Probability question of rolling 6 dices and getting required events

Question

Question:

Samuel Pepys wrote Isaac Newton to ask which of these events is more likely: that a person get (a) at least $1$ six when $6$ dice are rolled. (b) at least $2$ sixes when $12$ dice are rolled, or (c) at least $3$  sixes when $18$ dice are rolled.  What is the answer?

My Approach:

When $6$ die are rolled total number of occurrences $=6*6*6*6*6*6=6^6=46,656$ since $1$ dice has $6$ faces

Considering A as the first event of at least $1$ six when all six die are rolled. It is possible that only only one face of a dice gets a face of six and also numerous other occurrences are possible. Can you state a specific way to solve this problem and find a solution without much complication? Thank you.

Answer 1

Use the converse probablility. Let $X,Y,Z$ the random variables for the number of sixes if you roll $6,12$ and $18$ dice, respectively. In general all variables are binomial distributed as $Bin\left(n,\frac16\right)$. Thus

$P(X\geq 1)=1-P(X=0)=1-\left(\frac56\right)^6$

$P(Y\geq 2)=1-P(X=1)-P(X=0)$

$=1-\binom{12}{1}\cdot \left(\frac16\right)^{1} \left(\frac56\right)^{11}-\left(\frac56\right)^{12}$

Similar for $P(Z\geq 3)$

Answer 2

Roll $6$ dice and record the number of $6$'s you get. That's a random variable. Call it $X$.

Now roll $6k$ dice and record the number of $6$'s you get. Call this random variable $X_k$. This is exactly the same thing as rolling $6$ dice $k$ times, recording the number of $6$'s you get for each 6-die roll, and adding up those $k$ numbers. So $X_k$ is a sum of $k$ copies of $X$---and hence (by the central limit theorem) close to normally distributed when $k$ is large.

Clearly the mean of $X$ is $1$, and more generally the mean of $X_k$ is $k$. But because $X_k$ is close to normal, it is close to symmetrically distributed around its mean. Therefore, $Prob(X_k\ge k)$ is close to $1/2$.

On the other hand, it's very easy to see that $Prob(X>1)=1-(5/6)^6\approx .67$. So if your goal is to toss the dice $6k$ times and throw at least $k$ 6's, you're definitely better off with $k=1$ than with $k$ large.

Unfortunately, this fails to address the cases of $k=2$ and $k=3$, which are admittedly not terribly large. This leaves two options:

1) Settle them with the obvious direct computations

or

2) Find some reason to believe that the convergence to the normal distribution is, in an appropriate sense, monotonic.

Obviously 2) would be more satisfying, but I don't see how to do it.