Day 4 · King of Hearts¶
Edited by Charles and Sophia
Probability · Conditional probability · Counting arrangements
Original question on OpenQuant
Problem¶
A standard 52-card deck is uniformly shuffled. The first ace appears at position 14. Given only this information, what is the probability that the card at position 15 is the king of hearts?
Core idea
Every ordering of the deck is equally likely, so a conditional probability is a ratio of counts: (orderings that satisfy the condition and put K♥ at 15) over (orderings that satisfy the condition).
Short solution¶
1. Count the valid orderings (the denominator). "First ace at position 14" means cards 1–13 are non-aces and card 14 is an ace. Fill the deck left to right:
| Positions | Choice | Count |
|---|---|---|
| 1–13 | ordered choice of 13 of the 48 non-aces | \(\binom{48}{13}\,13!\) |
| 14 | one of the 4 aces | \(\binom{4}{1}\) |
| 15–52 | the remaining 38 cards in any order | \(38!\) |
2. Count the valid orderings with K♥ at position 15 (the numerator). K♥ must not appear in positions 1–13, so those 13 cards come from the 47 non-aces other than K♥:
| Positions | Choice | Count |
|---|---|---|
| 1–13 | ordered choice of 13 of the 47 non-ace, non-K♥ cards | \(\binom{47}{13}\,13!\) |
| 14 | one of the 4 aces | \(\binom{4}{1}\) |
| 15 | K♥ | \(\binom{1}{1}\) |
| 16–52 | the remaining 37 cards in any order | \(37!\) |
3. Take the ratio. The \(13!\) and \(\binom41\) cancel, leaving two simple ratios:
Answer
Watch out
The tempting answer is \(1/38\), treating K♥ as certainly still in the deck. But the condition says nothing about which non-aces filled positions 1–13, so K♥ may already be gone. Excluding it from those 13 slots is exactly what produces the extra factor \(35/48\).
Why the two ratios simplify so cleanly
Each ratio has a meaning. \(35/48\) is the probability that K♥ is not among the first 13 non-aces, and \(1/38\) is the probability that it then lands first among the 38 remaining cards.