Two cards are drawn successively without replacement from a standard deck of 52 cards. What is the probability that both cards are of the same suit and at least one of them is an ace?

Step-by-step solution

1. Step 1: Count unordered pairs that satisfy the condition.\n• Choose the suit: $4$ ways.\n• In each suit there is exactly one ace.\n• To have at least one ace, one card must be that ace and the other any of the remaining $12$ cards in the same suit.\n• So, number of favourable pairs per suit = $12$.\n• Total favourable pairs = $4\times 12=48$.
2. Step 2: Total number of unordered card pairs from the deck = $\left(\genfrac{}{}{0ex}{}{52}{2}\right)=1326$.
3. Step 3: Probability = $\frac{48}{1326}$. Simplify by dividing numerator and denominator by $6$: $\frac{8}{221}$.
4. Step 4: The same result is obtained if we consider ordered draws: favourable ordered pairs = $48\times 2=96$, total ordered pairs = $52\times 51=2652$, giving $\frac{96}{2652}=\frac{8}{221}$.

Concept tested: Counting combinations – probability with multiple conditions

Published 11 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked