The math of five-letter word games: counting, odds and bits

Guessing a five-letter word in six tries has become a daily ritual for millions of Americans, and “what’s the best starting word?” is a debate that never dies. Underneath it sits real mathematics: counting, probability and information theory. Here the game is just a jumping-off point for those three ideas, which are useful long after the puzzle is solved. Every list and number in the examples is hypothetical, picked to keep the arithmetic clean.

Start with the multiplication principle: if one choice has $a$ options and an independent one has $b$, the pair has $a \cdot b$ outcomes. A five-letter string over a 26-letter alphabet, with repeats allowed, has 26 options in each slot.

If all five letters must be different, the first slot has 26 options, the second 25, and so on. That is the number of permutations of 26 objects taken 5 at a time.

Almost none of these strings are actual words. A typical answer list for this kind of game runs to a few thousand words, a tiny slice of the 11.9 million strings. That is what makes the game playable: language has already ruled out nearly everything before your first guess.

One more count: each tile of feedback gets one of three colors (right letter, right spot; letter elsewhere; letter absent). Five tiles give $3^5$ = 243 possible color patterns. Some never occur in practice, but 243 is the ceiling on how many groups a single guess can create.

If N candidates remain and each is equally likely, picking one at random wins with probability 1/N. With 200 candidates that is 1/200, or 0.5%. With 2 candidates, 50%. The whole game is about shrinking N as fast as possible, and every guess is a question that partitions the list, one group per color pattern.

Suppose the answer is one of N equally likely options. A well-designed yes-or-no question cuts the list in half. How many such questions do you need? The number $k$ with $2^k = N$, that is, $k = \log_2 N$. That quantity is measured in bits.

With 200 candidates, $\log_2 200$ is about 7.64 bits. A guess in the game is not a yes-or-no question: it has up to 243 answers, so at best it could deliver $\log_2 243$, about 7.92 bits. No real word comes close, but the comparison shows the theoretical limit.

When a guess splits N candidates into groups of sizes $n_1, n_2, \dots, n_k$, the probability of landing in group $i$ is $p_i = n_i/N$. Two measures grade the guess:

The first is intuitive: you land in a group with probability proportional to its size, and that size is what you’re left with. The second, Shannon entropy, measures how many “halvings” the guess delivers on average. Lower expected remaining and higher entropy mean a better guess.

Say 200 candidates remain (hypothetical numbers). Guess A splits them into four groups of 80, 60, 40 and 20. Guess B splits them into four groups of 50. Which is better?

Check. A’s groups add up to 80 + 60 + 40 + 20 = 200 and B’s to 4 × 50 = 200, so no candidate went missing. A’s probabilities add to 0.4 + 0.3 + 0.2 + 0.1 = 1. Four groups can never yield more than $\log_2 4$ = 2 bits, and that maximum happens only with equal groups, which is exactly case B. Finally, after B you still have 50 candidates, about 5.64 bits to go, which is why the game takes more than one guess.

• Confusing strings with words. 26 to the 5th counts strings; the real answer list is far smaller, and that list is what sets N.
• Using permutations when repeats are allowed. Real words repeat letters; the 7,893,600 figure only applies when all letters are distinct.
• Assuming more greens on guess one is always better. What matters is the expected size of the group you land in, not one lucky round.
• Taking a plain average of group sizes. The average of 80, 60, 40 and 20 is 50, but the expected value is 60, because bigger groups are more likely. Each group is weighted by its own size.
• Using the wrong log base. Bits mean base 2. With the natural log the unit becomes “nats”; convert by dividing by $\ln 2$.