Playing Dice and probabilities

Question

The dice (numbers from 1 to 6 are written on the edges) was thrown 7 times. Determine the probability that the sum of the dropped numbers will be a multiple of 3.

I'm writing to clarify the answer. Statistically, the probability is 1/3(checked by the script). But I can't solve this analytically. Please help.

Answer 1

First approach:

Prove by induction that the probability that the remainder when the sum of $n$ dropped numbers is divided by 3, is 0, 1, or 2 is $ \frac{1}{3}$ (for each of the 3 cases).

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Second approach:

Consider the outcomes of rolling the dice 7 times. Partition them into 3 sets based on the remainder when the sum is divided by 3.

Create a bijection between these 3 sets.

Hint: There is a bijection that merely cyclically permutes the first rolll.

Answer 2

Hint:

Let $n $ denote the summation of the results of the first 6 throws.

Then the set $\{n+i\mid i=1,\dots ,6\} $ contains exactly two elements that are multiples of 3.

The point is that this is true for every integer $n $.