抽样分布疑问：为何需乘以组合/排列？附大样本圆盘抽样案例

Great question—this is a super common sticking point when first working through sampling distributions, so let’s break it down using your exact disk example to make it concrete.

First, let’s ground the basics: since your bag of disks is enormous, each draw is independent. Drawing a 1 doesn’t shift the probability of drawing a 1 next, so we can treat each pick as a repeatable, independent trial with ( P(\text{draw 1}) = 0.25 ) and ( P(\text{draw 2}) = 0.75 ).

The core issue: multiple sample sequences lead to the same statistic #

When you’re calculating the probability that the median is 1, you can’t just compute the probability of one specific sequence like 112—you need to account for all sequences that result in a median of 1.

Let’s list every sequence that gives a median of 1:

• Any sequence with 2 or 3 ones: For 3 disks, the median is the middle value after sorting. If you have 2 ones, sorted sequences look like [1,1,2] (median 1); if you have 3 ones, sorted sequences are [1,1,1] (median 1).

Now let’s calculate the total probability for these cases, using combinations to count valid sequences:

1. Exactly 2 ones, 1 two: How many unique sequences fit this? We need to choose 2 positions out of 3 to be 1s (the rest will be 2s). That’s what the combination formula C(3,2) calculates—there are 3 such sequences: 112, 121, 211. Each sequence has the same probability: 0.25 * 0.25 * 0.75 = 0.25² * 0.75. To get the total probability for this case, we multiply the single-sequence probability by the number of valid sequences: 3 * (0.25² * 0.75).
2. Exactly 3 ones: There’s only 1 sequence here (111), which is C(3,3) = 1. Its probability is 0.25³, so multiplying by 1 keeps the value the same (but it’s still part of the combination logic).

Adding these together gives the total probability that the median is 1:
(3 * 0.25² * 0.75) + (1 * 0.25³) = 0.140625 + 0.015625 = 0.15625

Why this multiplication is non-negotiable #

Sampling distributions describe the probability of getting a specific statistic (like the median) across all possible samples. If you skip multiplying by the number of valid sequences, you’re only calculating the probability of one specific sample—not the total probability of the statistic itself. That would undercount all the other ways to get the same median, leading to a wrong result.

Quick note on combinations vs permutations #

• We use combinations here because the order of draws doesn’t affect the median (e.g., 112 and 121 have the same median). Combinations count unique groups (ignoring order) that produce the statistic.
• If we cared about ordered outcomes (e.g., the exact probability of drawing 1, then 1, then 2), we’d use permutations—but since our statistic is order-independent, combinations are the right tool.

内容的提问来源于stack exchange，提问作者Clive Long