Let us imagine that the letters are written on slips of paper, one letter to a slip. So we have $13$ slips of paper. It is convenient to paint letters that repeat using different colours, so we have $2$ distinct $E$'s, $E_r$ (blue) and $E_r$ (red).
It is not clear whether the question asks for the probability of exactly $1$ $E$, or the probability of at least $1$ $E$. I would lean to the exactly $1$ interpretation.
We interpret "at random" to mean that all selections of $4$ slips of paper from the $13$ are equally likely. There are $\binom{13}{4}$ ways to choose $4$ slips of paper from the $13$.
Of these, $\binom{11}{3}$ have $E_b$ but not $E_r$, and $\binom{11}{3}$ have $E_r$ but not $E_b$. Thus the required probability is
$$\frac{2\binom{11}{3}}{\binom{13}{4}}.$$
Remark: At least $1$ is easiest to do by finding first the probability of the complement, no $E$. There are $\binom{11}{4}$ ways to choose $4$ slips, none of which is $E_b$ or $E_r$.
You are assuming that order doesn't matter in the second case, but this is the wrong assumption. Order definitely matters. You have, in the case of $n=6$ a sample space of $6^2$ Even though when you finally pull both tags out, a $(2,1)$ is the same as $(1,2)$, these are still different events and must be treated differently. Since this is the case, let's look at $n=6$. All consecutive numbers then would be;
$$(1,2),(2,1),(2,3),(3,2),(3,4),(4,3),(4,5),(5,4),(5,6),(6,5)$$
So here you have 10 events that are possible, not 5. And your sample space is 36... Therefore, your probability is
$$\frac{10}{36}=\frac{2\cdot5}{6^2}=\frac{2(6-1)}{6^2}$$
And this makes sense. You can only have $n-1$ consecutive pairs, since the $n$-th pair would be $(n,1)$ which are not consecutive, and our sample space consists of $n^2$ events. Since there are two ways to get consecutive integers, the formula is
$$P(\text{consecutive numbers with replacement of n tags})=\frac{2(n-1)}{n^2}$$
Also, in the first case, again, you are making the faulty assumption that $(1,2)$ is the same as $(2,1)$ and I think that you make this assumption because the probability is correct under both assumptions, i.e., yours, and the correct assumption. Why?
The sample space under sampling without replacement is in the case of $n=6$ is $6\cdot5=30$. This is the case because pairs such as $(1,1), (2,2)$ are impossible without replacement. However, you can still get the ten pairs of consecutive numbers listed above, so therefore, under the correct assumption
$$P(\text{consecutive numbers without replacement of n tags})=\frac{2(n-1)}{n(n-1)}=\frac2{n}$$
If $n=6$, you get your probability is $\frac1{3}$, which is what you got under your faulty assumption. Hope this helps shed a little light.
Best Answer
This may not be the most efficient way of doing this, but you can consider it as separate cases.
If all toppings are distinct, then you have $C_4^{15}$ combinations.
If there are three distinct toppings, you have $3 \cdot C_3^{15}$ combinations (because we have $C_3^{15}$ choices for toppings and then $3$ choices for which of those three toppings is doubled).
If there are two distinct toppings, you have $3 \cdot C_2^{15}$ combinations (because there are $C_2^{15}$ choices for topping and $3$ possibilities: either both toppings doubled, the first is tripled, or the second is tripled).
If there is only one distinct topping, you have $15$ possibilities.
Add these up to get your total.
Let's enumerate them with $5$ options instead of $15$ for illustrative purposes
Case 1:
Case 2:
Case 3:
and
Case 4: