# Roll two dice. What is the probability that one die shows exactly two more than the other die? [closed]

Two fair six-sided dice are rolled. What is the probability that one die shows exactly two more than the other die (for example, rolling a $1$ and $3$, or rolling a $6$ and a $4$)?

I know how to calculate the probabilities of each event by itself, but I do not know how to proceed with this problem.

• All you need to do is compute them separately, and use the probabilistic meaning of "or" and "and." Mar 28, 2016 at 3:57
• The part that throws me off is the "exactly two more than the other die" Mar 28, 2016 at 3:59
• That means they show 1 and 3, or 2 and 4, or 3 and 5, or 4 and 6. Mar 28, 2016 at 4:00
• Just count cases to handle that. Then multiply two to handle the fact that the dice are independent. Mar 28, 2016 at 4:00
• You know that standard six-sided dice have six faces, each with a number (or number of pips) being one of $\{1,2,3,4,5,6\}$. You have $4$ is exactly two more than $2$, so one die showing $4$ while the other die showing $2$ is allowed. Similarly, $5$ is exactly two more than $3$, so one die showing $5$ and the other a $3$ is allowed. $5$ is not exactly two more than $1$ however, so $5$ for one die and $1$ for the other is not allowed. Mar 28, 2016 at 4:01

To get yourself started, you could draw a table. The rows could be one roll, and the columns could be the other roll. Then the checkmark shows where the rolls are "two away" from each other.

\begin{array}{r|c|c|c|c|c|c} &1&2&3&4&5&6\\\hline 1&&&\checkmark&&&\\\hline 2&&&&\checkmark&&\\\hline 3&\checkmark&&&&\checkmark&\\\hline 4&&\checkmark&&&&\checkmark\\\hline 5&&&\checkmark&&&\\\hline 6&&&&\checkmark&& \end{array} Notice that, since all pairs are equally likely, we have a $8/36 = 2/9$ chance of being "two away".

• Very nice graph! This was exactly the way I was going to explain it. "When in doubt, just do an exhaustive count of all the possible solutions." Mar 28, 2016 at 4:26
• Especially if you "don't know how to start". Thanks.
– Em.
Mar 28, 2016 at 4:30
• Side remark: MSE is usually understood to mean Meta Stack Exchange in the “SE universe”, while this side would be called Math.SE or similar. Mar 28, 2016 at 9:07
• @AntonSherwood quick "cheat" : when you see some interesting formatting in an answer, click "edit" to see the source code. Mar 28, 2016 at 14:18
• @user There is only one distinct ordered pair (1,1), while (1,3) & (3,1) are different. Per Casey, pretend one die is red and the other white, and our ordered pairs are in (r, w) format. There are six equally-probable outcomes for each die, and the events are independent, making 36 also-equally-probable outcomes for the two dice. Red 1 and white 3 is different from red 3 and white 1, making two ways to roll {1,3} (unordered pair) but red1 and white 1, aka {1,1}, can only be done one way. With two dice of the same color, we can't tell by sight (1,3) from (3,1), but we know they both exist. Mar 28, 2016 at 15:35

Total possible results: $6\times6=36$

Favorable results: $1-3,2-4,3-5,4-6$ and opposites, $8$.

Then the probability is $8/36=2/9$.

The probability of rolling a 1 and 3 is 1/18. Same for the probability of 2&4, 3&5, and 4&6.

So the overall probability of the dice being two apart equals 4/18 = 2/9.

Any result will do as long as the other die can score the same number plus two, that gets us with n-2 per die (n being number of sides). This gets us 2(n-2) posible results over n^2 (as we have two identical dice)

then the probability is: 2(n-2)/n^2

Just for fun, I counted eight. Could use the multiplication rule:

The probability of Die 1 landing on 1-4 is 4/6. The probability of the Die 2 landing on the number that's Die1+2 is then 1/6.

(4/6) * (1/6) = 4/36

We multiply this by 2 to account the scenario where Die 2 is the 1-4 die, and then Die 1 is two higher than Die 2. So, 8/36.

If the first die is 1, the other can only be 3, probability = 1/6

If the first die is 2, the other can only be 4, probability = 1/6

If the first die is 5, the other can only be 3, probability = 1/6

If the first die is 6, the other can only be 4, probability = 1/6

If the first die is 3, the other can only be 1 or 5, probability = 2/6

If the first die is 4, the other can only be 2 or 6, probability = 2/6

Total probability is (1+1+1+1+2+2)/(6+6+6+6+6+6) = 8/36 = 2/9

• Sorry but 1/6 + 1/6 + 1/6 + 1/6 + 2/6 + 2/6 = 4/3, not 2/9.
– Did
Mar 29, 2016 at 12:45
• @Did You are absolutely right, I corrected it with what I meant. Thanks!
– user
Mar 29, 2016 at 13:28
• But now the ratio by (6+6+6+6+6+6) is an even greater mystery. Where does it come from?
– Did
Apr 2, 2016 at 13:28
• Not a mistery, (6+6+6+6+6+6) is the sum of all the possible outcomes.
– user
Apr 2, 2016 at 15:36
• Then none of the preceding computations is relevant. If one wants to count cases, then one should count cases from the start, not jump from probabilities to numbers of cases.
– Did
Apr 2, 2016 at 15:38