Herb Susmann CV About

Bananagrams Probabilities

There was a fabled game of Bananagrams in which my Dad drew his initial 11 tiles, and immediately spelled:


If you haven’t played the game, it’s like a free-form version of Scrabble. You start by drawing a number of tiles (typically 11 or 21), and try to form a word or words out of them.

The story made me wonder how likely it is to spell an 11 letter word on the first draw.

The first step is to calculate the probability of drawing a particular word. Consider a bananagrams bag filled with only two letters, S for success and F for failure. Start pulling out tiles from the bag at random, without replacing each tile back in the bag after drawing it, and count how many S tiles you get. The hypergeometric distribution models the probability that you will get a certain number of S tiles for a given number of draws. The multivariate hypergeometric distribution extends this to the multivariate case; that is, it models the probability you’ll draw a certain number of As, Bs, Cs, etc. after drawing a number of tiles from the bag.

Fortunately, the R package extraDistr provides an R version of the multivariate hypergeometric probability mass function. Here’s a function that, given a word of length $N$ and the number of each letter tile in a bag, gives the probability of drawing that word in $N$ draws:

word_probability <- function(w, freqs) {
  # Count the number of times each
  # letter appears in the word
  letters = table(str_split(toupper(w), ""))

  # Assign a frequency of zero to any letter
  # not used in the word
  letter_freqs <- rep(0, 26)
  names(letter_freqs) <- LETTERS
  letter_freqs[names(letters)] <- letters

  dmvhyper(x = t(as.matrix(letter_freqs)),
           n = freqs,
           k = sum(letter_freqs))

Using this function, we can find the probability of drawing

bananagram_freqs <- c(13, 3, 3, 6, 18, 3, 4, 3, 12, 2,
                      2, 5, 3, 8, 11, 3, 2, 9, 6, 9, 6,
                      3, 3, 2, 3, 2)

word_probability("RASTAFARIAN", bananagram_freqs)

And the result is $4.28\times10^{-6}\%$. Pretty lucky!

Now, what is the probability of drawing any valid 11 letter word to start the game? Note that in most cases, spelling a word using all your 11 tiles excludes the possibility of spelling another word. This suggests the the probability of spelling word $A$ OR word $B$ is given by $P(A \cap B)=P(A) + P(B)$.

However, there is a special case: what if word $A$ and word $B$ are spelled with the same letters? In order to avoid double counting, we need to only want to include words with the same letters once.

I downloaded a list of words in the SOWPODS scrabble dictionary from a GitHub repository and loaded them into R. To deduplicate words with the same letters, I sorted the letters in each word and removed duplicates:


sowpods <- read_csv("./sowpods.txt", col_names = c("word")) %>%

sort_word <- function(x) {
  paste0(sort(str_split(x, "")[[1]]), collapse = "")

sowpods <- sowpods %>%
  mutate(length = nchar(word)) %>%
  mutate(word_sorted = map(word, sort_word)) %>%
  unnest(word_sorted) %>%
  distinct(word_sorted, length)

I then used the word_probability function to calculate the probability of drawing each 11 letter word, and then summed them all up:

sowpods11 <- sowpods %>%
  filter(length == 11) %>%
  mutate(prob = map(word_sorted, word_probability, bananagram_freqs)) %>%


Which computes the probability of drawing a valid 11 letter word in the opening tiles to be $~0.28\%$.

Now, suppose you start the game by drawing a different number of tiles. We can compute the probability of starting with a valid word for a range of starting tile numbers:

sowpods_probs <- sowpods %>%
  mutate(prob = map(word_sorted,
                    bananagram_freqs)) %>%
  unnest(prob) %>%
  group_by(length) %>%
  summarize(prob = sum(prob))

ggplot(sowpods_probs, aes(x = length, y = prob)) +
  geom_col() +
  xlab("Word Length") +
  ylab("Probability of drawing complete word")

Bananagram word probabilities

Drawing 3 letters has the highest probability of forming a word, at $53.7\%$. This validates my strategy of dumping early in the game to get new tiles when I get stuck, because the new letters often help me get out of the rut.

Jeopardy! Survival Analysis

Long streaks are rare in Jeopardy. Most winners only win one game, and slightly less than 40% win two games in a row. In fact, only 6 contestants have won more than ten games in a row.

This Kaplan-Meier survival plot visualizes the survival function of Jeopardy winners:

Kaplain-Meier survival plot of Jeopardy! winning streaks

The data includes seasons 1-33 (aired 1984-2016), and does not include any championship or tournament games. The point on the extreme right is due, of course, to Ken Jennings and his 74 game winning streak.

Here’s the R code for the figure.

Who was in each episode of Star Trek?

There is a website with scripts for every episode of Star Trek, so for fun I downloaded them and generated a visualization of which characters were in each episode of Star Trek.


Star Trek Characters

Build a Crystal Radio

I had fun leading a workshop a few weeks ago on building a crystal radio. We used a simple design that incorporates a loop antenna (doubling as an inductor), a variable capacitor, a germanium diode, and an earpiece.

Two pieces of wood make a frame for the antenna. We nailed in picture hanger hooks to the ends to wind wire around to form an antenna. Then we used a few nails to hold the variable capacitor in place, and soldered all the components together.

The walls of the library we were working in dampened outside radio waves a lot, but once we stepped outside we could all hear some nice strong AM stations.

Crystal Radio Workshop

Head of the Charles