# Why probability of unordered samples is not equally likely?

Consider a survey where a sample of size k is collected by choosing people from a population of size n one at a time, with replacement and with equal probabilities. Then the n^k ordered samples are equally likely, making the naive definition applicable, but the "n+k-1 choose k" unordered samples (where all that matters is how many times each person was sampled) are not equally likely.

[Source: Introduction to probability by joseph k. blitzstein]

Could you please explain why is it like that?

## 1 Answer

You can see what is happening by choosing a very simple case: The population has only two people, $$A$$ and $$B$$. And your sample size will be two sample people. Thus $$n=k=2$$.

Then looking at the ordered samples: $$P(s_1 = A \wedge s_2 = A) = \frac12 \cdot \frac12 = \frac14 \\ P(s_1 = A \wedge s_2 = B) = \frac12 \cdot \frac12 = \frac14 \\ P(s_1 = B \wedge s_2 = A) = \frac12 \cdot \frac12 = \frac14 \\ P(s_1 = B \wedge s_2 = B) = \frac12 \cdot \frac12 = \frac14 \\$$

But looking at"unordered samples" where all we care about is $$n_A$$ and $$n_B$$, the number of selected $$A$$'s and $$B$$'s, we have: $$P(n_A = 2, n_B = 0) = P(s_1 = A \wedge s_2 = A) = \frac14 \\ P(n_A = 1, n_B = 1) = P(s_1 = A \wedge s_2 = B)+P(s_1 = B \wedge s_2 = A) = \frac14 + \frac14 = \frac12\\ P(n_A = 0, n_B = 2) = P(s_1 = B \wedge s_2 = B) = \frac14$$ We see that one of the unordered possibilities has a different probability than the others.

What has happened is that some of the unordered possibilities comprise just one ordered possibility, but some of them (just one in this case, but most of them in other cases) comprise multiple ordered possibilities.

• Thanks! It should be P(nA = 1, nB = 1) in second case. Also third case P(na = 0, nb = 2). – Majeed Siddiqui Oct 16 '16 at 21:37