Consider the problem of a rolling a three-sided die six times (independently). The probability of seeing 1 is 0.5, 2 is 0.25, and 3 is 0.25. With this model, I have been given the claim that:

We know that there are $6\choose{2}$ ways of rolling exactly two 1's.

This does not look right to me. If it was a coin-tossing model with binary outcomes, that would be right.

But since we have 3 outcomes for each roll, we can count the set of all length-6 rolls with two 1's by first specifying the index position of the two 1's $6\choose2$ ways, and then the outcome of the remaining four rolls (which has $2^4$ ways), for a total of $2^4\times {6\choose2}$ ways.

Is this reasoning correct?

Note: The motivation is question 2 from this problem and solution on OpenCourseWare.

  • 1
    $\begingroup$ What does a three-sided die look like? A Toblerone chocolate bar still in its wrapping? $\endgroup$ – Dilip Sarwate Sep 15 '14 at 22:04
  • $\begingroup$ Depends on what the sample space is. With the "natural" sample space of all words of length $6$ over the alphabet $\{1,2,3\}$, your answer is right. If we use a sample space of words made up of the letters Y (getting a $1$) and N (not a $1$) then $\binom{6}{2}$ is right. Disadvantage of this sample space is that the elements are not equiprobable. $\endgroup$ – André Nicolas Sep 15 '14 at 22:31

The key seems to be in the sentence (from the linked-to solution) that precedes what you quoted:

For the purposes of solving this problem we treat obtaining a 2 or 3 as an equivalent result.

In other words, they are reducing things to the coin-tossing model.


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