Probability Of Rolling 2 Identical Numbers In 3 Rolls

Question

A single fair die is rolled 3 times.

What is the probability of rolling exactly two identical numbers?

I've done the working for if a die is rolled 2 times: There's a 1/6 * 1/6 chance of getting the same number. Then there are 6 different pairs of numbers (1,1) (2,2) etc. This gives a 1/6 chance of getting the same number.

However when another roll is added, obviously the chances would go up, but im getting stuck on how the 1st and 3rd roll could be the same, and then you would need to equate those instances. Sorry if I'm unclear!

Thanks for simplifying the question to: Possible outcome of same numbers being:

1 of the same number, +

2 of the same number, +

3 of the same number,

and then just subtract 1 and 3.

Answer 1

What's the probability they're all equal? ($6$ times ${\frac{1}{6}}^3$, why?)

What's the probability the three throws are all distinct? ($\frac{5}{6} \times \frac{4}{6}$ do you see why?)

What remains is the probability you're looking for: the three throws have outcome set of size $1,2$ or $3$...

Answer 2

There are ${3\choose 2}\cdot 5$ ways to get exactly two $1$s: e.g.: $\underbrace{11\bar{1}}_{5}, \underbrace{1\bar{1}1}_{5},\underbrace{\bar{1}11}_{5}$.

There are $6$ numbers, so: ${3\choose 2}\cdot 5\cdot 6=90$ ways to get exactly two identical numbers.

There are total $6^3=216$ different outcomes.

Hence: $$P(\text{2 same numbers in 3 rolls})=\frac{90}{216}=\frac5{12}.$$