Here's the situation: I have a (fair) die, which I roll successively until I get 5 consecutive ones, in which case I stop rolling. My questions are:

  1. What is the probability that I stop after exactly 11 throws? (or $k$ throws in general?)
  2. What is the probability that I roll the die at least 9 times? (or $k$ times in general?)
  3. More generally, is there a name for the kind of probability distribution involved?

For the first question, I'm thinking of the following string: $$ABCDEZ11111$$ where $ABCDE$ can be anything other than $11111$, and $Z$ can be anything other than $1$. The probability is therefore $$\left(1-\left(\frac{1}{6}\right)^5\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)^5$$ by this reasoning (or is it?).

I have no idea how to approach the second question. Is it valid to use the inclusion-exclusion principle there?

Edit: After a bit of searching, I found this in Enumerative Combinatorics Volume 1, exercise 44 in Chapter 4:

Exercise 44 question

and the solution:

Exercise 44 answer

  • $\begingroup$ You did question one for general $k$? Use it for question 2! The point is: You roll at least nine times, if you roll exactly nine times or exactly 10 times or ... (and these events are mutually exclusive!). $\endgroup$ – martini Oct 26 '12 at 7:32
  • $\begingroup$ @martini: No, I didn't do question 1 for general $k$. I think there's something special about the fact that $11 = 2 \cdot 5 + 1$, which gives you the nice $Z$ in the middle of the 11-letter string (if that line of reasoning is even valid). If $k = 1000$ then I know that the last 5 letters must be 1s, but beyond that I'm lost (maybe tedious inclusion-exclusion?). $\endgroup$ – wj32 Oct 26 '12 at 7:35
  • $\begingroup$ You're right ... seems like it's inclusion-exclusion ... $\endgroup$ – martini Oct 26 '12 at 7:48
  • $\begingroup$ You might find some usefull information in this question: math.stackexchange.com/questions/112726/… $\endgroup$ – Jean-Sébastien Oct 27 '12 at 1:30
  • $\begingroup$ @Jean-Sébastien: Thanks. I think this case is a bit more complicated though, because you cannot simply force the string to alternate between two symbols :( $\endgroup$ – wj32 Oct 27 '12 at 1:33

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