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Suppose n dice are rolled, yielding numbers between 1 and 6(inclusive) with equal probability. What is the probability that the sum of the numbers appearing is divisible by 3?

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  • $\begingroup$ did you check it for n=1,2,3? $\endgroup$ – miracle173 Feb 4 '17 at 10:13
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Hint: $n=1$ gives us two "successes" (3 and 6) and thus a probability of 1/3.

$n=2$ gives us $2 + 5 + 4 + 1 = 12$ successes (corresponding to 3,6,9,12). There are a total of 36 outcomes. This gives a probability of $1/3$.

$n=3$ gives us $1 + 10 + 25 +25 +10 + 1 = 72$ successes (corresponding to 3,6,9,12,15,18). There are a total of 216 outcomes. This gives a probability of $1/3$.

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Hint:

For every integer $m$ the set $\{m+1,m+2,m+3,m+4,m+5,m+6\}$ contains exactly $2$ elements that are divisible by $3$.

Now for $m$ take the sum of $n-1$ dice.

Also have a look this question and the corresponding answers.

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