Is P(both aces) just (4/52)²?

Here is a question that tests this misconception head-on, in the spirit of JEE Advanced 2021 Paper 1 Q13:

The misconception answer multiplies the same chance twice: $\left(\tfrac{4}{52}\right)^2 = \tfrac{1}{169}$. The correct move is to look at the deck the second card is drawn from — after one ace is gone there are 51 cards and 3 aces — so the second factor is the conditional $\tfrac{3}{51}$:

About 0.45%, smaller than the $\tfrac{1}{169} \approx 0.59\%$ the plain product gives — because removing the first ace makes the second one rarer.

Multiplying $P(A)$ by $P(B)$ for "and" is a genuine rule, and students apply it reflexively whenever they see "both", even when nothing has told them the events are independent. The product rule for independent events has quietly become, in their head, the product rule for any two events. They are already using a multiplication rule; what they have lost is the condition attached to it.

Kelly and Zwiers, and Sánchez and Hernández, document this independence misconception: students over-apply the product rule, conflating "I have several events" with "they are independent". Restating the multiplication rule does not help, because they are already multiplying. What dislodges it is looking at the deck after the first card has gone — 51 cards, 3 aces — so the changed sample space becomes a thing they see, and the $\tfrac{4}{52}$ they reused is visibly the probability for a deck that no longer exists.

For "and", the general rule is $P(A \cap B) = P(A)\,P(B \mid A)$ — the chance of the first times the chance of the second given the first has happened. Only when $B$ does not depend on $A$, so $P(B \mid A) = P(B)$, does this collapse to the plain product $P(A)\,P(B)$.

So before multiplying the individual probabilities, ask one question: does the first event change the second? Drawing with replacement leaves the deck unchanged, so the draws are independent and the plain product is fine. Drawing without replacement — or being handed only a joint or conditional value — means the second factor is $P(B \mid A)$, read off the situation after the first event. Independence must be given or structurally true, never assumed.

Draw two cards without replacement and find $P(\text{both aces})$ properly.

1. First draw. Four aces in 52 cards, so $P(\text{first ace}) = \tfrac{4}{52}$.
2. Look at the deck for the second draw. One ace is in your hand and not put back, so the deck now holds $51$ cards and $3$ aces. The second draw is $P(\text{second ace} \mid \text{first ace}) = \tfrac{3}{51}$, not $\tfrac{4}{52}$.
3. Multiply, with the conditional second factor.
   $$P(\text{both aces}) = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221} \approx 0.45\%.$$
4. Compare with the unwarranted product. Assuming independence gives $\left(\tfrac{4}{52}\right)^2 = \tfrac{1}{169} \approx 0.59\%$ — larger, because it pretends the second ace is as likely as the first when in fact one ace has already gone.
5. Conclude. The independence assumption inflated the answer by a definite amount. Use $P(A)\,P(B \mid A)$, and only drop to $P(A)\,P(B)$ when the first draw leaves the second untouched.

The same framework settles JEE Advanced 2021 Paper 1 Q13: with only the joint $P(E \cap F \cap G)$ given, a student who assumes independence and multiplies $P(E)P(F)P(G)$ contradicts the data — the correct route bounds the intersection via inclusion–exclusion (Boole's inequality), never the unwarranted product. And a box of 5 balls, 2 red, drawn without replacement gives $P(\text{both red}) = \tfrac{2}{5} \times \tfrac{1}{4} = \tfrac{1}{10}$, not $\left(\tfrac{2}{5}\right)^2 = \tfrac{4}{25}$.