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Considering 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.)

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