sided die (with sides labelled 1, 2, 3, 4) and let $X$ be the number rolled. Then take a fair 8-sided die (with sides labelled $1,\ldots,8$) and roll it repeatedly until you roll a number strictly larger than $X$. Let $Y$ be the number of times you roll the 8-sided die, not including the last roll (i.e., the number of times you roll a number less than or equal to $X$).

i. Write down the conditional distribution of $Y$ given $\{X = x\}$, including parameter(s). ii. Find $E(Y | X)$

My answer

Given the rolls of the second die are trials with "successes" defined to mean rolling higher than $X$ conditional on $\{X = x\}$, the trails have a constant success probability $P = 8-x/x$ per trail, each roll is independent, thus $Y$ follows a geometric distribution with $P = (8-x)/x$.

$$E(Y|X=x) = 1 - p / p = x / 8 - x$$ $$E(Y|X) = X / 8 - X$$

I'm struggling to find $E(Y)$.

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    $\begingroup$ $p=1-x/8$. $8-x/x =7$. $(8-x)/x > 1$ given $x = 1,2,3,4$. So both of them are not probability. So try to fix $p$ first. $\endgroup$
    – user158565
    May 31, 2017 at 22:53
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    $\begingroup$ Add the self study tag. $\endgroup$ May 31, 2017 at 23:11
  • $\begingroup$ Small detail. I believe you need to use $p=\frac{8-x}{8}$ as succes probability instead of $p=\frac{8-x}{x}$. $\endgroup$ Nov 22, 2022 at 8:17

1 Answer 1

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On any roll of the 8-sided die, the success probability is 1/2 that is getting 5,6,7 or 8 when $X=4$. So let $n$ equal the number of rolls before a success which is 1 less than the number of rolls to a success. Then $P(n=0)=1/2$, $P(n=1)=(1/2)^2$, $P(n=2)=(1/2)^3$ and so on.

To get $E(Y|X=4)$ take $y P(n=y)$ and sum over all $y$ from 0 to infinity. This is conditional on $X=4$. If $X=1$, $p=7/8$. If $X=2$, $p=3/4$. If $X=3$, $p=5/8$.

Repeat the process for 1, 2, and 3 and you will get all the conditional probabilities and expectations. To get the unconditional probability take $P(Y=y|X=x)P(X=x)$ summed for $x=1,2,3$ and 4. Note that $P(X=x)=1/4$ for each $x$ between 1 and 4. Then take the sum $y P(Y=y)$ for $y$ between 0 and infinity.

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  • $\begingroup$ You lost me at the first sentence: it appears to assume $X=4$. What if $X$ is not $4$? $\endgroup$
    – whuber
    May 31, 2017 at 22:55
  • $\begingroup$ Sorry I forgot about the conditioning on X. I will edit the answer. $\endgroup$ May 31, 2017 at 22:59

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