One in Three, and the Card That Was Never in the Problem

Four cards labelled water, earth, wind and fire are shuffled and turned over one at a time. You win if both water and earth appear before fire. What is the chance?

The popular answer is one in four, and it is a correct probability for a slightly different event. The answer to the question asked is $1/3$, and the gap between the two is a clean lesson about which parts of a random ordering you are allowed to ignore.

The card that does not matter

Wind appears nowhere in the winning condition. Whether you win depends only on where fire sits relative to water and earth, so wind can be removed from the deck before you start thinking. That deletion is the whole solution, and it deserves a moment of care because it is the step people skip and then get wrong.

For our problem $n = 4$ and $k = 3$. Each of the six relative orders of water, earth and fire covers exactly four of the twenty-four full orderings, which is what makes the reduced picture uniform, and it was checked group by group rather than assumed.

Six orders, two wins

With three cards there are $3! = 6$ possible orders, all equally likely. Winning means both water and earth come before fire, which is the same thing as fire coming last of the three. Two of the six put fire last, so

There is an even shorter route. Among the three relevant cards, fire is equally likely to occupy any of the three positions by symmetry, so the chance it lands last is $1/3$ and you are done. Or as a chain of two conditions: the first of the three cards to appear is not fire with probability $2/3$, and given that, the next one is not fire with probability $1/2$, giving $\tfrac23 \cdot \tfrac12 = \tfrac13$.

The same answer with nothing deleted

If deleting a card still feels like a liberty, count all twenty-four orderings. Exactly eight of them have water and earth both before fire:

Why one in four is so convincing

The wrong answer has two independent derivations, which is unusual and explains its popularity.

The first is a miscount of what has to come last. Fire is last of all four cards with probability $6/24 = 1/4$, since fixing the final position leaves $3! = 6$ arrangements of the rest. That is a correct calculation of a real event, and that event does imply a win. It is just narrower than winning: wind is allowed to come after fire, and the two orderings where it does are wins that $1/4$ throws away. Exactly: $\tfrac13 - \tfrac14 = \tfrac1{12}$, which is $2/24$.

The second derivation is an independence slip, and it lands on the same number by coincidence. Water beats fire with probability $1/2$, earth beats fire with probability $1/2$, so multiply:

The two events are positively correlated. Learning that water arrived before fire is evidence that fire is late in the deck, which makes it more likely that earth also beat it. Multiplying probabilities assumes that evidence does not exist.

The general shape

Suppose $r$ named cards all have to appear before fire, and any number of irrelevant cards are shuffled in. Delete the irrelevant ones and you have $r+1$ cards in uniform random order, so

With $r = 2$ that is our $1/3$, and with $r = 3$ it is $1/4$, which is precisely why the trap looks right: it is the correct answer to the same question asked about one card too many. Note also how badly the independence slip degrades as $r$ grows. It predicts $2^{-r}$ while the truth is $1/(r+1)$, so at $r = 5$ it says $1/32$ where the answer is $1/6$.

Two questions one word apart

Change "both" to "either" and the answer moves a long way. You win if at least one of water and earth beats fire, which fails only when fire is first of the three, so

Counted out over all twenty-four orderings that is 16 wins, twice the 8 from before. The reduced picture in figure 1 handles both questions at once: the two orders with fire last are the "both" wins, and the four orders where fire is not first are the "either" wins. Reading the winning condition off the six-row list is faster and safer than re-deriving anything.

What the argument needs

Uniformity is the only assumption, and it is doing more work than it looks. Every ordering must be equally likely for the induced order of a subset to be uniform. A shuffle that is merely "random enough" can break the symmetry argument, and the standard failure is a deck cut rather than shuffled, where cards keep their neighbours.

The deletion step also needs the event to genuinely not mention the deleted cards. Change the winning condition to "water and earth before fire, and wind not first" and wind is back in the problem. The answer is then no longer $1/3$, and the six-order picture in figure 1 cannot see the difference, which is why the condition has to be read before the picture is drawn.

Nothing here depends on the cards being elements. They are four interchangeable labels, and the same computation answers any question of the form "do these two arrive before that one".

Sources and further reading

• The object being shuffled — Random permutation
• The counting behind equations (2) and (4) — Permutation
• The failure diagnosed in equation (3) — Independence

The answer was checked four ways before publication: exhaustive enumeration of all 24 orderings, which gives 8 wins with no formula involved; the reduced six-order picture; the chain form; and ten million simulated shuffles, which returned 0.333360 against the exact 0.333333. The uniformity of the reduced sample space was verified directly, with each of the six relative orders covering exactly four of the twenty-four.