Wordle (https://www.powerlanguage.co.uk/wordle/) has recently become a well-known word guessing game. The rules are simple:

… a five-letter word is chosen that players aim to guess within six tries. After each guess, the letters are either marked as green, yellow, or gray; green means that the letter is correct and in the correct position, yellow means that the letter is in the answer but not in the right position, and gray means the letter is not in the answer at all
(source: Wikipedia).

Users can post their daily game progress as an emoji block graphic on social media channels like this:


🟩 = correct letter, correct position
🟨 = correct letter, wrong position
⬜ = wrong letter

My questions to you:

  1. How many valid Wordle images are there? Take into consideration the respective meaning of the color code, and that a game ends either with a correct guess (🟩🟩🟩🟩🟩) or after six incorrect guesses.

  2. If every Twitter user (assumed 330 million) posts one Wordle image per day, how long does it take at least until every image has been posted once?


  1. Default difficulty, 'Hard Mode' turned off. You can guess anything at anytime. (Hard Mode would mean that any revealed hints must be used in subsequent guesses).

  2. Word lists do not need to be considered.

  • 1
    $\begingroup$ To start, if you can literally guess anything six times, the answer is 3^5^6 = 2.1e14. Some thoughts ... Some grids are impossible, like any row with 4 greens and a yellow. Further, if you constrain playing after winning (so you cannot have an entirely green grid) that will be reduced. If you disallow non-existent words like ZZZZZ that will eliminate lots more ... hence "no word lists" needs a little more elaboration. $\endgroup$
    – jay613
    Commented Jan 16, 2022 at 22:07
  • $\begingroup$ Again, as simple as possible. No word list means no word list. In theory, any combination of five letters. And no playing after winning, of course. $\endgroup$ Commented Jan 16, 2022 at 22:15
  • 1
    $\begingroup$ I don't think you can say "of course" in this situation. :). Creating a poorly-defined challenge to a poorly-defined set of rules. I like the question, even without an answer. But any attempt to create an answer needs a lot of arbitrary assumptions. Does no playing after winning imply optimal play? Or does it just imply that your challenge does implement one constraint of the original game, that you cannot make ANY guess after winning? But until winning, you can enter anything at all ? If the word is HELLO you can enter HELLZ ZZZZZ ZZZZZ ZZZZZ etc. $\endgroup$
    – jay613
    Commented Jan 16, 2022 at 22:19
  • 3
    $\begingroup$ Without a word list this isn't Wordle $\endgroup$ Commented Jan 17, 2022 at 0:46
  • $\begingroup$ It appears that you haven't actually defined what a Wordle image is. It should be obvious that a Wordle image is something that the game can generate, and this is very much dependent on the word lists embedded in the game. $\endgroup$ Commented Jan 17, 2022 at 1:01

3 Answers 3


Consider a Single "guess row":
It can have 5 Wrong entries - 1 Possibility
It can have 4 Wrong entries - (5/1) x 2 Possibilities (the remaining 1 is either Correct or Partially Correct)
It can have 3 wrong entries - (5x4/2x1) x 2x2 Possibilities (the remaining 2 are either Correct or Partially Correct)
It can have 2 wrong entries - (5x4x3/3x2x1) x 2x2x2 Possibilities (the remaining 3 are either Correct or Partially Correct)
It can have 1 wrong entry - (5x4x3x2/4x3x2x1) x 2x2x2x2 Possibilities (the remaining 4 are either Correct or Partially Correct)
Possibilities in total = 1 + 10 + 40 + 80 + 80 = 211.

Single "guess row" can have 0 wrong entries:
0 Correct, 5 Partially Correct - 1 Possibility.
1 Correct, 4 Partially Correct - 5 Possibilities.
2 Correct, 3 Partially Correct - 5x4 / 2x1 Possibilities.
3 Correct, 2 Partially Correct - 5x4x3 / 3x2x1 Possibilities.
The total here is 1 + 5 + 10 + 10 = 26 Possibilities.

4 Correct, 1 Partially Correct - not Possible.
All Correct - 1 Possibility - the winning guess.

Hence guesses total 211+26 = 237.

