# 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?

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.