Intuitive explanation of Newton - Pepys problem Quoting Wikipedia for description of problem:

In 1693 Samuel Pepys and Isaac Newton corresponded over a problem
posed by Pepys in relation to a wager he planned to make. The problem
was:
Which of the following three propositions has the greatest chance of success?

*

*Six fair dice are tossed independently and at least one “6” appears.

*Twelve fair dice are tossed independently and at least two “6”s appear.

*Eighteen fair dice are tossed independently and at least three “6”s appear.

Pepys initially thought that outcome C had the highest probability,
but Newton correctly concluded that outcome A actually has the highest
probability.

I know how A has highest probability mathematically. But it feels kind of unintuitive to me.
Newton's explanation from Wikipedia

Although Newton correctly calculated the odds of each bet, he provided
a separate intuitive explanation to Pepys. He imagined that B and C
toss their dice in groups of six, and said that A was most favorable
because it required a 6 in only one toss, while B and C required a 6
in each of their tosses. This explanation assumes that a group does
not produce more than one 6, so it does not actually correspond to the
original problem

So my question is what is the intuition behind the result?
 A: Generalising the question
Let $X \sim Bin(\frac{n}{p},p)$, so $\mu = n$
Does $\mathbb{P}(X \ge n)$ increase or decrease with $n$?
If $\mathbb{P}(X \ge n)$ is decreasing with $n$, then Proposition 1 is more likely than 2 is more likely than 3, and vice versa.
Intuition

*

*$\mathbb{P}(X = \mu) > 0  \therefore \mathbb{P}(X < \mu) + \mathbb{P}(X > \mu) < 1$

*In fact, $\mathbb{P}(X < \mu) < \frac{1}{2}$ and  $\mathbb{P}(X > \mu) < \frac{1}{2}$.

Think of this as some probability being used by $\mathbb{P}(X = \mu)$, and the remaining probability being shared roughly equally between $\mathbb{P}(X > \mu)$ and $\mathbb{P}(X < \mu)$.


*As $n$ becomes large, $\mathbb{P}(X = \mu) \approx 0$ and $\mathbb{P}(X < \mu) \approx \mathbb{P}(X > \mu) \approx \frac{1}{2}$.

As n increases, think of $\mathbb{P}(X < \mu)$ and $\mathbb{P}(X > \mu)$ consuming the probability occupied by $\mathbb{P}(X = \mu)$.
In particular, $\mathbb{P}(X < \mu)$ is growing with $n$.
Equivalently, $\mathbb{P}(X \ge n)$ is decreasing with $n$.
Summary
$X \sim Bin(6n,\frac{1}{6})$
For $n=1$, $\mathbb{P}(X=n)$ is large enough that both $\mathbb{P}(X < n) < \frac{1}{2}$ and $\mathbb{P}(X > n) < \frac{1}{2}$.
As $n$ increases, $\mathbb{P}(X < n)$ increases strictly, converging to $\frac{1}{2}$.
Therefore, $\mathbb{P}(X \ge n)$ decreases strictly $w.r.t.n$.
Therefore, $\mathbb{P}(\ge 1$ six out of 6 dice) > $\mathbb{P}(\ge 2$ sixes out of 12 dice) > ...
