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Task: We roll three (fair) dice, find the probability of their median being $k$ with $1\leq k\leq 6$?

My solution: Let $X$ be the median of the three rolled dice. In order to get $X=k$, we need to roll one $k$, one $j$ with $1\leq j\leq k$ and one $\ell$ with $k\leq\ell\leq 6$. For $k$ there's just one possibility, for $j$ there are $k$ possibilities and for $\ell$ there are $6-k+1$ possibilities. So the probability for $X=k$ is $$P(X=k)=\frac{\displaystyle\binom k1\binom11\binom{6-k+1}1}{\displaystyle\binom63}=\frac{k(7-k)}{20}~.$$

Problem: With my formula for $P(X=k)$ the probability of $1\leq X\leq6$ is $$P(1\leq X\leq6)=\sum_{1\leq k\leq6}P(X=k)=\sum_{k=1}^6P(X=k)=\frac{14}5$$ and not $1$, which brings me to my question(s): What did I do wrong? Can I even use the urn model for this problem?

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    Just to point out, in order for the median to be $1$ you need two one's (and the third can be anything). If the third isn't $1$ there are $5$ choices for it and three places to put it, so $15$ cases. adding $1$ for $(1,1,1)$ and noting that there are $6^3=216$ ways to roll the three dice (assuming the dice are distinguishable, so the possible rolls are equiprobable), we get $P(1)=\frac {16}{216}=\frac 2{27}$ while you appear to get $\frac 6{20}=\frac 3{10}$. – 2017-01-19

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The denominator in your calculation is going to be $6^3=216$, not $20$.

The ways to get a median of $k$ are divided into 3 cases, depending whether you have all three, two or one dice at $k$: $$ 1 \\ + 3\cdot 5 \\ +6\cdot (k-1)(6-k) $$ In the above, the 3 is because we are free to choose which of the three dice is the different one. THe 5 is because that different die can have any value but $k$. The 6 is because if all three dice are different, we can slot any die in the middle and then either of the others as the higher number. Then there are $(k-1)$ possible values for the lowest die (remember, all three are different in the third row above) and there are $(6-k)$ possible values of the highest die.

Then we have $$P(k=1) = P(k=6) = \frac{1 +15+0}{216} = \frac{2}{27}\\ P(k=2) = P(k=5) = \frac{1 +15+6\cdot 1\cdot 4}{216} = \frac{5}{27}\\ P(k=3) = P(k=4) = \frac{1 +15+6\cdot 2\cdot 3}{216} = \frac{13}{54} $$ and these probabilities indeed add to $1$.