**C**onsidering a binary random walk, starting at zero, what is the probability of being almost sure of winning at some point only to lose at the end? This is the question set by the post-election Riddler, with almost sure meaning above 99% and the time horizon set to n=101 steps (it could have been 50 or 538!). As I could not see a simple way to compute the collection of states with a probability of being positive at the end of at least 0.99, even after checking William Feller’s Random Walks fabulous chapter, I wrote an R code to find them, and then ran a Monte Carlo evaluation of the probability to reach this collection and still end up with a negative value. Which came as 0.00212 over repeated simulations. Obviously smaller than 0.01, but no considerably so. As seen on the above picture, the set to be visited is actually not inconsiderable. The bounding curves are the diagonal and the 2.33 √(n-t) bound derived from the limiting Brownian approximation to the random walk, which fits rather well. (I wonder if there is a closed form expression for the probability of the Brownian hitting the boundary 2.33 √(n-t). Simulations with 1001 steps give an estimated probability of 0.505, leading to a final probability of 0.00505 of getting over the boundary and loosing in the end, close to the 1/198 produced by The Riddler.)

## Archive for gambler’s ruin

## the riddle(r) of the certain winner losing in the end

Posted in Books, Kids, R, Statistics with tags gambler's ruin, Monte Carlo experiment, R, random walk, US elections 2020, William Feller on November 25, 2020 by xi'an## another riddle

Posted in Books, Kids, Statistics with tags École Polytechnique, gambler's ruin, The Riddler, William Feller on August 9, 2016 by xi'anA riddle from The Riddler, a few weeks ago, to ponder while waiting for ~~Godot~~ another plane:

Consider a game played with a coin, between you and a friend, on the integer line. You are assigned a negative integer -X, and your friend is assigned a positive integer, +Y. A marker is placed at zero on the number line. Then the coin is repeatedly flipped. Every time the coin lands heads, the marker is moved one integer in a positive direction. Every time the coin lands tails, the marker moves one integer in a negative direction. You win if the coin reaches -X first, while your friend wins if the coin reaches +Y first.

What is the expected number of coin flips in a complete game?

Not much to say there since this is the classical Feller’s gambler ruin problem that I used to analyse in my practical classes at Polytechnique. The winning probability is X/(X+Y). Assuming the coin is fair. And the distribution of the hitting times (win or loose) is easily analysed by the reflection principle. But a simpler recursion shows that the expected number of steps till hitting one boundary is XY, which is rather surprising as a multiplicative formula.