Bagel probability intuition; unordered vs ordered selection I was solving the following problem:

The bagel shop serves 4 types of bagels; plain, poppy seed, asiago, and everything. If 6 bagels were randomly selected for a breakfast meeting, what is the probability that no poppy seed bagels were selected?

My thought processes was; if I were to reach in and grab a bagel, there is a $\frac{3}{4}$ chance that it is not a poppy seed bagel. If I repeated this experiment 6 times, then I would have a $\left(\frac{3}{4}\right)^6 \approx 18\%$ chance of not drawing a poppy seed bagel.
However, when reviewing the solution it was calculated as
$$\frac{\text{Number of unordered sets not containing poppy seed}}{\text{Total number of unordered sets}}$$
or
$$\frac{\binom{6+3-1}{6}}{\binom{6+4-1}{6}} \approx 33\%$$
I understand intuitively why the solution makes sense, calculating the fraction of the number of unordered sets that do not contain poppy seed, my question is
Why does my intuition fail me in the first case?
 A: Your solution is correct under the most sensible reading of the problem.  The official solution depends on a non-standard reading.
Your solution assumes that each bagel is of a given type with probability $\frac 14$.  This is the case if, say, you are reaching into a big bag filled with dozens of each type of bagel.  That assumption leads to the calculation you performed.
The official solution assumes that each configuration of bagel types is equally likely.  That is, they assume that $6$ plain is as probable as $2$ plain, $2$ asiago, and then one each of the other types.  That's a very different probability distribution and it's hard to see a simple mechanism for implementing it.  Of course, it's fair to assume that this distribution is the relevant one here, but that should have been spelled out explicitly.  To stress, under the assumption that each bagel is of a given type with equal probability, the different configurations have different probabilities.
To illustrate the difference, suppose you had only two types, plain and poppy, and that you are choosing two bagels.  If each bagel is independently likely to be of either type then a mixed selection has probability $\frac 12$ while two plain or two poppy each have probability $\frac 14$.  Of course, if each configuration is assigned the same probability, then each of these selections has probability $\frac 13$.
A: Your answer is the correct answer.
The solution given to you fails to take account that the various unordered sets of bagels are not equally likely.
As a quick example, the probability of pulling all plain bagels is less than pulling five plain bagels and an asiago.
There's only one way to pull all plain bagels, but there's $6$ ways we could pull five plain and one asiago. The asiago could be the first bagel, the second, etc.
