Tails I win

Here’s a problem from Alexander Bogomolny on Twitter:

a counterintuitive property of coin-tossing: If Alice tosses a coin until she sees a head followed by a tail, and Bob tosses a coin until he sees two heads in a row, then on average, Alice will require four tosses while Bob will require six tosses

Can I demonstrate this in R?

Let’s make Alice and Bob flip coins:

# returns the number of flips needed till the start of the pattern
flips_till_pattern <- function(pattern) {
  a <- rbinom(n=50,size=1,prob = 0.5)  # simulate tossing a coin 50 times.
  # make a string of the form "Heads-Tails..." showing the sequence of random results
  str <- paste(stringr::str_replace_all(stringr::str_replace_all(a,"1","Heads"),"0","Tails"), collapse = '-')
  
  # a list of all the indices in the string where pattern holds
  inds <- c(1, unlist(gregexpr(pattern, str)))
  
  m <- substring(str, head(inds, -1), tail(inds, -1))
  ln <- lengths(strsplit(m, '-'))
  cumsum(c(ln[1], ln[-1] - 1))[1]+1
}

Playing this game 1000 times, here’s the average number of flips needed to get “Heads-Tails”

mean(replicate(1000,flips_till_pattern("Heads-Tails")))

and the average number needed for “Heads-Heads”

mean(replicate(1000,flips_till_pattern("Heads-Heads")))

More general solution

Now, you could say the above solution is terribly inelegant. There’s gotta be a way to implement this in a single line. But one advantage of my crude, string-based approach is that I can easily test other versions of the game.

For example, here’s the number of flips needed to get the pattern “Heads-Heads-Heads”

mean(replicate(1000,flips_till_pattern("Heads-Heads-Heads")))

Or how about “Heads-Tails-Heads”

mean(replicate(1000,flips_till_pattern("Heads-Tails-Heads")))

Interesting that a sequence where you throw the same result multiple times in a row is rarer than one in which you throw a pattern.

Simpler solutions

In this solution, 1 = Heads, 0 = Tails, and I can compute all the locations in a vector of 50 tosses where I find the pattern 1,1:

v = rbinom(n=50, size = 1, prob = 0.5)
x = c(1,1)
idx <- which(v == x[1])  # two heads in a row
idx[sapply(idx, function(i) all(v[i:(i+(length(x)-1))] == x))]

That’s a little shorter than the string-based solution above, but it’s not the one-line (or close) that I’d like. Any hints?

Update Yes, there are much better solutions.

from David Robinson tweet:

library(tidyverse)

crossing(trial = 1:10000,
         flips = 1:100) %>%
  mutate(heads = rbinom(n(), 1, 0.5)) %>%
  group_by(trial) %>%
  mutate(next_flip = lead(heads),
         hh = heads & next_flip,
         ht = heads & !next_flip) %>%
  summarize(first_hh = which(hh)[1] + 1,
            first_ht = which(ht)[1] + 1) %>%
  summarize(first_hh = mean(first_hh),
            first_ht = mean(first_ht))

and, if you’re so inclined, math.stackexchange.com has a mathematical solution (with fancy equations)

Thanks to R4ds and Twitter for the excellent solutions!

edit Now Correlaid has posted another great solution, with Markov diagrams.