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Rolling a fair dice $7$ times, what is the probability that the total sum is divisible by $3$, and no odd value was rolled?

I know that there are $3^6$ possible ways for the first $6$ dice to give only even values, but how many of them are divisible by $3$ ?

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    $\begingroup$ Hint: $2,4,6$ cover the residues $\pmod 3$. Therefore the first $6$ rolls could be any even numbers, and then the last one is determined by congruence. $\endgroup$ – lulu Jan 15 '17 at 0:21
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If we forget about this "no-odd" thing for a second, and just focus on "divisible by $3$", imagine having thrown the die six times already, and just waiting for the seventh roll to decide everything. How many different results on the seventh die will make the total result divisible by $3$? Does it matter what the sum of the first $6$ are?

Now adapt this to the case where you're only throwing evens. Does that really change anything of the above argument?

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As you have pointed out, the probability of the first six dice being all even is $\frac{3^6}{6^6}=\frac{1}{2^6}$.

The sum of the first six dice will be some number. No matter what this number is, the probability that the seventh roll is both even and makes the total sum divisible by $3$ is $1/6$.

To see this explicitly, you can do some casework. I will do one case for you. Suppose the sum of the first six dice is one more than a multiple of $3$. Then the seventh roll must be a $2$ to make the total sum a multiple of $3$. The seventh roll could be $5$ as well, but you only want even rolls.

It turns out that no matter what the sum of the first six rolls is, there is exactly one "good" seventh roll that will give you what you want.

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