# What statistical distribution would best capture a set of Wordle outcomes?

Wordle is a simple and popular word game. It is based on an older game, and has recently gone viral and attracted attention. It is available in several languages and the original site (now owned by The New York Times is here)

The rules are simple. Each day, there is a secret target five letter word. You guess a five letter word, and you get hints of which letters in your guess belong to the target word and whether the letter is in the same position. You have six tries after which the letter is revealed. That is, the outcome is the number of tries, which we can index by $$1, ..., 6$$ and assign the $$7$$ to the failure to guess the word.

Let's say I play Wordle 100 times and I want to model the distribution of outcomes. Each game yields an outcome $$y \in \{1,...,7\}$$. In order to simplify the question, I am assuming that failing and succeeding in seven tries is the same. I am not so much interest in Wordl but on the statistical distribution of this outcome.

If we were just flipping a coin until seeing a Head, we would have a geometric distribution. But in Wordle each try gives hints and changes the probability: it is close to impossible to guess the word in one try and pretty common to guess the word in 4 to 6 tries, and the geometric distribution does not capture this.

How would you model this distribution? It seems like "cheating" to try to fit something more flexible like a Negative Binomial just to fit the data. Is there a distribution that generalizes the geometric distribution with a probability parameter that changes monotonically after ever try?

Just to clarify, I don't care about linking the peculiar characteristics of World itself. I Just want something more flexible than the geometric distribution, without having to go full non-parametric and model the whole thing using seven parameters.

• (1) Failing to guess the word in 6 tries might reasonably be equated to guessing it in 7 or more tries; not to guessing it in precisely 7. (2) You seem to be undecided between asking about a mechanistic model of the outcome of a stochastic process, in which case you'd need to describe how the word to guess is selected, & the algorithm that's playing the game; & asking about an empirical model of outcomes achieved by actual (human) players, in which case a (censored) negative binomial distribution doesn't sound like cheating at all. Commented Mar 1, 2022 at 9:35
• Also it'd be helpful to at least outline the game & link to the detail of the rules, rather than assume all potential answerers are already familiar with it. Commented Mar 1, 2022 at 10:16
• Maybe something like a Dirichlet-Multinomial might help... Doesn't quite get to the monotonicity that you mentioned.
– nick
Commented Mar 1, 2022 at 13:15
• (3) "Is there a distribution that generalizes the geometric distribution with a probability parameter that changes monotonically after ever try?" - any distribution with support over the integers having monotonically increasing hazard. A negative binomial distribution with 'shape'/'size' parameter greater than one would in fact meet this criterion - though now I've played the game a little it's clear you don't want a distribution more dispersed than a Poisson. Commented Mar 1, 2022 at 17:03
• @Scortchi a good strategy in the game is to have a second and possibly also third guess with ruled out words in order to gather more information about additional letters. Say the first guess 'STORY' and you get the 'ST...' correct. Then instead of guessing something with the same letters again e.g. 'STEEP' it would be better to replace the already known first 'ST' with other letters like 'BLEEP'. So I would say that the 'hazard' in the second and third steps is probably lower. (Unless there would be an incentive to guess the word as fast as possible, but the goal is to guess within six). Commented Mar 3, 2022 at 10:46

The categorical distribution is the most general and has parameters to assign a probability to every single outcome.

$$\begin{array}{rcl} P(y = 1) &=& p_1\\ P(y = 2) &=& p_2\\ P(y = 3) &=& p_3\\ P(y = 4) &=& p_4\\ P(y = 5) &=& p_5\\ P(y = 6) &=& p_6\\ P(y = 7) &=& p_7\\ \end{array}$$

When you are looking for something like a geometric distribution or other distribution with less parameters, then there is some continuous mechanism, like a coin flip for each turn, and you can use the properties of that coin flip to parameterize the distribution.

As Scortchi mentions in the comments, such a mechanism is not necessary in order to use a parametric distribution. But these parametric distributions tend to relate to some mechanism and when you apply the parametric distribution then you could see the related mechanism as a model for the underlying mechanism of the distribution.

