Sum of dice adding to less than $5$

Question

I have seen some places saying that the sum of $2$ dice being thrown being less than $5$ can occur for the cases

$$(1,1),(2,2),(2,1),(3,1),$$

giving a probability of $\frac4{36}$ for the event.

However, I have also seen people counting the numbers twice, as there are two different ways to throw the dice, giving the cases

$$(1,1),(1,1),(2,2),(2,2),(2,1),(1,2),(3,1),(1,3),$$

and a probability of $\frac8{36}$.

Which way is correct?

Answer 1

Neither is correct. The second approach is the correct one – you list all the ordered possibilities. Nevertheless, you have listed $(1,1)$ and $(2,2)$ twice.

Answer 2

Neither.

There is a single way to roll a sum of 2: roll $(1,1)$.

There are two ways to roll a sum of 3 and two ways to roll a sum of 4, as you pointed out, where the numbers are different. Notice that $(2,2)$ also gives a sum of 4.

So the total probability is $6/36$.

One nice way to see this visually is to make a table, one with the value of the first die along the columns and the value of the second die along the rows, and record their sum at each cell in the table. The table will have $6 \times 6 = 36$ cells and you'll see that the diagonal rolls $(1,1)$, $(2,2)$, etc. all only appear once, while all the others appear twice.

Answer 3

Neither method is correct.

• In the first method, your outcomes are not equally likely

• In the second method, you're counting some outcomes twice.

In order to use the formula $$ \frac{\text{favorable outcomes}}{\text{total outcomes}}, $$ you need your events to be equally likely. In the first method, the outcomes $(1,1)$ and $(1,2)$ are not equally likely. The first method requires both dice to have a value $1$ while the second method has two situations for the dice.

To make this clearer, suppose that the dice are red and blue. Then, $(1,1)$ means that both the red die and the blue die show $1$. On the other hand, in the first method, $(1,2)$ represents the two possibilities ($1$-Red and $2$-Blue) or ($2$-Red and $1$-Blue). Since there are two possible ways to get a $1$ and a $2$, this $(1,2)$ has double the chances of occurring when compared to $(1,1)$.

For the second formulation, you're double counting the pairs of the form $(1,1)$. In this case, you're trying to describe $(1,1)$ for $1$-Red and $1$-Blue as well as $(1,1)$ for $1$-Blue and $1$-Red, but these are exactly the same situation.

Therefore, in the second case, you shouldn't duplicate the pairs that are identical under reversing the coordinates.

To calculate the probability correctly, the list should be $$ (1,1),(1,2),(1,3),(2,1),(2,2),(3,1). $$ Or, in other words, for the red and blue dice, $$ (1R,1B),(1R,2B),(1R,3B),(2R,1B),(2R,2B),(3R,1B). $$ Since there are $6$ possibilities for the red die and $6$ possibilities for the blue die, this results in $36$ total possible outcomes. Putting this all together, the probability is $6/36=1/6$.

Answer 4

$$P = \begin{bmatrix} \color{red} 2 & \color{red}3 & \color{red}4 & 5 & 6 & 7 \\ \color{red} 3 & \color{red} 4 & 5 & 6 & 7 & 8 \\ \color{red}4 & 5 & 6 & 7 & 8 & 9 \\ 5 & 6 & 7 & 8 & 9 & 10 \\ 6 & 7 & 8 & 9 & 10 & 11 \\ 7 & 8 & 9 & 10 & 11 & 12 \\ \end{bmatrix}$$ The entry $P_{i,j}$ is the sum of dice $1$ that shows value $i$, whereas dice $2$ shows value $j$. How many entries are less than $5$ ?

Answer 5

First, your denominator is $36,$ which means that when you're counting all possible outcomes, you're counting $3,4$ and $4,3$ separately. So you'd better do the same in your numerator.

The answer to your question in "neither." The list of possible outcomes less than $5$ is

$$(1,1), (1,2), (2,1), (2,2), (3,1), (1,3).$$

If you imagine one die is red and the other blue, then you can see that $red=1, blue=2$ is different from $red=2, blue=1,$ so both of those are listed.

But $red=1, blue=1$ is the same as $blue=1, red=1$, so it's counted only once.