-1

Suppose you have a fair six-sided die. What is the expected value of rolls until a $1$ face up, conditioned on all of the rolls landing face up on odd number?

If $A$ represents: rolling a die $n$ times, at $n$th time we get $1$ faced up, $(n-1)$th we have odd number faced up $B$ represents: we get odd number for all $n$ times $P(A|B) = \frac{[(\frac{1}{2})^{n-1}(\frac{1}{6})]}{(\frac{1}{2})^n}.$

I think $P(A|B)$ represents the probability of rolling a die until obtain the face $1$, conditioned on all of the rolls landing face up on odd number.

I calculate the limit of $P(A|B)$ and get $\frac{1}{3}$;

Based on my undertstanding from this post: What is the expected value of the number of die rolls necessary to get a specific number? $E = 1 + \frac{2}{3} E$ then I guess the expected value is $3$? I am not sure about the answer. Please help. Thanks in advance!

Alessio K
  • 10,599
Squirrel K
  • 1

1 Answers1

1

The only numbers that matter are $1,3,5$, so we can discount any rolls in which these numbers do not arise. Of course of the remaining, admissible, rolls, there is a $1/3$ chance each of these arises.

It is unclear whether the conditions admit a $1$ on the first roll as "acceptable" (since no prior rolls were of odd numbers). Assuming an initial $1$ is not admissible, the only rolls that need be considered start with a $3$ or $5$.

Given that condition, there is a $1/3$ chance any particular odd number arises on a given roll.

So there is a $1/3$ chance the game ends with a $1$ on the second roll.

What is the chance that it ends on the third roll? That arises only if a $3$ or $5$ appears on the second roll (chance = $2/3$) and a $1$ appears on the third roll $1/3$ or $2/9$.

Can you continue to the next roll? Can you generalize from here?

David G. Stork
  • 29,774