Aces Before Kings: Probability via Exchangeability
You shuffle a standard 52-card deck uniformly at random and reveal cards one by one without replacement. Let $T$ be the first time you have seen all four aces or all four kings (whichever group is completed first). Compute $P(\text{aces finish first})$ -- that is, the probability that all four aces appear before all four kings are complete -- and justify your answer rigorously using a symmetry or exchangeability argument. No lengthy enumeration.
Hints
- Strip away the 44 non-special cards -- they cannot affect which group completes first. Focus only on the relative ordering of the 8 aces and kings.
- Reframe the event: aces finish first exactly when the last of the 8 special cards to appear is a king, not an ace.
- By exchangeability every one of the 8 special cards is equally likely to occupy the last position. Four of them are kings, giving probability $4/8 = 1/2$.
Worked Solution
How to Think About It: Two heuristics crack this without any enumeration. First, *discard irrelevant randomness*: the 44 cards that are neither aces nor kings only pad the sequence — they never change which of the 8 special cards comes before which. By exchangeability of a uniform shuffle, the relative order of the 4 aces and 4 kings is a uniformly random arrangement of 4 A's and 4 K's. Second, *reframe a race as a statement about the last runner*: "aces finish first" means the 4th ace appears before the 4th king — equivalently, the very last of the 8 special cards is a king. The stopping time $T$ in the problem statement is a red herring: you never need its distribution, only the relative order of the specials.
Quick Estimate: Pure symmetry, runnable in one breath. Swapping the labels ace $\leftrightarrow$ king is a bijection on shuffles that preserves the uniform measure (relabeling cards doesn't change "uniformly random") and maps the event "aces finish first" onto "kings finish first." So the two events have equal probability. Ties are impossible — cards are revealed one at a time and the 4th ace and 4th king are different physical cards — so the two events partition everything: each has probability $\tfrac12$. Sanity check on the smallest case: 1 ace vs. 1 king in a deck — the ace "group" completes first iff the ace precedes the king, which is $\tfrac12$ by symmetry. Consistent.
Approach: Reduce to the 8 special cards by exchangeability, then ask which group owns the last special card.
Formal Solution:
Step 1 — reduction to the specials. Condition on the set of 8 positions in the 52-card deck occupied by the aces and kings. Under a uniform shuffle, every assignment of the 4 ace labels and 4 king labels to those 8 positions is equally likely — there are $\binom{8}{4} = 70$ equally likely assignments. The other 44 cards are irrelevant to which group completes first.
Step 2 — reframe via the last special card. "Aces finish first" holds iff the 4th (last) ace occurs strictly before the 4th (last) king among the specials, which holds iff the special card in the last of the 8 special positions is a king.
Step 3 — symmetry. By exchangeability, the last special position is equally likely to hold any of the 8 special cards, and 4 of them are kings: $$P(\text{last special card is a king}) = \frac{4}{8} = \frac{1}{2}$$
Ties are impossible (the two groups are disjoint and cards are revealed one at a time), so "aces first" and "kings first" partition the sample space, and the label-swap bijection confirms each has probability $\tfrac12$.
$$\boxed{P(\text{aces finish first}) = \frac{1}{2}}$$
Generalization: If one group has $a$ cards and the other $b$, the group of size $a$ finishes first with probability $\dfrac{b}{a+b}$ — the identical argument asks whether the last of the $a+b$ specials belongs to the *other* group. (Check: $a = b = 4$ gives $4/8 = 1/2$.)
Answer: $P(\text{aces finish first}) = \dfrac{1}{2}$.
Intuition
The power of exchangeability arguments is that they let you ignore irrelevant randomness. The 44 non-special cards are just noise that gets interspersed randomly -- they do not affect the relative order of aces and kings. Once you condition on the positions of the 8 specials and note that all internal orderings of those 8 are equally likely, the problem collapses to a trivial symmetry argument: which of the 8 equally likely positions is last?
This style of reasoning appears constantly in probability puzzles and in market microstructure. For example, the probability that the bid-ask spread closes on the ask side (buyer arrives last among $n$ buyers and $m$ sellers) has the same structure: condition on the set of arrivals, use exchangeability, ask which side has the last arrival. Getting comfortable with "strip away the irrelevant randomness and use symmetry on what remains" is a core interview skill. The common mistake is to attempt a full enumeration over all $\binom{52}{4}$ positions of the aces, which is correct but tedious -- the symmetry argument gives you $1/2$ in two sentences.