For the wordle outcomes it is unlikely that you can generate a distribution based on a single parameter or few parameters. There are too many factors and effects in order for the distribution to be easily modelled by a few parameters.

You have many mechanisms that are not much related

• The probability to make the first guess

• The probability to make the second and third guess based on how the first few hints worked out.

(In my personal strategy these probabilities are zero. Even if I would have only one letter left over to guess, then I would still try a different wordt that incorporates multiple letters)

• The probability to make it in the fourth or fifth guess with more hints. Depending on how well the initial words match one can guess in the fourth or fifth. But I don't think that these can be modelled with only a single parameter.

• The probability to not guess after six.

If the algorithm to guess the word fails. With a good strategy the probability should be nearly zero. This probability requires a seperate parameter and is unrelated to the probabilities to guess in 4 or 5 turns.

These mechanisms are much unrelated and can not be easily combined and modelled by a smaller set of parameters.

It is also difficult because guessing words leads to an algorithm that is not very uniform. It depends on what you get. If you would instead have something like mastermind then you could create something like a parametric distribution (e.g. the probability to guess in turn $$k$$ as function of the length and number of colours).

• Thanks for the answer. This is what I meant with "without having to go full non-parametric". There is a bit more structure to the problem that somehow should constrain the probabilities and would not require seven parameters. Commented Mar 3, 2022 at 7:35
• @Hernan: There are in effect six parameters: as the probabilities must sum to one, $p_7$ can be written as $1-\sum_{i=1}^6 p_i$. The structure you've specified amounts not to a further reduction in parameters but to a constraint on the values they can take $\frac{p_i}{\sum_{j=i}^7 p_j}$ < $\frac{p_{i+1}}{\sum_{j=i+1}^7 p_j},\ i=1, \ldots, 6$. Commented Mar 3, 2022 at 9:44
• @SextusEmpiricus: I wouldn't necessarily have qualms about modelling these outcomes with a categorical distribution, but the assertion that a model with fewer parameters would require a mechanistic justification seems too strong. People use models all the time on no stronger grounds than that they're constrained/flexible in roughly the right ways & fit the observed data pretty well. Commented Mar 3, 2022 at 10:15
• @Hernan for such a small amount of categories I see little problems in using a full parametric model. But maybe you mean to do something else. You say "I don't care about linking the peculiar characteristics of Wordle itself", but could you you explain what you are trying to do. Is your problem about Wordle? Why do you want to have "something more flexible than the geometric distribution". What is the point? Commented Mar 3, 2022 at 10:18
• @Hernan: As I've already commented, any discrete distribution can be looked at as "a sequential fail/success trial with changing probabilities". You could stipulate how they change, say $\log \Pr(X=x) - \log \Pr(X>x) = \alpha + \beta x, x=0,1,\ldots$, & the distribution thus defined could be a good or a bad fit to the observed data. Commented Mar 3, 2022 at 14:49

The probability models being suggested seem to depend on considering the selection of letters as equally likely. This does not seem very realistic to me. The word choice for #1 comes from the 26 characters in the alphabet, letters 1, 2, 3, 4, 5. Vowels a,e,i,o,u and y have an order from most to least likely and consonants have a similar order. This to me would be the starting point.

• It's hard to square this with what's known about the game. The target word is chosen from a relatively small dictionary and the frequencies of letters in the probe words are determined as much by the game strategy and history as anything; they are rather unlike the frequencies of letters in common English texts, for instance.
– whuber
Commented Jun 25, 2022 at 1:17
• "The probability models being suggested" which are these and how do they make the consideration of equally likely letters? Commented Jun 25, 2022 at 5:45
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
– mkt
Commented Jun 25, 2022 at 6:48
• I have played Wordle 162 times with 6 fails. My distribution is (1,0), (2,9), (3,36), (4,57), (5,41), (6,13), (7,6). Mean = 4.19, Mode = 4.00, Median = 3.64. Chi-Square? Looks like it could be and model parameter might be function of attempts. Commented Jul 9, 2022 at 14:41