The Wordle Solver, Explained

After reading this you will understand how the solver ranks next guesses by information: why SOARE beats ADIEU, how each color pattern splits the field, and when a word you cannot win with is still the smartest play.

What the solver does, with one example

Wordle gives you six tries to find a hidden five-letter word. After each guess, every tile turns green (right letter, right spot), yellow (right letter, wrong spot) or gray (letter not in the word, subject to a duplicate rule). The solver takes the guesses and colors you have entered, computes the full list of answers still consistent with them, and then ranks candidate next guesses by how much they are expected to narrow that list.

Here is the hook. Before you guess anything, all 2315 original Wordle answers are possible. Guess SOARE and the solver reports an expected information gain of 5.89 bits. That number is not decoration. It says that on average this one guess cuts the space of possibilities by a factor of about 2^{5.89} \approx 59, from 2315 down to roughly 39 candidates. A weaker opener like FUZZY reveals far less, because it repeats a rare letter and leaves most common answers untouched.

When to lean on it, and when not

The solver is at its best in two situations: choosing an opener, and untangling a hard mid-game where many answers still fit. Once the field is down to two or three words, the math matters less than reading the board yourself. Entropy tells you how to split a large set efficiently; it cannot know that today's answer is a word you happen to recognize.

Do not use it to cheat casually and then wonder why the game feels flat. Use it to study. The most useful habit is to make your own guess first, then check whether the solver would have suggested something that reveals more bits. Over a few weeks you will start to feel which letters carry information.

The first-guess table is precomputed over all 2315 original answers. SOARE leads at 5.89 bits, with ROATE and RAISE just behind. All three cover the same high-frequency letters (R, A, S, O, E, T) in different arrangements.

The formula, and the intuition behind it

The ranking uses Shannon entropy, from Claude Shannon's 1948 paper A Mathematical Theory of Communication. The idea: a guess sorts the remaining answers into buckets, one bucket per color pattern it could produce. A guess that spreads answers evenly across many buckets tells you more than one that dumps almost everything into a single bucket.

For a candidate guess, let each possible color pattern i occur with probability p_i, where p_i is the fraction of remaining answers that would produce that pattern. The expected information is:

H = -\sum_{i} p_i \log_2 p_i

Here H is measured in bits. Each p_i is the count of answers giving pattern i divided by the total number of remaining answers. The sum runs over every pattern the guess can actually produce (at most 3^5 = 243 patterns, though far fewer occur in practice). The minus sign makes H positive, since \log_2 p_i is negative for p_i \lt 1.

The intuition in one line: -\log_2 p_i is the surprise of seeing pattern i, and H is the average surprise. High average surprise means the guess usually lands you in a small, informative bucket.

A worked example you can reproduce

Run the demo, which uses the default first-guess state (an empty board over all 2315 answers). Consider a toy version first so the arithmetic is checkable by hand, then scale up.

Splitting a field of eight answers

Suppose only eight answers remain and a candidate guess sorts them into color patterns as follows.

How one guess buckets eight remaining answers
Pattern bucket	Answers in bucket	p_i	-p_i \log_2 p_i
A	4	0.5	0.5
B	2	0.25	0.5
C	1	0.125	0.375
D	1	0.125	0.375

Compute each probability: 4/8 = 0.5, 2/8 = 0.25, and two buckets at 1/8 = 0.125.
For bucket A: -0.5 \times \log_2 0.5 = -0.5 \times (-1) = 0.5.
For bucket B: -0.25 \times \log_2 0.25 = -0.25 \times (-2) = 0.5.
Each of C and D: -0.125 \times \log_2 0.125 = -0.125 \times (-3) = 0.375.
Add them: 0.5 + 0.5 + 0.375 + 0.375 = 1.75 bits.

So this guess reveals 1.75 bits. A guess that put all eight answers into one bucket would score 0 bits: you learn nothing. A guess that split them into eight buckets of one would score \log_2 8 = 3 bits, the maximum for eight answers. The 5.89 bits for SOARE over 2315 answers is the same computation, just over hundreds of buckets instead of four.

