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We roll a regular and fair die repeatedly until two fours in a row come up, then we stop. What is the probability of halting after n rolls?

My attempt at solution:

Let $A_n$ be the event that the $n^{th}$ roll is the first roll where two fours occur consecutively. Let $E_n$ be the event that the $n^{th}$ roll results in landing on a four.

By conditioning on the $(n-1)^{th}$ roll, we see that for $n \ge 2$:

$P(A_n) = P(A_n|E_{n-1})P(E_{n-1}) + P(A_n|E_{n-1}^c)P(E_{n-1}^c)$

Now, $P(A_n|E_{n-1})$ is the probability of rolling a 4 on roll n which is $1/6$ and $P(A_n|E_{n-1}^c) = 0$ since there is no way for $A_n$ to occur if $(n-1)^{th}$ roll is not a four. Thus,

$P(A_n) = (1/6)(1/6) + (0)(5/6) = 1/36$

But this answer does not depend on n and I'm sure it should. I'm not sure where I went wrong

Alborz
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    I think I saw the same question before somewhere but I can't find it... – bobeyt6 Oct 30 '22 at 03:19
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    $P(A_n | E_{n-1})$ is not the probability of rolling a $4$ on roll $n$. What about the fact that before $n$, you should not have had any consecutive fours? That information is not captured in your logic, hence the mistake. – Sarvesh Ravichandran Iyer Oct 30 '22 at 03:25
  • @bobeyt6 There seem to be plenty of questions asking for the expected value of the number of rolls (which I think is $42$), but for the distribution, even in the slightly complex case that one waits for three consecutive rolls of a $4$, the answer requires Markov chains like here. It may be possible that this question can be argued using Markov chain-like reasoning. I don't think it's a duplicate because if we're not allowed to use Markov chains then I actually haven't found a post discussing this question. – Sarvesh Ravichandran Iyer Oct 30 '22 at 03:30
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    @SarveshRavichandranIyer That's what I was afraid of. Thanks makes sense now – Alborz Oct 30 '22 at 16:11

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