The grid can have 0 to 6 "guess rows" with a terminating "winning guess" (which would even be the seventh row), hence total is 237^0 + 237^1 + 237^2 + 237^3 + 237^4 + 237^5 + 237^6 = 1 + 237 + 56169 + 13312053 + 3154956561 + 747724704957 + 177210755074809 = 177,961,648,104,787

It would take ~~ 539277 Days ~~ 1477 Years to tweet these grids!

  • 1
    $\begingroup$ Please edit your first answer appropriately, instead of posting a new answer, and delete this one, if possible. $\endgroup$ Commented Jan 16, 2022 at 15:42
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    $\begingroup$ Either I pressed some wrong keys or some bug, my incomplete answer was posted earlier, in the middle of my typing ! I will be deleting that incomplete answer , @friedemann_bach , once Jaap Scherphuis sees my reply to his helpful comment. $\endgroup$
    – Prem
    Commented Jan 16, 2022 at 15:55
  • 2
    $\begingroup$ See my earlier comments under Gareth's answer. $\frac{237^7-1}{236}=177961648104787$ is only theoretical. The greatest number of distinct responses from any solution word is $192$ (for SAINT), though this alone provides enough images to keep Twitter users busy for centuries. $\endgroup$ Commented Jan 16, 2022 at 18:14
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    $\begingroup$ @DanielMathias , Yes, the GP Summation will work here to give the answer more easily. Nice catch with SAINT ! $\endgroup$
    – Prem
    Commented Jan 16, 2022 at 19:17
  • $\begingroup$ Looks good, your numbers are exact! $\endgroup$ Commented Jan 29, 2022 at 21:26

I count 238 possible row-colorings:

⬜⬜⬜⬜⬜ x1
⬜⬜⬜⬜🟨 x5
⬜⬜⬜⬜🟩 x5
⬜⬜⬜🟨🟨 x10
⬜⬜⬜🟨🟩 x20
⬜⬜⬜🟩🟩 x10
⬜⬜🟨🟨🟨 x10
⬜⬜🟨🟨🟩 x30
⬜⬜🟨🟩🟩 x30
⬜⬜🟩🟩🟩 x10
⬜🟨🟨🟨🟨 x5
⬜🟨🟨🟨🟩 x20
⬜🟨🟨🟩🟩 x30
⬜🟨🟩🟩🟩 x20
⬜🟩🟩🟩🟩 x5
🟨🟨🟨🟨🟨 x1
🟨🟨🟨🟨🟩 x5
🟨🟨🟨🟩🟩 x10
🟨🟨🟩🟩🟩 x10
🟩🟩🟩🟩🟩 x1

(Okay, a much easier way to see this is: 35 = 243, minus the 5 variations on 🟨🟩🟩🟩🟩 which are impossible, makes 238.)

So a naΓ―ve upper bound on the number of images is 2386 β‰ˆ 181 trillion; and a slightly less naΓ―ve upper bound is 1 + 237 + 2372 + 2373 + 2374 + 2375 + 2376 β‰ˆ 178 trillion (because as soon as you see 🟩🟩🟩🟩🟩, the game is over).

However, I think the real answer is probably much less than that. There are a lot of images that are impossible to attain due to vagaries of the word list. For example, these two rows can never appear in the same image:


There's nothing logically wrong with such a combination; e.g. if the target is ABCDE, then guessing ABEDX CBDAE will produce those colorings. But it happens that actually there are no words on the Wordle target/dictionary lists that satisfy that particular three-way cryptogram. So that knocks down our naΓ―ve estimate by roughly 6Γ—5Γ—2384 β‰ˆ 96.2 billion right there. A billion out of hundreds of trillions is just a drop in the bucket, but there are lots of drops, and they'll add up.

A naΓ―ve lower bound is 1 + 191 + 1912 + 1913 + 1914 + 1915 + 1916 β‰ˆ 48.8 trillion, since the target word PARSE permits 192 different row-colorings.

Another 1 + 2Γ—187 + 3Γ—1872 + 4Γ—1873 + 5Γ—1874 + 6Γ—1875 β‰ˆ 1.38 trillion images include the row 🟨🟨🟨🟩🟨 somewhere, which isn't possible if the target word is PARSE, but is possible if the target word is SAINT (which permits 189 different row-colorings). We could keep chipping away, but it becomes difficult to avoid double-counting any images.

