Intuitive explanation of Newton - Pepys problem

Question

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?

1. Six fair dice are tossed independently and at least one “6” appears.
2. Twelve fair dice are tossed independently and at least two “6”s appear.
3. 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?

Answer 1

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

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

3. 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) > ...