Question

Which is more likely, to throw at least 1 six with 6 dice, or at least 2 sixes with 12 dice, or at least 3 sixes with 18 dice?

Answer 1

Try it out and see what happens.

Answer 2

For the first one it is: \[
\binom{6}{1}\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^5
\]The second one is: \[
\binom{6}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{10}
\]The third one is: \[
\binom{6}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^{16}
\]

Answer 3

I don't get it. How did you get this?

Answer 4

And I think you are wrong, first one equals about .4, second one equals about .067, last one equals about .006 . this is what I got on calculator. It doesn't make sense.

Answer 5

Oooh wait

Answer 6

Should be \[
\binom{12}{2}
\]for the second and \[
\binom{18}{3}
\]for the third...

Answer 7

Basically, the fraction parts are finding out the probability that the 6's come up consecutively at the start and then the rest are not 6's.

Answer 8

Then the binomials count how many ways you could shuffle the 6's around, so you account for every sequence they could appear in.

Answer 9

Hmmm, it's looking a lot like the binomial distribution but...

Answer 10

But it seems like it actually is the binomial distribution.