Which event has a higher probability? Which event has a higher probability? 

$24$ rolls of 2 dice at once we get at least 2 $1$s

or

one roll of 4 dice at once we get at least one $1$?

 A: Hint. $1$ minus the probability that in 24 rolls of 2 dice we never get 2 ones (at once): $1-\left(\frac{6^2-1}{6^2}\right)^{24}$.
1 minus the probability that in 1 roll of 4 dice we never get one: $1-\left(\frac{6-1}{6}\right)^{4}$.
Which number is greater? 
P.S. Both numbers are quite close to $0.5$.
A: For a roll of four dice getting a single 1:

$$\frac{1}{6}+\frac{5}{6^2}+\frac{5^2}{6^3}+\frac{5^3}{6^4}=\frac{671}{1296}$$, since I am doing the chance of it not giving a one all times, then invert that (take it away from one)

For 24 rolls of 2 dice (at once), or 48 rolls of a single dice you get at least 2 'ones':

Well, chances of it happening is , $1-(\frac{25}{6^2})^{24}=$ something very close to $1$. I got these numbers by saying the chance thee is not a one, and taking it to the power of $24$ and finally inverting it.



Well, these results show that $$\frac{671}{1296}< \approx1$$, so we know that:

The chance of getting at least two $1$s in 24 rolls of two dice is greater than getting four rolls of a dice, and we get a single one.


