Two Kings Off the Top: One in 221

Take two cards off the top of a shuffled deck. The deck is standard, fifty-two cards, four of them kings, no jokers. What is the chance both cards are kings?

The instant answer is $(4/52)^2 = 1/169$, and it is not a sloppy answer. It is the exact answer to a slightly different question, which is why it survives being checked. The right answer here is $1/221$, about $0.4525\%$, and the two differ by exactly $13/17$.

The deck changes between the draws

The first card is a king with probability

and now the deck is different in two ways at once. It holds fifty-one cards instead of fifty-two, and three kings instead of four. Given that the first card was a king,

and the two multiply:

The reduction is clean because $221 = 13 \times 17$, which is (1) and (2) reappearing as factors. If the arithmetic in (3) had come out ugly, that would have been a signal to check it.

The multiplication in (3) is a genuine conditional probability statement rather than a pattern: $P(\text{both}) = P(\text{second} \mid \text{first}) \cdot P(\text{first})$, and the conditioning is what turns $4/52$ into $3/51$. Saying it in that form is worth the extra breath, because it is the step the reflex answer omits.

The same number without any conditioning

A second route avoids the sequence entirely. Two cards drawn off the top form an unordered pair, and every pair is equally likely, so count pairs. There are

and $6/1326 = 1/221$ again. This matters more than it looks. The usual objection to (3) is that it smuggles in an order the problem never mentioned, and (4) settles that objection by never mentioning order at all. Two methods, no shared step, one answer.

Which way the error goes, and by how much

The truth is smaller than the reflex, and knowing the direction in advance is worth as much as the number. Taking a king out of the deck makes the next king harder to find, so the events are negatively dependent and the product of the marginals overstates the joint probability.

The size of the overstatement has a closed form. With $k$ special cards in a deck of $n$,

For four kings in fifty-two cards that is $156/204 = 13/17 \approx 0.765$, so the reflex answer is about thirty percent too large. The right-hand form of (5) says where the error lives: it is dominated by $1/k$, the scarcity of the special cards, and barely affected by the size of the deck. With four kings the correction is severe. With four hundred kings in a deck of five thousand two hundred, the same calculation gives $0.9976$ and treating the draws as independent is fine.

That is the general lesson worth carrying out of this problem. Sampling without replacement differs from sampling with replacement by a factor governed by how few of the interesting items there are, not by how big the population is.

When one in 169 is the right answer

Put the first card back, shuffle, and draw again. Now the second draw sees a full deck with four kings, the two events are independent, and

is exact. So the reflex answer is not wrong arithmetic, it is the correct answer under a different sampling protocol, and that is precisely what makes it hard to catch. Nobody double-checks a calculation that came out clean.

This is also why the wording deserves a moment. "Two kings in a row" could describe two consecutive draws from a shoe that gets reshuffled, in which case (6) is right. "Two cards off the top" cannot, and fixing the protocol before computing anything is the habit that separates the two.

The same argument, other targets

Nothing in (3) or (4) used the word king. Any rank works, since every rank has four cards, so the chance of two matching a named rank is $1/221$ for all thirteen of them. For a group of $m$ named cards,

so "both cards are aces or kings" with $m = 8$ gives $56/2652 = 14/663$, roughly $2.1\%$. Note this is not the same as "both aces or both kings", which is only $12/1326 = 2/221$, because the group version allows one of each.

Deeper into the deck the numbers fall off fast. Three kings in three cards is $\binom{4}{3}/\binom{52}{3} = 4/22100 = 1/5525$, and all four kings in four cards is $1/\binom{52}{4} = 1/270725$.

There is one variant worth doing in your head, because it comes out beautifully. What is the chance the two cards match each other in rank, without naming which rank? Summing (7) over the thirteen ranks gives $13 \times 6 = 78$ favourable hands out of $1326$, which is $1/17$. And that is exactly the factor from equation (2), for a reason that needs no counting at all: whatever the first card is, the second has to be one of the three cards left that share its rank, out of fifty-one. The first card is free.

I mention it because it is the cleanest illustration of when to condition and when to count. Conditioning made the matched-rank question a one-liner and would have made no progress on "two aces or two kings". Counting handles both and takes longer on the first. Neither method is the right one; the skill is noticing which question you have.

One remark on rounding

As a percentage, $1/221$ is $0.4525\%$. It is tempting to write $0.45\%$ and move on, and a reader who does the division will find that slightly wrong. A little under half of one percent is the honest short form. The precision matters here only because the whole problem is about a small correction to an obvious answer, and rounding away a quarter of a percentage point in a number that size is the same kind of carelessness the problem is testing.

Sources and further reading

• The statement behind (3) — Conditional probability
• The law in the definition above — Hypergeometric distribution
• The distinction (5) and (6) turn on — Sampling with and without replacement
• The counting in (4) and (7) — Combination

Every number above was checked before publication four ways: exact rational arithmetic, an exhaustive enumeration of all $1326$ two-card hands built from real rank-and-suit pairs, an ordered-pairs recount that settles the objection about ordering, and a seeded five million deal simulation which landed on $0.00459$ against the exact $0.0045249$.