, 2020

Solution:

Note that $2002 = 2 \cdot 7 \cdot 11 \cdot 13$, so that each positive factor of $2002$ is included on exactly one card. Each card can be identified simply by whether or not it is divisible by each of the 4 primes, and we can uniquely achieve all of the $2^{4}$ possibilities. Also, when considering the product of the values on many cards, we only care about the values of the exponents in the prime factorization modulo $2$, as we have a perfect square exactly when each exponent is even.

Now suppose Anne-Marie has already drawn $k$ cards. Then there are $2^{k}$ possible subsets of cards from those she has already drawn. Note that if any two of these subsets have products with the same four exponents modulo $2$, then taking the symmetric difference yields a subset of cards in her hand where all four exponents are $0 \pmod{2}$, which would cause her to stop. Now when she draws the $(k+1)$th card, she achieves a perfect square subset exactly when the exponents modulo $2$ match those from a subset of the cards she already has. Thus if she has already drawn $k$ cards, she will not stop if she draws one of $16 - 2^{k}$ cards that don't match a subset she already has.

Let $p_{k}$ be the probability that Anne-Marie draws at least $k$ cards. We have the recurrence
$$p_{k+2} = \frac{16 - 2^{k}}{16 - k} p_{k+1}$$
because in order to draw $k+2$ cards, the $(k+1)$th card, which is drawn from the remaining $16 - k$ cards, must not be one of the $16 - 2^{k}$ cards that match a subset of Anne-Marie's first $k$ cards. We now compute

$$\begin{aligned} & p_{1} = 1 \\ & p_{2} = \frac{15}{16} \\ & p_{3} = \frac{14}{15} p_{2} = \frac{7}{8} \\ & p_{4} = \frac{12}{14} p_{3} = \frac{3}{4} \\ & p_{5} = \frac{8}{13} p_{4} = \frac{6}{13}, \\ & p_{6} = 0 \end{aligned}$$

The expected number of cards that Anne-Marie draws is
$$p_{1} + p_{2} + p_{3} + p_{4} + p_{5} = 1 + \frac{15}{16} + \frac{7}{8} + \frac{3}{4} + \frac{6}{13} = \frac{837}{208}$$