What went wrong in this probability calculation inspired by Magic the Gathering? The problem: I have a deck of $40$ cards, all distinct from one another. I draw $3$ hands of $7$ cards each, with replacement. What is the probability that I will see the same card in all three hands?
Experimentally, the probability seems to be about $0.21$. This aligns with the result I get from the following reasoning:

You have a $7/40$ chance of drawing any particular card in a hand. So the probability would be $\left(\frac{7}{40}\right)^3$ of seeing any particular card in all three hands. Since you do not care which card it is, multiply by $40$ to get $40 \times \left(\frac{7}{40}\right)^3 \approx 0.21$.

But here's another way of doing it:

The number of all possible sequences of hands is $\binom{40}{7}^3$. To find the number of sequences of hands where we have a card in common, draw your first hand. There are $\binom{40}{7}$ ways to do this. Then choose a card; there are $7$ ways to do this. Then draw two more hands; there are $\binom{40}{6}^2$ ways to do this, since one of your cards has been chosen for you. Computing (desired sequences)/(all possible sequences) = $$\frac{\binom{40}{7} \times 7 \times \binom{40}{6}^2}{\binom{40}{7}^3} \approx 0.29$$.

What went wrong with this second way?
 A: Not to be a party pooper, but both your approaches end up overcounting the number of desired sequences. The reason is that some other card besides the one you're keeping track of can still appear in all three hands. As an extreme case, consider drawing three hands of seven from a deck of seven cards: both of your lines of reasoning yield a probability of $7$.
The way to avoid overcounting the number of favorable sequences is to use inclusion-exclusion. Let $C_i$ denote the event that card $i$ is seen in all three hands. You want $P(\cup_i C_i)$, which is readily evaluated:
$$
\begin{align}
P(\bigcup_i C_i)&=\sum_i P(C_i) - \sum_{i<j}P(C_i\cap C_j) + \cdots +\sum_{i_1<i_2<\cdots<i_7}P(C_{i_1}\cap\cdots\cap C_{i_7})\tag1\\
&={40\choose1} P(C_1) - {40\choose 2}P(C_1\cap C_2) + \cdots +{40\choose 7}P(C_1\cap\cdots\cap C_7),\tag2
\end{align}
$$
where line (2) replaces the $k$th sum of line (1) with $40\choose k$ copies of $$P(C_1\cap\cdots\cap C_k)=\frac{{40-k\choose 7-k}^3}{{40\choose 7}^3}.$$ By my calculation the end result is $0.1995872$. (Your approaches are computing just the first term in the inclusion-exclusion expansion.)
A: To flesh out the intuition behind the error that grand_chat points out in each case even more explicitly (insufficient reputation to just put this in a comment.): 
In the first calculation, $\left({7 \over 40}\right)^3$ is indeed the probability of $C_i$ (any given Card $i$ appearing in all three hands), but $P(C_1 \cup C_2) < 2 \times P(C_i)$ because $C_1$ and $C_2$ are not disjoint (Card $2$ also appears in all three hands in some of the sets of hands in which Card $1$ is always present). So, you can't just sum $P(C_1)$ and $P(C_2)$ without subtracting the nonzero $P(C_1,C_2)$ and, analogously, you can't just multiply by $40$ without doing inclusion-exclusion. 
In the second calculation, the logic is similar. There are indeed $\binom{40}{7} \times \binom{39}{6}^2$ equally likely sets of three hands in which the first card from the first hand is in all three hands, but in some of those sets of hands the second card (or third card, etc.) from the first hand is also in all three hands, so you count some hands more than once if you then just multiply by $7$ to get the number of sets of hands in which a card is present in all three. 
