Q.Four cards are successively drawn without replacement from a deck of $52$ playing cards. What is the probability that all the four cards are kings?

The key idea is conditional probability — the probability of each successive king depends on the previous draws since there is no replacement.

Step 1: Probability the first card is a king:

$\frac{4}{52}$

Step 2: Given the first is a king, probability the second is also a king:

$\frac{3}{51}$

Step 3: Given the first two are kings, probability the third is a king:

$\frac{2}{50}$

Step 4: Given the first three are kings, probability the fourth is a king:

$\frac{1}{49}$ …

This problem is a classic example of conditional probability in action. When we draw cards without replacement, the outcome of each draw changes the deck for the next draw. The probability of getting a king on the second draw depends on whether we got a king on the first draw — that's the "conditional" part.

The key insight: instead of trying to count all possible sequences of four cards and then count how many are all kings, we can think step by step. At each draw, we ask: "Given what has already happened, what's the chance of drawing a king now?" Multiplying these conditional probabilities gives the overall probability.

Let's walk through it.

1. First draw. The deck has 52 cards, and 4 of them are kings. The probability of drawing a king is simply:

2. Second draw, given the first was a king. Now the deck has only 51 cards left, and only 3 kings remain (since we already took one). So the conditional probability is:

3. Third draw, given the first two were kings. The deck now has 50 cards, with 2 kings left. So:

4. Fourth draw, given the first three were kings. Only 49 cards remain, and just 1 king is left. So:

Now, the probability that all four events happen is the product of these conditional probabilities (this is the multiplication rule for dependent events):

Let's simplify step by step. First, reduce the fractions where possible:

So the product becomes: …

Method: Multiplication Theorem for Dependent Draws (without replacement)

Use this for the probability that a whole sequence of draws all succeed, when each draw changes what remains.

Steps

Step 1: Write the chain of conditional probabilities.

For a run of successes, each factor conditions on all previous successes:

Step 2: Update both counts after each draw. …

Common Mistakes

Mistake 1: Treating the draws as independent, i.e. computing $\left(\tfrac{4}{52}\right)^4$.

Why it's wrong: there is no replacement, so after a king is drawn only $3$ kings remain in $51$ cards; the probability changes every step. Correct approach: multiply the conditional probabilities $\tfrac{4}{52}\cdot\tfrac{3}{51}\cdot\tfrac{2}{50}\cdot\tfrac{1}{49}$. …