Repeatedly rolling a fair 6 sided die

Question

Suppose you repeatedly roll a fair 6 sided die. What is the probability that you obtain at least one 5 before the first 6? Will this involve an infinite sum?

Answer 1

It is easier to calculate the complementary event, namely $A$, that is: the probability that you don't roll a five before the first six. Then this probability is an infinite sum, that is, the sum over $n$ that the probability that you get the first six in the $n$-th roll and no five in the previous rolls, namely the event $A_n$, i.e.:

$$ \Pr [A]= \sum_{n\geqslant 1}\Pr [A_n]=\sum_{n\geqslant 1}\left(\frac46\right)^{n-1}\frac16=\frac16\cdot \frac1{1-\frac46}=\frac12 $$ Then the probability that you want is $1-\Pr[A]=\frac12$.

Alternatively, following the comment of @David you can notice that the probability to get a five before a six is the same probability to get a six before a five, and each one is the complementary event of each other.

Answer 2

Rolling a 5 or a 6 are equally likely. In this very simple joint probability space, that makes the events of rolling a 5 or 6 essentially interchangeable. You could put a sticker on the 5 and 6 sides that read "6" and "5" respectively, and nothing at all would change about your experiment. Because of this symmetry, the chance of rolling a 5 before a 6 must be the same as rolling a 6 before a 5, which is therefore 50%.

Answer 3

If you roll anything other than a $5$ or a $6$ then the process effectively starts over, with no effect whatsoever on the probability of rolling a $5$ before a $6$. On the other hand, if you roll a $5$, you might as well stop rolling since you know that, whenever you do manage to roll a $6$, it will have happened after rolling a $5$. We therefore only really care about a die roll that stops the process, i.e., if we roll either a $5$ or a $6$. These are equiprobable, so...