The ceiling for any guess is \log_2 N, where N is the number of remaining answers. With 2315 answers that ceiling is \log_2 2315 \approx 11.18 bits. No single guess reaches it, because no five-letter word splits 2315 words into 2315 singleton buckets.

Watch entropy fall as the field shrinks

The most instructive thing about entropy is that its ceiling drops as you learn more. The chart shows the maximum possible bits, \log_2 N, against the number of remaining answers. Early on there is a lot to gain; near the end there is almost nothing left to extract.

The ceiling on information grows only logarithmically: going from 8 answers to 2315 raises the ceiling from 3 bits to about 11.18 bits, not by a factor of 289.

Reading the results

The solver shows you two things: the list of answers still possible, and a ranking of next guesses by expected bits. Read them together.

Remaining answers: Every word consistent with all your colors so far. If this list has one entry, you have solved it. If it has two or three, guess one of them directly.
Expected bits: The H value above, averaged over the remaining answers. Higher is better because it means a smaller expected field next turn.
Can-answer flag: Whether the guess is itself still a possible answer. Late in the game, prefer a guess that could win. Early on, a non-answer guess often scores more bits.

A guess that cannot be the answer can still top the ranking. If the field is WATCH, MATCH, BATCH, PATCH, LATCH, CATCH, guessing any one of them wins only if you are lucky and eliminates one word otherwise. Guessing CLAMP, which shares no template with the field but contains C, L, M, P, tests four of the distinguishing first letters at once and can isolate the answer in a single move. Sacrificing a turn to split the field is often worth it.

Common mistakes

The duplicate-letter rule trips people up when entering colors. If the answer is CABLE and you guess KEBAB, the middle B turns green but your second B goes gray even though B is in the word, because the answer holds only one B. Enter exactly the colors the game showed, not the colors you expected.

Three other errors recur. First, chasing green tiles too early: locking letters in place feels like progress, but a guess that reveals more bits usually shrinks the field faster. Second, ignoring the ranking's tie-breaks: when two guesses show the same bits, the tool prefers one that could be the answer, and so should you late in the game. Third, treating entropy as a guarantee. 5.89 expected bits is an average across all possible answers; a specific game can land you in an unlucky large bucket. The number describes the long run, not one puzzle.

Related tools on this site

If you enjoy pulling words apart, the Anagram & Word Finder builds words from your letters and supports patterns like c?t, which is handy when you already know a few positions. The Word Frequency & N-gram Analyzer shows which letters and words dominate a body of text, echoing why common letters carry more information. To see exactly which codepoints a word contains, including anything invisible, use the Unicode Inspector. For counting and comparison work, there is a Word & Character Counter and a Text Diff Checker.

Frequently asked questions

Is SOARE really the best first word?

By expected information over the 2315 original answers, yes: 5.89 bits, ahead of ROATE and RAISE. Other metrics, such as minimizing the worst-case number of remaining words, can favor a different opener. The gap between the top handful is small, well under a tenth of a bit.

Why does the solver sometimes suggest a word I cannot win with?

Because splitting the field can matter more than a slim chance of a lucky win. When six answers differ only in one letter, a guess outside that set can test several distinguishing letters at once and reveal more bits than any single member.

Does more bits mean I will solve it faster?

On average, yes. Each bit halves the expected field, so 5.89 bits cuts 2315 to about 39. But it is an average. A single game can put you in a large bucket where progress stalls for a turn.

Does it use word frequency or just the answer list?

The rankings here are computed over the original answer list with each word equally likely. Some solvers weight by real-world word frequency, which can shift choices slightly. Equal weighting keeps the math transparent and matches the game's fixed answer set.

Does anything I type get uploaded?

No. The solver runs entirely in your browser. Nothing you enter leaves your device.

Try it yourself in the tool