<s>[INST] Probability that the minimum is $2$ and the maximum is $5$ after rolling a die $4$ times 
Suppose that a fair $6$-sided die is rolled $4$ times. Letting $N$ be
  the minimum number of spots observed and $X$ be the maximum number of
  spots observed, give the value of $P(N= 2, X= 5)$

For this to happen, we need all $4$ die to be greater than or equal to $2$ but we need at least one to be $2$. Similarly, we need all $4$ die to be less than or equal to $5$ but we need at least one to be $5$. So I need to find the probability of getting a $2$, a $5$, and a $2,3,4$, or $5$, twice. I have 
$${4\choose{1}} \cdot{3\choose{1}} \cdot{2\choose{2}}\cdot {1\over{6}} \cdot{1\over{6}}\cdot {4\over{6}}^2 =.148$$
since we are choosing one spot of $4$ to be a $2$, one spot of the remaining $3$ to be a $5$, and $2$ spots of the remaining $2$ to be a $2,3,4,$ or $5$.
I do not think this is right because I did an excel simulation and got a probability of a little over $.09$.
 [/INST] Your enumeration counts some rolls repeatedly.  For instance a 2 2 4 5 roll is counted twice, on account of the two different choices for 2.
I prefer an inclusion-exclusion-style argument for this, which yields a probability of 
$$
\frac{4^4-2*3^4+2^4}{6^4} = \frac{55}{648} \approx 0.0849
$$
Here the $4^4$ is all rolls with all values from 2 to 5, the two copies of $3^4$ are the rolls with all values from 2 to 4 and all rolls from 3 to 5, and the $2^4$ are the rolls with all values either 3 or 4.  You want to exclude values without a 5 and without a 2, but after subtracting the $2*3^4$ representing both of these possibilities, you need to add in the intersection of these two situations since those got subtracted twice.
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