If the number of caramel candies is $a$, then the probability that the first 6 drawn will not be caramel and the 7th drawn will be caramel (assuming that we do not put back the drawn candies) is
$P(\text{7th}|a)=\frac{93!(100-a)(99-a)(98-a)(97-a)(96-a)(95-a)a}{100!}$. Now, given that this has occurred, the probability $P(a|\text{7th})$ of any particular value of $a$ given that the first caramel is the 7th drawn should be $P(\text{7th}|a)$ for that particular $a$ divided by the sum of all possible $P(\text{7th}|a)$. $\sum_{a=0}^{100}P(\text{7th}|a)=\frac{101}{56}$, so
$$P(a|\text{7th})=\frac{P(\text{7th}|a)}{\frac{101}{56}}=\frac{56\cdot 93!(100-a)(99-a)(98-a)(97-a)(96-a)(95-a)a}{101!}.$$
Bashing out some values and adding things up, the probability that $a\le 19$ is slightly less than 50% (49.673%) and the expected value of $a$ is $\frac{65}{3}=21\frac{2}{3}$.
edit: (I've slightly altered my original answer above, mostly in the notation, to better accommodate the work below; I believe that the work above assumed that, without knowing how long it took to draw the first caramel, each possible number of caramels was equally likely.)
Suppose that $P(a)$ is the probability that there are $a$ caramels. As above, for any particular value of $a$, the probability $P(\text{7th}|a)$ that the first caramel drawn is the 7th candy drawn is
$$P(\text{7th}|a)=\frac{93!(100-a)(99-a)(98-a)(97-a)(96-a)(95-a)a}{100!}.$$
So, the probability that there are $a$ caramels and that the first caramel drawn is the 7th candy drawn is $P(a\text{ and 7th})=P(a)\cdot P(\text{7th}|a)$. By Bayes's Theorem:
$$\begin{align}
P(a|\text{7th})&=\frac{P(a\text{ and 7th})}{P(\text{7th})}=\frac{P(a\text{ and 7th})}{\sum_{k=0}^{100}P(k\text{ and 7th})}
\\
&=\frac{P(a)P(\text{7th}|a)}{\sum_{k=0}^{100}P(k)P(\text{7th}|k)}
\end{align}$$
Now, if $P(a)=\frac{1}{100}$ for all $a$, this yields the results in my original answer. If $P(a)={100 \choose a}\frac{1}{2^{100}}$ (a binomial distribution with caramel and not equally likely for each individual candy when the bag is originally filled), the expected value of $a$ is 47.5.
If $P(a)={100\choose a}p^a(1-p)^{100-a}$ (a binomial distribution where the probability of each single candy being caramel is $p$ when the bag is originally filled), the expected value of $a$ is $1+93p$. If this expected value of $a$ given that the first caramel drawn was the 7th candy drawn is to equal the expected value of $a$ without having drawn any candies, which is $100p$, then $p=\frac{1}{7}$, so the expected value of $a$ is $\frac{100}{7}=14\frac{2}{7}$.