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So I have stumbled upon this question recently and though I solved it I'm not sure if it's correct or not.

A fair $6$-sided die is rolled repeatedly until a $6$ is obtained. Find the expected number of rolls conditioned on the event that none of the rolls yielded an odd number.

My attempt:

Let $X$ be the random varibale denoting the no of rolls required until a $6$ is obtained. Without any given condition $X \sim Geom(\frac16)$ But $$\mathbb{P}(\text{obtaining a }6 | \text{roll does not yield an odd number})= \frac13$$ so, $$X | \text{none of the rolls yielded an odd number} \sim Geom\bigl(\frac13\bigr)$$ and hence $$\mathbb{E}(X | \text{none of the rolls yielded an odd number})= 3$$

Is this the correct answer? Is there any other approach? Any help would be very much appreciated!

Prof. Shanku
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1 Answers1

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Define random variable $X$ to be the number of rolls until the first 6. Let $A$ be the event that all rolls yield even numbers. We want to compute $E(X|A)$. By conditioning on the first roll $k$ that yields an odd number (if any), we have $$E(X) = E(X|A) P(A) + \sum_{k \ge 1} E(X|\text{first odd at roll $k$}) P(\text{first odd at roll $k$}).$$ Because $X \sim \text{Geom}(1/6)$, we have $E(X) = 6$. By conditioning on whether the first roll is in $\{6\}$, $\{2,4\}$, or $\{1,3,5\}$, we have $P(A) = 1(1/6) + P(A)(2/6) + 0(3/6)$, so $P(A) = 1/4$. Now $$E(X|\text{first odd at roll $k$}) = k + E(X) = k + 6,$$ and $$P(\text{first odd at roll $k$}) = (2/6)^{k-1} (3/6).$$ Substituting everything into the first equation yields $E(X|A) = 3/2$.

RobPratt
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