# 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.

-

## closed as off-topic by RecklessReckoner, USER91500, Claude Leibovici, G. Sassatelli, user21820Mar 29 at 16:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – RecklessReckoner, USER91500, Claude Leibovici, G. Sassatelli, user21820
If this question can be reworded to fit the rules in the help center, please edit the question.

All you need to do is compute them separately, and use the probabilistic meaning of "or" and "and." – Alfred Yerger Mar 28 at 3:57
The part that throws me off is the "exactly two more than the other die" – J. Doe Mar 28 at 3:59
That means they show 1 and 3, or 2 and 4, or 3 and 5, or 4 and 6. – Christopher Carl Heckman Mar 28 at 4:00
Just count cases to handle that. Then multiply two to handle the fact that the dice are independent. – Alfred Yerger Mar 28 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. – JMoravitz Mar 28 at 4:01

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.

-

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." – Alan Thompson Mar 28 at 4:26
Especially if you "don't know how to start". Thanks. – probablyme Mar 28 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. – chirlu Mar 28 at 9:07
@AntonSherwood quick "cheat" : when you see some interesting formatting in an answer, click "edit" to see the source code. – Carl Witthoft Mar 28 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. – Monty Harder Mar 28 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.

-

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 at 12:45
@Did You are absolutely right, I corrected it with what I meant. Thanks! – user Mar 29 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 at 13:28
Not a mistery, (6+6+6+6+6+6) is the sum of all the possible outcomes. – user Apr 2 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 at 15:38