Three fair cubical dice are rolled.What is the probability that exactly two dice have the same score?Suppose the three dice are rolled one after another.If two dice have different scores,what is the probability that two of three dice have the same score?

See below:

Explanation:

Probability of 2 with the same value rolled

Since the dice are identical, we can't simply apply a rule such as ${6}^{3} = 216$ possible rolls because this will include duplicates. Instead, we can take the identical dice and place them in baskets (the first basket being the value 1, the second basket being the value 2, etc). This leads to using the stars and bars method of solving the question.

Here we'll denote the dice as d and the box sides as |.

A sample of the arrangement is:

d||||dd|

For each grouping of 2 of the d, there are 5 ways to achieve it. There are 6 possible bins, so we have $5 \times 6 = 30$ ways.

The total number of ways to arrange the 3 d's and the 5 |'s is:

$\left(\begin{matrix}8 \\ 3\end{matrix}\right) = 56$

which means the probability of getting two identical numbers and one different is:

$\frac{30}{56} = \frac{15}{28} \approx 0.5357$

To check this, let's look at the number of triples and then the number of 3 unique numbers:

Triples:

$\frac{6}{56}$

Three Unique:

1, 2, 3
1, 2, 4
1, 2, 5
1, 2, 6

4 ways. By extension, there are $4 + 3 + 2 + 1 = 10$ ways with the first die being a 1.

This progression of "losing one way" for each digit that moves to the right continues, so with the first die being:

• a 2, we have 6 ways,
• a 3, we have 3 ways
• a 4, we have 1 way

Or in total, 20 ways total.

Our check then is that we have $20 + 30 + 6 = 56$ total ways to arrange the 3 dice.

2 rolled different, then probability of having two the same

In this case, we have 2 values already (say a 1 and a 2). This means that on the third roll we'll be able to see if we have two numbers the same. That's 2 acceptable rolls within a possible 6 results, so that's $\frac{2}{6} = \frac{1}{3}$.

If we're interested in the probability of the entire series happen (roll first die, roll the second die getting a different value, rolling a third and getting two values to match), then we first need to look at the probability of the first two rolls not being the same, which is $\frac{5}{6}$. In total, then, we'd have $\frac{1}{3} \times \frac{5}{6} = \frac{5}{18}$