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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.

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    $\begingroup$ All you need to do is compute them separately, and use the probabilistic meaning of "or" and "and." $\endgroup$
    – A. Thomas Yerger
    Mar 28, 2016 at 3:57
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    $\begingroup$ The part that throws me off is the "exactly two more than the other die" $\endgroup$
    – J. Doe
    Mar 28, 2016 at 3:59
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    $\begingroup$ That means they show 1 and 3, or 2 and 4, or 3 and 5, or 4 and 6. $\endgroup$
    – Christopher Carl Heckman
    Mar 28, 2016 at 4:00
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    $\begingroup$ Just count cases to handle that. Then multiply two to handle the fact that the dice are independent. $\endgroup$
    – A. Thomas Yerger
    Mar 28, 2016 at 4:00
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    $\begingroup$ 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. $\endgroup$
    – JMoravitz
    Mar 28, 2016 at 4:01

7 Answers 7

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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".

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    $\begingroup$ 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." $\endgroup$
    – Alan Thompson
    Mar 28, 2016 at 4:26
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    $\begingroup$ Especially if you "don't know how to start". Thanks. $\endgroup$
    – Em.
    Mar 28, 2016 at 4:30
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    $\begingroup$ 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. $\endgroup$
    – chirlu
    Mar 28, 2016 at 9:07
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    $\begingroup$ @AntonSherwood quick "cheat" : when you see some interesting formatting in an answer, click "edit" to see the source code. $\endgroup$
    – Carl Witthoft
    Mar 28, 2016 at 14:18
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    $\begingroup$ @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. $\endgroup$
    – Monty Harder
    Mar 28, 2016 at 15:35
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Total possible results: $6\times6=36$

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

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

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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.

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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

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Just for fun, I counted eight.
Dice image courtesy of Google[1]

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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.

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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

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  • $\begingroup$ Sorry but 1/6 + 1/6 + 1/6 + 1/6 + 2/6 + 2/6 = 4/3, not 2/9. $\endgroup$
    – Did
    Mar 29, 2016 at 12:45
  • $\begingroup$ @Did You are absolutely right, I corrected it with what I meant. Thanks! $\endgroup$
    – user
    Mar 29, 2016 at 13:28
  • $\begingroup$ But now the ratio by (6+6+6+6+6+6) is an even greater mystery. Where does it come from? $\endgroup$
    – Did
    Apr 2, 2016 at 13:28
  • $\begingroup$ Not a mistery, (6+6+6+6+6+6) is the sum of all the possible outcomes. $\endgroup$
    – user
    Apr 2, 2016 at 15:36
  • $\begingroup$ 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. $\endgroup$
    – Did
    Apr 2, 2016 at 15:38

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