May cooler heads prevail

Method 1: Out of $40$ outcomes of equal probability, $4*2 + 14*1 = 22$ result in an observation of heads. In $8$ of these, the other side of the coin will show heads as well. Thus the desired probability is $\dfrac{8}{22} = \dfrac{4}{11},$ and so $a + b = 4 + 11 = \boxed{15}.$

Method 2: Using conditional probability, with $A$ being the event that the "other" side shows heads and $B$ being the event that the observed side shows heads, we wish to calculate $P(A|B) = \dfrac{P(A \cap B)}{P(B)}.$

Now $P(B) = \left(\dfrac{4}{20}\right)*1 + \left(\dfrac{14}{20}\right)\left(\dfrac{1}{2}\right) = \dfrac{11}{20},$ and $P(A \cap B) = \dfrac{4}{20} = \dfrac{1}{5},$

since only the $4$ double-headed coins will show heads on the "other" side. Thus

$P(A|B) = \dfrac{\frac{1}{5}}{\frac{11}{20}} = \dfrac{4}{11},$ and so $a + b = 4 + 11 = \boxed{15}$ as before.

Since we can only be dealing with coins with heads on them, we only care about the 14 fair coins and the 4 two-headed coins. I first thought that the answer is $\frac{4}{4+14}=\frac{2}{9}$ .

However, I soon realized that this solution is false because the probability that a two-header was picked originally is twice the probability that a fair coin was picked. Therefore, the probabilty is $\frac{2 \cdot 4}{2 \cdot 4 +14}=\frac{4}{11}$ , yielding an answer of $4+11=\boxed{15}$

In order for the other side of the coin to show heads, the coin must be one of the 4 coins with heads on both sides. This means we know the coin was not any of those with tails on both sides.

We also know that if we were to take 14 fair coins and randomly observe one side of each, on average, 7 of them would show heads.

This means the coin selected is one of those 7, or one of the 4 coins with heads on both sides. So of these 11 possibilities, 4 have the desired outcome, giving us a probability of $\displaystyle \frac{4}{11}$ , and $a+b=15$

Any side of any coin has the same probability of being chosen. There are $14 + 4 \cdot 2 = 22$ sides that show heads. $8$ of those sides have the quality that the other side is also heads.

$\frac {8}{22} = \frac {4}{11}$

$4 + 11 = \boxed {15}$