Why does it seem to decrease probability to roll more times?

Question

I am looking how to find the probability of rolling 10 sixes in 60 rolls. After reading Calculate probabilities. 40 sixes out of a 100 fair die rolls & 20 sixes out of a 100 fair die roll, it seems that I can use the following formula: $60 \choose 10$$(\frac{1}{6})^{10}$$(1-\frac{1}{6})^{60-10}$

That comes to about 13.7%. So far, so good.

But then I decided to test it. What about getting 10 sixes in 120 rolls? $120 \choose 10$$(\frac{1}{6})^{10}$$(\frac{5}{6})^{120-10}$

That's only 0.37%! I'm getting something wrong, but what?

Answer 1

The formula you quote is for getting exactly (not at least) $10\ 6$'s in $60$ rolls. The factor $(1-\frac 16)^{60-10}$ ensures that the other dice roll something other than $6$. Your extension is correct for getting exactly (not at least) $10\ 6$'s in $120$ rolls. The expected number is $20$, so it is not surprising that the probability has decreased-you are farther from the expected number.

Answer 2

What you've calculated (correctly) are the probabilities that you get exactly $10$ sixes in $60$ rolls and $120$ rolls, respectively. If you think about this, it makes sense that the probability of the latter should be lower. Intuitively, there are more possibilities for how many sixes show up in the second case.