Permalink Submitted by Anonymous on October 17, 2012

As pointed out in the comment above, the analysis of when to offer a double is not correct.

Assume that (1) you double when your probability of winning reaches d, (2) your opponent follows the same strategy, and (3) all doubles are accepted (which is the favored strategy if d<4/5). Then it is easy to show (in the Brownian motion model of the authors) that for d<2/3, the probability that your opponent will get an opportunity to redouble (before you win) is >1/2. In this case, an escalating series of doubles is likely, and your expected return is given by an infinite series of growing, alternating-sign terms (which does not converge). On the other hand, for d>2/3, the series converges (to a finite positive value). In this case, offering a double increases your expected return (by a factor of 2), and so is the favored strategy. Thus d=2/3 is the correct threshold for doubling.

## