Winning at Wordle

Understanding the data

Let’s start by getting the ~500 most commonly-occurring five-letter words (that I downloaded as a csv file from the internet).

suppressPackageStartupMessages(library(tidyverse))

Warning: package 'lubridate' was built under R version 4.4.1

words <- 
  read_csv(
    file = "five-letters.csv", 
    col_names = FALSE
    ) |> 
  rename("word" = X1) |> 
  mutate(word = str_to_lower(word)) |> 
  mutate(
    l1 = str_sub(string = word, start = 1, end = 1),
    l2 = str_sub(string = word, start = 2, end = 2),
    l3 = str_sub(string = word, start = 3, end = 3),
    l4 = str_sub(string = word, start = 4, end = 4),
    l5 = str_sub(string = word, start = 5, end = 5)
  )

words

ABCDEFGHIJ0123456789

word <chr>	l1 <chr>	l2 <chr>	l3 <chr>	l4 <chr>	l5 <chr>
about	a	b	o	u	t
above	a	b	o	v	e
abuse	a	b	u	s	e
actor	a	c	t	o	r
acute	a	c	u	t	e
admit	a	d	m	i	t
adopt	a	d	o	p	t
adult	a	d	u	l	t
after	a	f	t	e	r
again	a	g	a	i	n

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For reference, I’ll also chart the popularity of each letter by their order in these five-letter words.

words_long <- words |> 
  pivot_longer(
    cols = -word, 
    names_to = "measure", 
    values_to = "values"
    ) |> 
  mutate(
    position = as.integer(
      str_sub(string = measure, start = 2)
      )
    ) |> 
  select(values, position)

words_long |> 
  count(values, position) |> 
  mutate(
    position = case_match(
      position,
      1 ~ "1st",
      2 ~ "2nd",
      3 ~ "3rd",
      4 ~ "4th",
      5 ~ "5th"
      )
    ) |> 
  ggplot(
    aes(
      x = values, 
      y = n, 
      fill = values %in% c("a", "e", "r", "s", "t")
      )
    ) + 
  geom_col() + 
  scale_y_continuous(
    limits = c(0, NA), 
    minor_breaks = NULL, 
    expand = expansion(mult = 0, add = 1)
    ) + 
  scale_fill_manual(values = c("#edd9c0", "#63431c")) +
  facet_wrap(~position, nrow = 1) + 
  theme_minimal() + 
  labs(
    title = "Frequency of letter by word order",
    subtitle = "Emphasis on the letters a, e, r, s and t\n",
    x = NULL,
    y = NULL
  ) +
  theme(
    legend.position = "none",
    plot.title.position = "plot"
    )

As you can see from the emphasis, some of these letters appear a lot more than others, and especially at the start and end of the word.

Finding the best first guess

As Wordle tells you if your letters are in the word and in the correct position, I’ll treat the latter as more important than the former. My train journey wasn’t long enough for me to delve deeper, so I assumed that it is doubly good to guess a letter in the right position than it is to guess a correct letter in any position. Given the way that my brain works, I find it easier to guess a word if I know the first and last letter, so I’ll award correct guesses in these positions with 50% more kudos than those in other positions.

If I incorporate all these preferences — and remove words where a letter occurs more than once — the following words become the best options for a first guess at Wordle.

multiplier_bonus <- 2
multiplier_edge <- 1.5

frequency <- words_long |> 
  count(values, position) |> 
  arrange(values, position) |> 
  pivot_wider(
    id_expand = TRUE, 
    names_from = position, 
    values_from = n, 
    values_fill = 0
    ) |> 
  select(values, "f1" = `1`, "f2" = `2`, "f3" = `3`, "f4" = `4`, "f5" = `5`) |> 
  mutate(f0 = f1 + f2 + f3 + f4 + f5)

results <- words |> 
  left_join(frequency |> select(values, f1, f0), by = c("l1" = "values")) |> 
  rename("a1" = "f0") |> 
  left_join(frequency |> select(values, f2, f0), by = c("l2" = "values")) |> 
  rename("a2" = "f0") |> 
  left_join(frequency |> select(values, f3, f0), by = c("l3" = "values")) |> 
  rename("a3" = "f0") |> 
  left_join(frequency |> select(values, f4, f0), by = c("l4" = "values")) |> 
  rename("a4" = "f0") |> 
  left_join(frequency |> select(values, f5, f0), by = c("l5" = "values")) |> 
  rename("a5" = "f0") |> 
  mutate(
    precise = f1 + f2 + f3 + f4 + f5,
    bonus = (multiplier_edge * f1) + f2 + f3 + f4 + (multiplier_edge * f5),
    general = a1 + a2 + a3 + a4 + a5,
    total = (multiplier_bonus * bonus) + general
    ) |> 
  arrange(desc(total))

