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My question

Lets say we have a fair 6-sided die that we roll continuously. We roll this same dice 10 times in a row. What is the probability that we get a 3 after our 4th roll but before our 10th roll?

In other words, given the events

$$R_1,R_2,R_3,R_4,R_5,R_6,R_7,R_8,R_9,R_{10}$$

What is the probability that

$$R_5,R_6,R_7,R_8,R_9$$

We get our first 3?

My attempt

$R_1,R_2,R_3,R_4$ cannot be 3s. This means that they have probabilities $\frac{5}{6}$.

The first 3 appearing somewhere here $R_5,R_6,R_7,R_8,R_9$ is probably something like $(5 \text{ choose } 1) * 6^4$ (I have no reasoning, just intuition).

I have no idea how to handle the last part of the problem with "before $R_{10}$"...

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I am making the assumption that you're going to roll the dice 10 times, even if you manage to roll a 3 at a desired location (e.g. if the first 3 appears at 7, you continue rolling to the 10th dice.)

To deal with $R_5, R_6, R_7, R_8, R_9$, observe that we can model this with the binomial distribution. Let X denote the number of 3's in the rolls $R_5, R_6, R_7, R_8, R_9$.

$$ X \sim Bin\Big(n=5, p=\frac{1}{6}\Big) $$

(Where we have 5 tries and the probability of rolling a $3$ on any of these tries is $\frac{1}{6}$)

We want the probability:

$$ Pr(X > 0) = 1- Pr(X=0) = 1 - \Big(\frac{5}{6}\Big)^5 $$

So, multiplying this with the probability that you arrived at for not rolling a 3 in the first 4 rolls, we obtain the final result: $$ P(\text{No 3's in the first four rolls} \cap \text{at least one 3 in rolls 5 to 9} ) = \Big(\frac{5}{6}\Big)^4 \Big( 1 - \Big(\frac{5}{6}\Big)^5 \Big) $$

Which I believe works out to be $\approx0.288$

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  • $\begingroup$ How does "but before our 10th roll?" factor into your answer? $\endgroup$ – AlanSTACK Jan 24 at 3:11
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    $\begingroup$ I'm assuming that you do not care about the 10th roll - whether or not the 10th roll is a 3 or otherwise, it's irrelevant. Only what happens in the first 9 rolls is relevant. $\endgroup$ – Sean Lee Jan 24 at 3:13
  • $\begingroup$ Does that part of the question not change the probability? $\endgroup$ – AlanSTACK Jan 24 at 3:15
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    $\begingroup$ To put it another way, success consists in getting a 3 in at least one of the trials $R_5,R_6,R_7,R_8,R_9$ after not getting a 3 in any of the trials $R_1,R_2,R_3,R_4$. So the solution doesn't require information about the outcomes of the tenth or later rolls. $\endgroup$ – hardmath Jan 24 at 3:19
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    $\begingroup$ @hardmath Thank you, that cleared it up for me. $\endgroup$ – AlanSTACK Jan 24 at 3:25

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