Problem: Roll a fair 6−sided die until a 6 appears.Given that the first 6 occurs before the first 5, find the expected number of times the die was rolled.
Note: A discussion on the solution already exists in the following link, but that doesn't discuss this alternative approach.
Correct solution using arithmetic-geometric progressions:
$E = \sum_{i=0}^\infty(4/6)^i*(2/6)*(i+1)$,
$E = \sum_{i=0}^\infty(2/3)^i*(1/3)*(i+1)$,
$E = 3$.
Here, $2/3$ is the probability of a throw resulting in a non-terminating number $i.e., 1, 2, 3, or, 4$. If a throw yields $5, or, 6$, we stop. From symmetry, the arrived value of expectation is also our answer.
Argument: Given that $5$ did not occur, since the die is fair, it should be deducible that the probability of occurrence of each of $1$, $2$, $3$, $4$, and, $6$, should be $1/5$.
Solving for $E$:
$E = \sum_{i=0}^\infty(4/5)^i*(1/5)*(i+1)$,
$E = 5$.
The answer is understandably $E = 3$. What exactly is wrong with the above argument, and why? (Which mathematical rule/property is violated?)
Also, consider the following:Given a fair 6-sided die, what is the probability of a throw yielding $1$ if it is known that the throw did not yield $5$?The answer is trivially $1/5$; and is the same for the numbers $1$, $2$, $3$, $4$, and, $6$.
How does the conditional probability of $5$ not occurring fit this example, but not the main problem?