results_tidy <- results |> 
  select(word, l1:l5) |> 
  pivot_longer(cols = l1:l5, names_to = "value_letter", values_to = "values") |> 
  select(word, values) |> 
  distinct() |> 
  summarise(duplicates = n() != 5, .by = word) |> 
  filter(!duplicates) |> 
  left_join(results, by = join_by(word == word)) |> 
  select(word, total, bonus, general, l1:l5)

results_tidy

ABCDEFGHIJ0123456789

word <chr>	total <dbl>	bonus <dbl>	general <int>	l1 <chr>	l2 <chr>	l3 <chr>	l4 <chr>	l5 <chr>
stare	1903	414.0	1075	s	t	a	r	e
share	1861	440.0	981	s	h	a	r	e
arise	1839	405.5	1028	a	r	i	s	e
trade	1802	394.0	1014	t	r	a	d	e
store	1792	389.0	1014	s	t	o	r	e
spare	1751	405.0	941	s	p	a	r	e
raise	1738	355.0	1028	r	a	i	s	e
stone	1729	400.0	929	s	t	o	n	e
stage	1701	386.0	929	s	t	a	g	e
suite	1701	404.0	893	s	u	i	t	e

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In other words, the best first guess at Wordle is stare. That shouldn’t be a surprise, as the chart above shows many occasions where s, t, a, r and e occur in five-letter words. Even better, s is the most frequent start to a word and e is the most common end letter in these words, giving the word a lot of ‘bonus’ points.

Finding the best subsequent guesses

Given that I’ve already chosen stare, what is the best second guess?¹

attempt1 <- c("s", "t", "a", "r", "e")

results_tidy |> 
  filter(
    !(l1 %in% attempt1),
    !(l2 %in% attempt1),
    !(l3 %in% attempt1),
    !(l4 %in% attempt1),
    !(l5 %in% attempt1)
    )

ABCDEFGHIJ0123456789

word <chr>	total <dbl>	bonus <dbl>	general <int>	l1 <chr>	l2 <chr>	l3 <chr>	l4 <chr>	l5 <chr>
child	1131	274.5	582	c	h	i	l	d
could	1114	270.5	573	c	o	u	l	d
blind	1091	268.0	555	b	l	i	n	d
bound	1069	278.0	513	b	o	u	n	d
pound	1066	269.0	528	p	o	u	n	d
doing	1051	232.5	586	d	o	i	n	g
found	1051	269.0	513	f	o	u	n	d
build	1021	249.0	523	b	u	i	l	d
wound	1015	255.5	504	w	o	u	n	d
would	1005	243.5	518	w	o	u	l	d

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In this case, it looks to be child. And I can use a similar approach to find the best third guess, which turns out to be funky.

attempt2 <- c("s", "t", "a", "r", "e", "c", "h", "i", "l", "d")

results_tidy |> 
  filter(
    !(l1 %in% attempt2),
    !(l2 %in% attempt2),
    !(l3 %in% attempt2),
    !(l4 %in% attempt2),
    !(l5 %in% attempt2)
  )

ABCDEFGHIJ0123456789

word <chr>	total <dbl>	bonus <dbl>	general <int>	l1 <chr>	l2 <chr>	l3 <chr>	l4 <chr>	l5 <chr>
funky	659	157	345	f	u	n	k	y

1 row

Whilst I could have continued with the analysis, my train journey didn’t permit it. I was therefore left with the following best guesses, expressed as the following mnemonic:

STARE at the CHILD that is FUNKY

👀 🧒🏻 😎

UPDATE in July 2023: It turns out that the mnemonic above is successful strategy, as it has given me a winning streak of 201 days (and counting).

Footnotes

1. These ‘best’ guesses might not be perfect, as my assumptions above and the approach in general could probably be improved, perhaps with Operational Research techniques. That said, I suspect that ‘stare’ and the next guesses are decent approximations to the ideal solution.