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So, the question is:
Calculate the probability that 10 dice give more than 2 6s.

I've calculated that the probability for throwing 3 6s is 1/216.

And by that logic: 1/216 + 1/216 + .. + 1/216 = 10/216.

But I've been told that this isn't the proper way set it up.

Anyone having a good way to calculate this?

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  • $\begingroup$ I don’t see your logic. The probability that three 6s appear when three dice are thrown is indeed $\frac{1}{216}$, but why would you just add that $10$ times to find the answer to the question? (The probability that four 6s appear when you throw three dice is zero. Do you think the probability that four 6s appear when you throw ten dice is $0+0+0\cdots+0$?) $\endgroup$
    – Steve Kass
    Sep 17, 2015 at 18:17
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    $\begingroup$ Hint: You can use the converse probability. The probability of having more than two sixes, is equal to 1 minus the probability of having no six, one six and two sixes. $\endgroup$ Sep 17, 2015 at 18:21

2 Answers 2

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The easiest way to calculate this is to note that $ P(\#6s \le 2) + P(\#6s \gt 2) = 1$ That is that there must either be two or less or more than two sixes. So we can calculate it as $1-P(no\ 6s) - P(one\ 6) - P(two\ 6s)$.

To get no sixes, we have to get the other 5 results on all other dice. To get one six we'll get a 6 on one die, but a different result on the other dice. Likewise we'll have to get two 6s, then 8 of the others.

This step is probably what you're getting confused about. We have to multiply these by the number of ways to 'pick' the die that gets the 6. So we'll be multiplying by $10\choose{n}$ for n 6s.

So the total sum is going to be $1 - {10\choose{0}}(5/6)^{10} - {10\choose{1}}(1/6)(5/6)^{9} - {10\choose{2}}(1/6)^2(5/6)^{8} \approx 0.22$

$10\choose 0$ is just for example, we're picking 0 dice so it's just 1

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  • $\begingroup$ For some reason had 10 sided dice in my head. It's fixed $\endgroup$
    – Jason Carr
    Sep 17, 2015 at 18:29
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The probability no 6s are thrown: $(\frac{5}{6})^{10}$.

The probability exactly 1 6 is thrown: $\binom{10}{1}(\frac{1}{6})^1 (\frac{5}{6})^9$

EDIT: $\binom{10}{2}(\frac{1}{6})^2 (\frac{5}{6})^8$

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    $\begingroup$ You also need to add the probability that exactly two sixes are thrown. $\endgroup$
    – joriki
    Sep 17, 2015 at 18:19
  • $\begingroup$ I'm sorry I didm't finish reading the question. I thought the OP wants the probability of '2 or more' $\endgroup$
    – Alex
    Sep 17, 2015 at 18:21

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