# Is my reasoning for this probability question correct?: Four fair dice are rolled. What is the probability of getting two different pairs?

Four fair dice are rolled. What is the probability of getting two different pairs?

I basically used the multiplication rule here: the first die can have $6$ possible outcomes, while the second one can have only $5$. The last two can only have one outcome each because they have to match the numbers on the first and second dice. Hence the numerator becomes $6 \cdot 5 \cdot 1 \cdot 1$. Then I divided by all possible outcomes, which is $6^4$. So the probability $= 30/6^4 = .023$.

• Why can't the first two dice show the same number? – saulspatz Jul 19 '18 at 3:53
• Suppose the dice are colored red, blue, green, white. The total $6\cdot 5\cdot 1\cdot 1$ refers to the number of outcomes where very specifically the red and green dice show the same number and the blue and white dice show the same number but different than the red die. This is a very restrictive scenario and is not required to be the case. You could have the red and blue be a pair, or you could have the red and white be a pair as well. – JMoravitz Jul 19 '18 at 4:14
• There are three different ways in which the dice can be paired together, not just the one way. As a result, your answer is off by a factor of three. – JMoravitz Jul 19 '18 at 4:16
• Please read this MathJax tutorial, which explains how to typeset mathematics on this site. – N. F. Taussig Jul 19 '18 at 8:54
• Im not sure what you mean by three ways: is it 6-5-1-1, 6-1-1-5, and 6-1-5-1? – David Jul 20 '18 at 2:03

I think that another way to solve the problem is to assume that the outcome of the four dice can be expressed as two pairs, one of them with 6 different outcomes (from 1 to 6) and the other one with just 5 different outcomes (because you can't repeat the outcome you got for the first pair). So, we get 30 results. Then, due to the fact that we will have outcomes that can be expressed as (XX)(YY) or (YY)(XX) which have the same results, we divide the total outcomes by 2, such that we have only outcomes composed by "unique pairs" (e.g. if you have an outcome (XX)(YY), you are not going to consider, by the moment, as a different outcome (YY)(XX)). In every outcome, you have 4 elements, but there are only two different pairs, so you will have $\frac{4!}{2!2!}=6$ different orderings. In the end, you will have $15\times6=90$ different outcomes that meet the rule, so the probability will be $\frac{90}{6^4}$ or 0.069.

• You can write 15 \times 6 = 90 when you are in math mode to obtain $15 \times 6 = 90$. Alternatively, you could write 15 \cdot 6 = 90 to obtain $15 \cdot 6 = 90$. – N. F. Taussig Jul 19 '18 at 8:39
• N.F. Taussig, thank you for the feedback. – Carlos Terrones Jul 19 '18 at 15:36

Fixing the places for one pair automatically fixes the places for the other pair, so

$\frac{[Choose\; places\; for\; one\; pair] \times [Choose\; two\; denominations\; for\; pairs\;]}{[Total\; ways]} = \frac{\binom42\times\binom62}{6^4}$

Hint:

The event of getting two different pairs have three 'types'.

• All four different

• Three same, one different

• Two same, other two different