I’m reading Fractals, Chaos, Power Laws by Schroeder right now. In one chapter, he mentions this:

One of the several counterintuitive facts of the fair-coin tossing game is that the expected number of times equals 1 that a player will increase his capital by any given amount G before the first return to his initial capital — independent of how large the gain G is!

A similarly counterintuitive result concerns the duration DkDkD_k of the fair game, obtained by solving a difference equation similar to equation 7:

Dk=K(B−K)D_k = K(B-K)

In other words, if the player has just a single dollar and the bank has $1 million, the expected duration of the game is 999,999 tosses!

I’m having trouble understanding why the duration is so long. Intuitively, it seems like it should be shorter. I ran a simple simulation of this game, and over 10,000 simulations, the average duration is 6,548 flips. What am I missing?

I also don’t understand the justification for the provided difference equation. Assuming K is the number of dollars the player has, and B the number of dollars the bank has, then I don’t get the equation. Doesn’t it imply that the length of a game where the player and bank have the same amount should last 0 flips ((B−K)=0(B-K) = 0)? Am I just wildly forgetting how difference equations work?

I apologize if I’m missing something obvious, but I haven’t been able to find an explanation for this result from looking at various pages on the gambler’s ruin problem.

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The distribution of duration has a heavy tail, which 1000010000 simulations won’t sample sufficiently.

– Brian Tung

Oct 20 at 22:44

Regarding the equation: That is not the difference equation (I don’t think); it’s the final result. As far as the average duration is concerned, the high prevalence of short games is countered by the extreme length of a small fraction of games. If the player gets to a bankroll of just 10001000, the expected time to the end of the game is already greater than 10000001000000.

– Brian Tung

Oct 20 at 22:48

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