So my answer is "Somewhere between 50.18 trillion and 178 trillion."

I'm using a computer to iterate over all the possible combinations of rows, of which there are 237 + (237 choose 2) + (237 choose 3) + (237 choose 4) + (237 choose 5) + (237 choose 6) β‰ˆ 237 billion, and tell me how many combinations are impossible.

62 combinations of 2 rows are impossible
23507 combinations of 3 rows are impossible
3732155 combinations of 4 rows are impossible
? combinations of 5 rows are impossible
? combinations of 6 rows are impossible

So the number of impossible pictures is at least 62Γ—114 + 23569Γ—732 + 3732155Γ—1824 + ?Γ—1920 + ?Γ—720... not that that helps narrow it down much.

  • $\begingroup$ The characters you use to represent row-colorings fail to resolve. Please replace them with characters that do resolve (ASCII ones would be nice). $\endgroup$
    – Rosie F
    Commented Apr 29 at 16:54

[EDITED to add:] No, this is wrong at present. For instance, my analysis assumes that you can have a row looking like GGGGY, which of course isn't possible because there's no other place for that last letter to go. That might possibly be the only way in which it goes wrong; I will think about that later, if someone else hasn't posted a more correct answer by then.

[EDITED to add:] As JaapScherphuis points out in comments, this also assumes that each answer is "better" than the one before. That was a kinda-deliberate assumption, on the basis of (1) the example in the question and (2) the fact that without any such assumption the question is a bit uninteresting -- but the specific assumption I made is definitely stupid, because there's no reason to expect yellow letters to be "stable". (I think it would be an interesting challenge to figure out how many images are possible with the constraint that after finding a green letter you leave it there and after finding a yellow one you keep that letter in your word but move it somewhere else. But it might be intractable without excessive brute force.)

So this is basically just a disastrously wrong answer in every respect. I'm leaving it here rather than deleting it because I don't believe in hiding my mistakes :-).

I assume there is no presumption that we are dealing with words from the real Wordle wordlist, either as target words or as guesses.

In that case:

fill in any cells past the first correct guess with green; then each column, independently, has 6 cells: zero or more grey cells, then zero or more yellow cells, then zero or more green cells, six in all.

How many ways are there to

fill a column like this? Add a thin horizontal line between the greys and the yellows, and between the yellows and the greens. (They may be in the same place; if so, put one just a little above the other. There are always some greys, some yellows, and some greens -- it's just that the number may be zero.) Then each column consists of eight symbols, two of which are horizontal lines; any two of the eight may be lines, and once we know which ones they are we can fill in the colours of the six cells. So there are (8 choose 2) = 28 configurations per column.

So there are

$28^7=13492928512$ legal diagrams. At 330M per day it would take a little under 41 days for them all to appear.

  • 1
    $\begingroup$ Yes, you're right; at the very least all permutations of GGGGY are impossible. Oops. $\endgroup$
    – Gareth McCaughan
    Commented Jan 16, 2022 at 14:38
  • 1
    $\begingroup$ I have confirmed that all of the other 238 lines are possible. A single-guess image can only be all green. A two-guess image can have any of 237 patterns for first guess. The math is easy, and the total is $\frac{237^7-1}{236}\approx1.78\times10^{14}$ $\endgroup$ Commented Jan 16, 2022 at 14:48
  • 1
    $\begingroup$ Previous comment is theoretical: Most target words cannot yield all 238 responses. $\endgroup$ Commented Jan 16, 2022 at 14:57
  • 2
    $\begingroup$ Progress in Wordle is not well-defined, though. You can probably do it with defining progress as "reducing the number of possibilities", but as you pointed out, not with each letter. $\endgroup$
    – justhalf
    Commented Jan 16, 2022 at 18:25
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    $\begingroup$ Indeed progress is not well defined, and also when you change the goal you should not assume that "progress" (even if it were defined) according to the original goal is still required. Perhaps the challenge is, "Playing real original Wordle constrained by its two original word lists, how many grids is it possible to build with real play?" Or perhaps the challenge is, "Unconstrained by guesses needing to be words, but using the real Wordle clue engine, how many grids is it possible to build?" etc etc. $\endgroup$
    – jay613
    Commented Jan 16, 2022 at 22:11

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