Question

The probability of winning a stake is $5. The chance of winning each stake is 50%. You plan to bet in increments of $5, $10, $20, $40, $80, $160 ($315 in total) for each consecutive loss. For each win, you start back at $5 and continue betting following the incremental bet scheme. This means you would need to win 64 times without losing 7 in a row. What is the probability of this happening?

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Starting with $100, making 1000 $1 bets on the outcome of a coin toss, what is the risk of losing all my money at any point during the 1000 bets?

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Jonathan Devor, Director of Research and Innovation at RoadMetric (2016-present)

Updated June 23, 2019

Q. Starting with $100, making 1000 $1 bets on the outcome of a coin toss, what is the risk of losing all my money at any point during the 1000 bets?

This is a nice question in combinatorics. As with many such questions, we’ll need to divide up the question into a series of special cases. In this case, we’ll assume that you lost your last dollar at the NN’th bet. We’ll want to count in how many ways is that possible.

The OP doesn’t state this explicitly, but let’s assume that with each (fair) coin flip, there’s a 50% chance you gain a dollar, and 50% chance you lose a dollar. Assuming you lose your last dollar after NN flips, this means that if you won nn time, you must have lost (n+100n+100) times. In other words, N=2n+100N=2n+100. And since N≤1000⟹n≤450N≤1000⟹n≤450.

Now we’re ready for the heavy lifting - finding the number of ways you can go broke exactly at the NN’th bet. The challenge is taking only the combinations where you have positive money until this point. This problem turns out to be equivalent to the famous Bertrand's ballot theorem. However, while the ballot problem has you start at zero and then reach some given positive end-point (without crossing zero), we want to start at with positive value, and then reach zero (again, without crossing zero). Since reversing the coin flip sequence order is completely symmetric, the number of combination is the same, and so we can directly apply the solution to the ballot problem. I’ll now discuss how this is done.

In the ballot problem, we have two candidates A and B, who receives aa and bb votes respectively (a>ba>b). In total there are (a+ba)(a+ba) possible arrangements of the votes. We’re then asked what fraction of these arrangements have the count of A’s votes always ahead of the count of B’s votes. There are several solutions to this, most notably using André's reflection method. Though the solutions are not difficult, I won’t repeat them here. The curious reader can easily look them up here and here and pretty much in any introduction to combinatorics textbook. The bottom line is that the number of sequences where A’s vote count is always ahead is:

a−ba+b⋅(a+ba)a−ba+b⋅(a+ba)

Translating that into the terms of the OP’s problem, we simply set the number of losses as a=n+100a=n+100 and the number of wins as b=nb=n. And then step backwards from the point of reaching $0. We require that the cumulative number of losses will always be larger than the number of wins, otherwise we would have had negative money at that point. In conclusion, if we flip 1000 times, the number of coin-flip combination that would have us go broke (for the first time) at iteration N=2n+100N=2n+100 is:

1002n+100⋅(2n+100n)⋅2900−2n1002n+100⋅(2n+100n)⋅2900−2n

Note that the final term is needed because it doesn’t matter what we flip after iteration NN, so we simply multiply by all 21000−N21000−N possibilities. Finally, we’ll sum this over all possible value: n=0…450n=0…450, and divide by all possible ways you can flip a coin 1000 time, giving us the final result: