How Math Can Improve Your Wordle Score | Quanta Magazine

Suppose you are trying to guess today’s 2-Wordle and you ask me for a hint. How would you feel if I told you, “The word does not contain a J”? You’d probably be annoyed, because none of the possible 2-Wordle words contain a J. In the context of information theory, this is a terrible hint because it contains no information. The probability of a 2-Wordle word not containing a J is 1, so the information associated with that event is $latex\log_{2}\frac{1}{1}=\log_{2}1=0$.

(Notice that the rule of exponents, $latex2^0=1$, becomes the rule of logarithms, $latex\log_{2}1=0$. Every rule of logarithms is really a rule of exponents in disguise.)

So you ask for another hint, and this time I tell you that the secret 2-Wordle doesn’t contain an A. This is more helpful. Among the 16 possible words, 12 of them don’t contain an A, so the information associated with this event is

$latex\log_{2}\frac{1}{\frac{12}{16}}=\log_{2}\frac{16}{12}=\log_{2}\frac{4}{3} \approx0.415$

bits.

This hint gives us some information, but how much information do we need? Well, there are 16 possible outcomes, so each word on the list has a $latex\frac{1}{16}$ chance of being the secret word. Plugging this into our formula gives us:

$latex\log_{2}\frac{1}{\frac{1}{16}}=\log_{2}16 = 4$.

This means there are four bits of information associated with knowing the identity of the secret word. How do we get all four bits of information? By adding up information.

Suppose I also tell you that today’s secret word doesn’t contain a T. As with A, there are 12 words that don’t contain a T, so by itself this hint provides

$latex\log_{2}\frac{1}{\frac{12}{16}}=\log_{2}\frac{16}{12}=\log_{2}\frac{4}{3} \approx0.415$

bits of information. But since these two events are independent — which means the probability of not containing an A and not containing a T is equal to the product of the probabilities of the individual events — we can add the information for a total of 0.415 + 0.415 = 0.83 bits. Combining the knowledge from the different hints means adding information together, which gets us closer to our goal of four bits.

A simple rule of thumb in information theory is that one bit of information is equivalent to cutting the possibilities in half, because half of the possibilities would be equivalent to an event with probability $latexp=\frac{1}{2}$, and this contains $latex\log_{2}\frac{1}{\frac{1}{2}}=\log_{2}2=1$ bit of information. In our current 2-Wordle game we know the word doesn’t contain an A or a T, which reduces the possibilities to nine out of the original 16, so slightly more than half. This corresponds to the 0.83 bits of information we’ve got, which is a little less than 1.

Suppose tomorrow brings a new 2-Wordle to solve and I tell you the word does contain an A. Since four of the 16 words contain an A, this amounts to

$latex\log_{2}\frac{1}{\frac{1}{4}}=\log_{2}4=2$

bits of information. Another way to think about it is that this cuts the possibilities in half twice: from 16 to eight to four.

Now watch what happens if I also tell you that the word contains a T. As with the A, the fact that four of the 16 words contain a T conveys

$latex\log_{2}\frac{1}{\frac{1}{4}}=\log_{2}4=2$

bits of information. And since these events are independent — meaning the probability of containing both an A and a T is equal to the product of the individual probabilities — we add the information. This means we now know 2 + 2 = 4 bits of information. As mentioned above, in 2-Wordle four bits of information should be sufficient to identify the secret word, and sure enough the only 2-Wordle word with both an A and a T is AT.

Related:

1. How Claude Shannon Invented the Future

2. What a Math Party Game Tells Us About Graph Theory

3. Think of a Number. How Do Math Magicians Know What It Is?

The secret to information theory is the clever mathematical choices Claude Shannon made in defining information. Information is high when probabilities are low, and information gets added up when outcomes are independent. The secret to Wordle is to use your guesses strategically to extract the maximum amount of information possible. If your first guess tells you that the secret word ends in S, don’t automatically guess another word that ends in S. Instead, choose a word that ends with a letter you don’t have any information about yet. After all, you don’t need confirmation that S is the last letter, you need more information about the other letters, so choose your guesses so the information adds up. You’ll find that a little information, like a little knowledge, goes a long way.

Exercises

1. In a game of 2-Wordle you guess AM and it comes back

How much information have you gained?

2. In a game of 2-Wordle, knowing that the secret word contains an N is equivalent to

$latex\log_{2}\frac{1}{\frac{2}{16}}=\log_{2}\frac{16}{2}=\log_{2}8 =3$

bits of information. Knowing that the word contains an O is equivalent to

$latex\log_{2}\frac{1}{\frac{4}{16}}=\log_{2}\frac{16}{4}=\log_{2}4=2$

bits of information. Why isn’t this apparent 2 + 3 = 5 bits of information enough to uniquely identify the secret word?

3. In a Wordle game where you know that the last four letters of the secret word are ATCH, there’s one situation where you definitely should guess a word like MATCH instead of CHIMP. What is it?

4. Suppose you are playing a Wordle-like game where you are trying to identify a secret two-digit number string (the first digit could be zero). Devise an optimal strategy for guessing the number.

Click for Answer 1:

The secret word contains an M but no A, so the only remaining possibilities are ME and MY. Knowing that the secret word is one of these two out of the original 16 possibilities is equivalent to

$latex\log_{2}\frac{1}{\frac{2}{16}}=\log_{2}\frac{16}{2}=\log_{2}8 =3$

bits of information.

Click for Answer 2:

You can’t add the information here because the events aren’t independent. In fact, the information is redundant: The only words with N in them are NO and ON, so if you know the word contains an N you already know it contains an O. You really have only three bits of information here, which makes sense because the possibilities have been halved three times, from 16 to eight to four to two.

Click for Answer 3:

If you only have one guess left. While CHIMP will guarantee a win in two guesses, you’ll never win in one guess. Guessing MATCH (or HATCH or CATCH or PATCH) at least gives you a chance to win in one guess.

Click for Answer 4:

You can guarantee a win in at most six guesses. The strategy is to guess 01, 23, 45, 67 and 89. These five guesses will tell you the two numbers and their positions (or that there is only one repeated number), guaranteeing a correct sixth guess. An interesting challenge is to devise a similar strategy for three-digit number strings!

Correction: May 26, 2022

Due to a production error, the inequality symbols in $latex16<25<32$ were formatted incorrectly when this article was initially published. The error has been corrected.