Here is my problem:

You have a standard dice, with possible rolls: $\{1, 2, 3, 4, 5, 6\}$. How many permutations exist in 10 rolls such that no two immediate rolls are the same?

For example:

$\{1, 2, 1, 2\}$ is a valid 4-roll pattern. It's okay to have repeated numbers. The only constraint is that no two adjacent numbers can be equal.

$\{1, 1, 2, 1\}$ is invalid because the first and second rolls equal 1.

So I thought of doing it this way:

$6^{10} - 6^9$

$6^{10}$ represents the number of total number of permutations of 10 rolls.

$6^9$ represents the number of rolls with at least 1 adjacent pair of numbers. My thinking is that, for 9 dice rolls, the result can be anything, but for one of the rolls it must be the same as the previous roll. Therefore it would be:

$6 * 6 * 6 * 6 * 6 * 6 * 6 * 6 * 6 * 1$

$= 6^9 * 1 $

$= 6^9$

However, I checked the answer to the problem and I got it wrong. (The correct answer was $6 * 5^9$. This answer comes with a different method.) I think it has something to do with my reasoning about the $6^9$ number. Can someone explain the flaw in my logic?


The easiest way to think about it is that the first roll can be any of the $6$ possible outcomes, and then subsequent rolls can be any outcome except the previous outcome, hence any of $5$ possible outcomes. Therefore there are $6 \cdot 5^9$ such permuations.

  • $\begingroup$ Right I see. But I still can't figure out the flaw in the way I was thinking about it before. $\endgroup$ – CowZow May 6 '13 at 17:51
  • $\begingroup$ The problem with your thinking is that you're double-counting a lot. When you let the other 9 rolls be anything, you're not excluding the possibility that they're all the same too. $\endgroup$ – ferson2020 May 6 '13 at 18:09

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