*Method 1** Using number of outcomes*

Rolling 3 dice has 6^{3} = 216 possible outcomes. You can find this by multiplying the number of possible outcomes for each roll, using a tree diagram or using combinations (see the chapter Permutations and Combinations).

Assume that rolling the dice happens one die at a time.

**a)**

Rolling the first die has 6 possible outcomes.

There are 5 outcomes that do not match the first value. There are 4 outcomes that do not match either of the first 2 values.

\dfrac{6 × 5 × 4}{6^{\displaystyle{3}}} = \dfrac{5 × 4}{6^{\displaystyle{2}}} = \dfrac{5}{9}\\[3ex]
The probability none of the dice match is 5/9.

**b)**

Rolling the first die has 6 possible outcomes.

There is 1 outcome that matches the first value for each of the other 2 dice.

\dfrac{6 × 1 × 1}{6^{\displaystyle{3}}} = \dfrac{1}{6^{\displaystyle{2}}} = \dfrac{1}{36}\\[3ex]
The probability all 3 of the dice match is 1/36.

**c)**

There are 3 pairings where the values can match:

first and second first and third second and third

So we will calculate the probability of one of these pairs occurring, and multiply that by 3 to get the total probability that any one of the three will occur.

Rolling the first die has 6 possible outcomes.

For the remaining two dice, one value must match and one value must not match that first die.

\dfrac{6 × 1 × 5}{6^{\displaystyle{3}}} = \dfrac{5}{6^{\displaystyle{2}}} = \dfrac{5}{36}\\[3ex]
The probability that one pair of the dice match is 5/36.

So the probability that any of the 3 pairings of the dice match is:

3(\dfrac{5}{36}) = \dfrac{3 × 5}{3 × 12} = \dfrac{5}{12}

*Method 2* Using probabilities

Assume that rolling the dice happens one die at a time.

**a)**

Any value can be the first value, so it does not affect the probability.

The probability that the second roll does not match the first is 5 values out of 6 possible values, so 5/6.

The probability that the third roll does not match either of the two first rolls is 4 values out of 6 possible values, so 4/6.

1 × \dfrac{5}{6 }× \dfrac{4}{6} = \dfrac{5 × 4}{2 × 3 × 2 × 3} = \dfrac{5}{9}

The probability none of the dice match is 5/9.

**b)**

Any value can be the first value, so it does not affect the probability.

The probability that the second roll matches the first is 1 value out of 6 possible values, so 1/6.

The probability that the third roll matches the two first rolls is also 1 value out of 6 possible values, so 1/6.

1 × \dfrac{1}{6} × \dfrac{1}{6} = \dfrac{1}{36}

The probability all 3 of the dice match is 1/36.

**c)**

There are 3 pairings where the values can match:

first and second first and third second and third

One die can have any value, so it doesn’t affect the probability.

Of the remaining two dice, one must match and one must not match that first die.

The probability of a match is 1/6. The probability of not matching is 5/6.

1 × \dfrac{1}{6} × \dfrac{5}{6} = \dfrac{5}{36}

The probability that any single pair of the dice match is 5/36.

There are 3 possible pairings.

So the probability of having exactly 2 dice match is 3(5/36) = 5/12.

**Note:** For (c), there is another method. If figuring out the options for the pairings is not clear to you, you can calculate the probabilities for (a) and (b) and subtract from 1.

probability that exactly 2 match = 1 – (probability that none match or 3 match)

probability that exactly 2 match =

1 – ( \dfrac{5}{9} + \dfrac{1}{36} ) = 1 \,-\, ( \dfrac{20}{36} + \dfrac{1}{36} ) = \dfrac{36}{36} \,-\, \dfrac{21}{36} = \dfrac{15}{36} = \dfrac{5}{12}