Relative Likelihood of n Sixes on 6n Dice

Problem

Which is more likely:

to throw at least $1$ six with $6$ dice

to throw at least $2$ sixes with $12$ dice

to throw at least $3$ six with $18$ dice?

The chance of getting $1$ six and $5$ other outcomes in a particular order is $\paren {\dfrac 1 6} \paren {\dfrac 5 6}^5$.

We need to multiply by the number of orders for $1$ six and $5$ non-sixes.

Therefore the probability of exactly $1$ six is:

$\dbinom 6 1 \paren {\dfrac 1 6} \paren {\dfrac 5 6}^5$

Similarly, the probability of exactly $x$ sixes when $6$ dice are thrown is:

$\dbinom 6 x \paren {\dfrac 1 6}^x \paren {\dfrac 5 6}^{6 - x}$

for $x = 0, 1, 2, 3, 4, 5, 6$.

The probability of $x$ sixes for $n$ dice is:

$\dbinom n x \paren {\dfrac 1 6}^x \paren {\dfrac 5 6}^{n - x}$

for $x = 0, 1, 2, \ldots, n$.

according to the binomial distribution.

The probability of $1$ of more sixes with $6$ dice is the complement of the probability of $0$ sixes:

$1 - \dbinom 6 0 \paren {\dfrac 1 6}^0 \paren {\dfrac 5 6}^6 \approx 0.665$

When $6 n$ dice are rolled, the probability of $n$ or more sixes is:

$\ds \sum_{x \mathop = n}^{6 n} \dbinom {6 n} x \paren {\dfrac 1 6}^x \paren {\dfrac 5 6}^{6 n - x}$

which equals:

$1 - \ds \sum_{n \mathop = 0}^{n - 1} \dbinom {6 n} x \paren {\dfrac 1 6}^x \paren {\dfrac 5 6}^{6 n - x}$

Calculating the probabilities, we get:

It is clear that the gambler does better by betting on $1$ six on $6$ dice than $2$ on $12$ or $3$ on $18$.

$\blacksquare$

According to David Wells in his Curious and Interesting Puzzles, this question was asked by the famous diarist Samuel Pepys of Isaac Newton in $1693$.