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The question is, what is the probability that, if you roll 5 standard six-sided dice, you get at least two 4's?

I know the correct answer, which is a form of the binomial distribution:

$1 - ((\frac{5}{6})^5 \cdot C(5,0) + (\frac{5}{6})^4 \cdot \frac{1}{6} \cdot C(5,1))$

That part makes sense to me. I want to understand why another line of reasoning is incorrect:

Since we need at least two 4's, we can just find the negation of the probability space, which is the probability that we have four dice turn up with something other than a 4. If that condition is met, then there is only one die left and it doesn't matter what it is, since we won't have at least two 4's.

So, the probability that a die does not turn up as 4 is $\frac{5}{6}$. We do not care about the ordering that this happens since all dice are the same, so we just have:

$(\frac{5}{6})^4$

I think an issue here is that $\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6}$ represents the case that we have 4 consecutive rolls, each of which are not 4. I don't see why we cannot just stop caring about the probabilities for what other dice are. What other issues are present with this line of reasoning?

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I think you posted your question incomplete... –  DonAntonio Mar 22 '13 at 7:32
    
Spot on the money. I finished typing it up now. –  dbmikus Mar 22 '13 at 7:36
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2 Answers 2

up vote 1 down vote accepted

If you select the 4 rolls at random, one of them may be a four, and so your logic doesn't hold.

If you select the four rolls on the basis that none contains a four, you have already made a selection which you haven't accounted for. Some combinations of four rolls may contain a four, some may not, you need to calculate the probability that you picked a combination with no fours. Which you haven't done.

Basically the logic flaw is that you select 4 rolls out of 5, but do not state exactly how you do that or calculate the probability that you pick a combination which meets your requirement (no fours). No sample space for this selection is determined.

It is possible to solve the problem in more or less the way you described, but its a lot harder and more work than the method you have (hopefully) been taught.

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The following is problematic in your reasoning:

"Since we need at least two 4's, we can just find the negation of the probability space, which is the probability that we have four dice turn up with something other than a 4. If that condition is met, then there is only one die left and it doesn't matter what it is, since we won't have at least two 4's"

Most probably, with the above argument, you either are going to count something twice or else something is not going to be counted at all. In your case it was the latter, since you say "all the dice are the same", but in fact you have two different possible outcomes for the negation of your event: exactly one single dice is $\,4\,$, or else none of the dies is 4, and this gives you exactly the probability that you already wrote down in your question.

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