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I understand one way to solve is as follows:

$2\cdot(\frac{5}{6})(\frac{1}{6})^2 + (\frac{1}{6})^3$

  • The first term captures the probability of rolling two sixes in the first and second or second and third terms
  • The second term captures the probability all three rolls as sixes

But I can't seem to solve it with counting in an elegant way.

I thought one way might be to set a denominator as:

${6\choose1}^3$

  • There are 6 possible values for every roll. We have three rolls. So by multiplication rule there are $6^3$ permutations.

Then the numerator gets a bit confusing.

We'd count

$666$
$66X$
$X66$

X here can be one of 5 values.

So it's $5*1 + 5*1 + 1$ as the numerator, again using multiplication law.

Is there a simpler way to get here? I feel like I've jumped through one too many hoops.

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  • $\begingroup$ Your first term of $\frac{5}{6}\cdot\frac{1}{6^2}$ is indeed the probability of rolling two sixes in the first and second slot. If it is being used for that purpose however you will need another $\frac{5}{6}\cdot\frac{1}{6^2}$ for the separate case of it being in the second and third slots. That is to say, the answer should have been $\color{red}{2}\cdot \frac{5}{6}\cdot\frac{1}{6^2} + \frac{1}{6^3}$ $\endgroup$
    – JMoravitz
    Sep 15, 2021 at 14:23
  • $\begingroup$ As for your approach from counting... that is exactly the same answer and logic as you should have had for your first approach... just organized differently. I wouldn't call it as having "jumped through hoops" to get there, it is really quite standard. As for multiplying by $1$... that is often omitted since $a\cdot 1 = a$, it saves effort to just write $a$ instead of $a\cdot 1$. Similarly, $\binom{6}{1}=6$ we don't usually write this as a binomial coefficient. $\endgroup$
    – JMoravitz
    Sep 15, 2021 at 14:26
  • $\begingroup$ I've edited the missing multiplication @JMoravitz $\endgroup$
    – Learning stats by example
    Sep 15, 2021 at 14:36
  • $\begingroup$ @JMoravitz That makes sense. I added those additional details as at least for me they clarify the underlying logic of the methods. I just wish I could make it feel a bit better. Feels so mechanical. $\endgroup$
    – Learning stats by example
    Sep 15, 2021 at 14:38

2 Answers 2

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Method 1

To get at least two consecutive sixes, the second roll must be six. Then you win if either the first or last roll is six as well. Using complement:

$$ \frac{1}{6}\cdot\left(1-\left(\frac{5}{6}\right)^{2}\right)=\frac{11}{216} $$

Method 2

Probability of getting consecutive sixes in the first two rolls is $\frac{1}{36}$, the same with consecutive sixes in the last two rolls. The probability of getting all sixes is $\frac{1}{216}$. Using Principle of Inclusion and Exclusion:

$$ \frac{1}{36}+\frac{1}{36}-\frac{1}{216}=\frac{11}{216} $$

Method 3

Say we have two sixes and one non six, there are $5$ possibilities for the non six. Notice that this non six must be the first or the third roll. If we have three sixes there is only one possibilities.

$$ \frac{2\times 5+1}{6^{3}}=\frac{11}{216} $$

Method 4

We will not get two same numbers in a row if both the first roll and third roll differ from the second roll, probability $\frac{5}{6}\times\frac{5}{6}$. The probability of getting two same numbers in a row is then the complement: $1-\frac{25}{36}$. The probability that the same numbers in a row is $6$ is then:

$$ \frac{1}{6}\cdot\left(1-\frac{25}{36}\right)=\frac{11}{216} $$

I believe there are other alternatives as well.

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$$\underbrace{\frac{1}{6}\times \frac{1}{6}\times\frac{5}{6}}+\underbrace{\frac{5}{6}\times\frac{1}{6}\times\frac{1}{6}}+\underbrace{\frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}}=\frac{11}{216}$$

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