# Rolling 3 dice and getting exactly 2 to be of the same number

I am reading about rolling a six-sided die and I came up with this equation myself. I know this is wrong but couldn't really figure out why. Could someone please help me understand?

The probability of rolling $$3$$ dice and getting $$2$$ to be the same number $$=\frac{|E|}{|S|}= \frac{6 \cdot 5 \cdot 3!}{6^3}$$

The first $$6$$ is for having $$6$$ different possible choices of our target(AAX results), $$5$$ is for $$5$$ possible choices for the non-pair number and $$3!$$ is the number of possible ways to order the three numbers we rolled (and I think this might be the incorrect part of my equation).

Finally, $$6^3$$ is just all possible outcomes.

• Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. Also, the plural of die is dice (which is an exception to the usual rule of forming the plural by adding an s). English has a lot of exceptions to the rules. Commented Jul 3, 2022 at 17:43
• Thank you so much for the edit! It's my first time posting and clearly, I've got a lot to learn. Will definitely take a look at the tutorial! Commented Jul 3, 2022 at 17:54
• Good question and you got good answers. One thing that you could have done yourself is to think this through with a two-sided die (usually called a coin :)) to see where your solution goes wrong. (You would get 12/8, which is obviously wrong, and listing all 8 possibilities is not a problem.) Commented Jul 4, 2022 at 18:15
• Your $6$ chooses for what the mathing faces and $5$ for what the non matching face for $6\times 5$ possible values. But you choosing the ways to arrange the dice is wrong. As the dice are not all different you must a) either divide by ways to arrange the two faces that are the same to get $\frac {3!}{2!} = 3$. Or b) come up with a simpler model. You only need to choose where the different face go-- the other two will be in the two remaining places. So $3$. The answer is $\frac{6\cdot 5\cdot\frac{3!}{2!}}{3^6} = \frac{6\cdot 5\cdot 3}{3^6}$. Commented Jul 5, 2022 at 3:12
• To illustrate: Suppose you have two $6$ and $5$ and the two sixes are red and blue. You have $3!=6$ ways to do this: $\color{red}6\color{blue}65;\color{blue}6\color{red}6;\color{red}65\color{blue}6;\color{blue}65\color{red}6;5\color{red}6\color{blue}6;5\color{blue}6\color{red}6$. but the switching the red an blue positions is supposed to be the same so we must divide by the numbers of way to place the blue and red die in their two positions (which is $2!$)... Alternatively if you pick where $5$ go and let the $6$s fall were they may there are $3$ choices. $xx5,x5x, 5xx\to 665,656,566$. Commented Jul 5, 2022 at 3:18

As you suspected, you made an error in counting the number of ways in which one number appears twice and another number appears once.

Since there are six possible outcomes for each of the three dice, the number of elements in our sample space is indeed $$6^3$$.

For the favorable cases, there are six ways to select the number which appears twice, five ways to select the number which appears once, and $$\binom{3}{2}$$ ways to select on which two of the three dice the number that appears twice will be displayed. The remaining die must display the other number. Hence, the number of favorable cases is $$\binom{3}{2} \cdot 6 \cdot 5$$

Therefore, the probability of obtaining exactly two of the same number when three dice are rolled is $$\frac{\dbinom{3}{2} \cdot 6 \cdot 5}{6^3}$$

Your error: Suppose the dice are blue, green, and red. Then we can express an outcome as the ordered triple $$(b, g, r)$$, where $$b, g, r$$ are, respectively, the numbers displayed on the blue, green, and red dice. Notice, for instance, that there are only three ways to obtain two threes and one five on three dice. They are: $$(3, 3, 5), (3, 5, 3), (5, 3, 3)$$

The factor of $$3!$$ would be correct if the dice displayed three different numbers. For instance, we get $$3! = 6$$ possible outcomes if the three dice display the numbers $$2, 4, 6$$ since there are three choices for the number on the blue die, two for the number on the green die, and one for the number on the red die. Those outcomes are: $$(2, 4, 6), (2, 6, 4), (4, 2, 6), (4, 6, 2), (6, 2, 4), (6, 4, 2)$$

• Thank you so much for the detailed and wonderful explanation. Yes, I did not take into account the distinctiveness of objects, and just as you said the 6! would only be the correct permutations for all distinct objects. Commented Jul 3, 2022 at 18:33

Another way to look at it

There are $$6^3=216$$ possible rolls

$$6$$ of them get $$3$$ equal numbers

$$6*5*4=120$$ of them get $$0$$ equal numbers

$$216-126=90$$ of them get $$2$$ equal numbers

Probability to get $$2$$ equal numbers is $$\frac{90}{216} = \frac{5}{12}$$

Yet another way

The $$6$$ rolls that get $$3$$ equal numbers can be changed into the rolls that get $$2$$ equal numbers in $$5+5+5=15$$ ways, giving $$6*15=90$$ such rolls.

caveat : these two alternative ways may be shorter in this particular variant of the puzzle (roll 3 dice) but are not trivially generalized (e.g.: select 4 cards of each type (clubs diamonds hearts spades)). So my answers are not generic.

There are actually only $$3$$ ways to order the dice. Since you have two of the same number, you need to divide $$3!$$ by $$2$$ to get the number of orders.$${}{}{}{}$$