A fair die is repeatedly tossed. What is the probability that a five is seen before any of the even numbers are seen?

I have my own solution below and just want someone to verify it.

According to the problem, it means that only 1's and 3's are allowed to appear before the first five. And now suppose there are $n$ rolls before the first five. For each of the $n$ rolls, there are two possibilities of 1 or 3. Therefore, there are totally $2^n$ different outcomes before the first five. And the probability for each outcome to happen is simply $\frac{1}{6}^n$. After all, the probability of getting the five at the last is 1/6, then the probability for one fixed $n$ is $$ 2^n \left(\frac{1}{6}\right)^n\frac{1}{6} $$ Finally we sum up all the scenarios for each possible $n$ from $0$ to infinity to get $$ \sum_{n=0}^{\infty} 2^n \left(\frac{1}{6}\right)^n\frac{1}{6} = \sum_{n=0}^{\infty} \left(\frac{1}{3}\right)^n\frac{1}{6} = \frac{1}{6}\frac{1}{1-\frac{1}{3}}=\frac{1}{4} $$ Did I do this right? Thank you very much!

  • 2
    $\begingroup$ Looks perfect to me. $\endgroup$ – Amateur Sep 7 '14 at 0:26
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    $\begingroup$ Another way to look at it is "Given that the last roll was one of $\{2,4,5,6\}$, what is the probability that it was a $5$?" It's not hard to see that the answer is $1/4$. $\endgroup$ – JimmyK4542 Sep 7 '14 at 0:28
  • $\begingroup$ or equivalently: the probability that the $k$th roll is $5$, conditional on the event $E_k$ that the $k$th roll is the first one without a $1$ or $3$, is clearly $\frac14$; and the events $E_1,E_2,\dots$ partition the set of all possible events. $\endgroup$ – Greg Martin Sep 7 '14 at 2:05

Community wiki answer so the question can be marked as answered:

As pointed out in the comments, your calculation is correct but unnecessarily complicated, as you can simply argue that the process ends when one of $\{2,4,5,6\}$ is rolled, and the probability for that roll to be a $5$ is $\frac14$.


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