If this reasoning were sound, all 100 people should have the same probability $1/100$ of getting the bag of quarters. But the probability for the first person who grabs a bag is easy to calculate; it's $2/101\
e1/100$.
The probability for the $n$-th person to get the bag of quarters is
$$ \frac{99}{101}\cdot\frac{98}{100}\cdot\dotso\cdot\frac{101-n}{103-n}\frac2{102-n}\;, $$
so the ratio of successive probablities is
$$ \frac{101-n}{102-n}\lt1\;, $$
so the probability is highest for the first person.