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?

Solution
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:


 * {| border="1"

! align="right" style = "padding: 2px 10px" | $6 n$ ! align="right" style = "padding: 2px 10px" | $n$ ! style = "padding: 2px 10px" | $\map P {\text {$n$ or more sixes} }$
 * align="right" style = "padding: 2px 10px" | $6$
 * align="right" style = "padding: 2px 10px" | $1$
 * align="right" style = "padding: 2px 10px" | $0.665$
 * align="right" style = "padding: 2px 10px" | $12$
 * align="right" style = "padding: 2px 10px" | $2$
 * align="right" style = "padding: 2px 10px" | $0.619$
 * align="right" style = "padding: 2px 10px" | $18$
 * align="right" style = "padding: 2px 10px" | $3$
 * align="right" style = "padding: 2px 10px" | $0.597$
 * align="right" style = "padding: 2px 10px" | $24$
 * align="right" style = "padding: 2px 10px" | $2$
 * align="right" style = "padding: 2px 10px" | $0.584$
 * align="right" style = "padding: 2px 10px" | $30$
 * align="right" style = "padding: 2px 10px" | $5$
 * align="right" style = "padding: 2px 10px" | $0.576$
 * align="right" style = "padding: 2px 10px" | $96$
 * align="right" style = "padding: 2px 10px" | $16$
 * align="right" style = "padding: 2px 10px" | $0.542$
 * align="right" style = "padding: 2px 10px" | $600$
 * align="right" style = "padding: 2px 10px" | $100$
 * align="right" style = "padding: 2px 10px" | $0.517$
 * align="right" style = "padding: 2px 10px" | $900$
 * align="right" style = "padding: 2px 10px" | $150$
 * align="right" style = "padding: 2px 10px" | $0.514$
 * }
 * align="right" style = "padding: 2px 10px" | $600$
 * align="right" style = "padding: 2px 10px" | $100$
 * align="right" style = "padding: 2px 10px" | $0.517$
 * align="right" style = "padding: 2px 10px" | $900$
 * align="right" style = "padding: 2px 10px" | $150$
 * align="right" style = "padding: 2px 10px" | $0.514$
 * }
 * }

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