Is it more likely to get $450-550$ or $379-479$ blue marbles in $1000$ draws? This is a follow up question to this: An urn contains $4$ blue and $4$ red marbles. What is $P(A\vert B)$?
An urn contains  $4$ blue and $4$ red marbles. At first a marble is drawn (without looking) and removed from the urn. Then, a marble is drawn from the urn, its color recorded and put back in the urn. This process is repeated $1000$ times. Let $D$ be the event that between $450$ and $550$ blue marbles are drawn and let $E$ be the event that between $379$ and $479$ blue marbles are drawn. Which event is more likely?
Intuitively I would say that event $D$ is more likely. This is because on average I would expect around $500$ blue marbles drawn. Since the event $D$ is centered around this average it seems more likely to me than event $E$ that is centered around a value of $429$. Is there a nice way to show this mathematically by maybe using some sort of approximation?
 A: The calculation is actually quite straightforward, but there is a fundamental flaw in your intuition that I feel is more important to address, because it illustrates a principle that one needs to understand thoroughly if one is to not be led astray.

The expected value is not necessarily near a mode of a distribution.

This statement, while simple on its face, may not seem to have immediate applicability to your reasoning.  However, the relationship becomes apparent once you observe that we are dealing with two distinct conditional distributions for the number of blue marbles drawn, once the unobserved marble is removed, and neither of them have their mode near $500$.  Therefore, the probability of event $D$ occurs over a range of unconditional outcomes that will not cover a large likelihood of occurrence, because in either case of the initial marble being removed, the conditional mean (and thus the mode) is too far away.
Explicitly, we note that if $X$ is the random number of blue marbles drawn, and $B$ is the event that the initial marble removed is blue, then $$X \mid B \sim \operatorname{Binomial}(n = 1000, p = 3/7),$$ and $$X \mid \bar B \sim \operatorname{Binomial}(n = 1000, p = 4/7).$$  Thus the conditional means and variances are $$\operatorname{E}[X \mid B] = \frac{3000}{7} \approx 428.57, \quad \operatorname{E}[X \mid \bar B] = \frac{4000}{7} \approx 571.43, \\ \operatorname{Var}[X \mid B] = \operatorname{Var}[X \mid \bar B] = \frac{12000}{49} \approx 244.898.$$
This in turn implies that the standard deviation is about $\sigma \approx 15.64$, and thus the outcome $X = 500$ is just over $4.5$ standard deviations from the means of either conditional distribution.  Thus the unconditional distribution of the number of blue marbles drawn is strongly bimodal.
The exact calculation is
$$\Pr[D] = \frac{1}{2} \sum_{x=450}^{550} \binom{1000}{x} \left( (3/7)^x (4/7)^{1000-x} + (4/7)^x (3/7)^{1000-x} \right), \\
\Pr[E] = \frac{1}{2} \sum_{x=379}^{479} \binom{1000}{x} \left( (3/7)^x (4/7)^{1000-x} + (4/7)^x (3/7)^{1000-x} \right).
$$
The normal approximation to the binomial would work here to evaluate the sum;
$$\Pr[D] \approx \frac{1}{2}\left(\Pr[1.369 \le Z \le 7.759] + \Pr[-7.759 \le Z \le -1.369]\right) \approx 0.085, \\
\Pr[E] \approx \frac{1}{2}\left(\Pr[-3.168 \le Z \le 3.222] + \Pr[-12.296 \le Z \le -5.906]\right) \approx 0.499.$$
This is not continuity-corrected but you can already see the dramatic difference.  In the case of event $E$, almost all of the probability is contributed by the scenario in which the initial marble removed is blue, thus making the conditional distribution have mean $428.57$, and in this case, which occurs with probability $1/2$, the chance of getting between $379$ and $479$ blue marbles in $1000$ draws is almost assured; this is why the resulting unconditional probability is so close to $1/2$.  But in the case of event $D$, you are highly unlikely in either case of the initial removed marble to get anywhere near $450$ to $550$ blue marbles because the conditional distributions don't have their modes near $500$.

The reason why I furnished an outline of the calculation in this case, when very minimal effort was shown in that regard, is because I consider it to be virtually irrelevant to the true pedagogical value of the question, which is to illustrate how we must be careful with using intuitive arguments.
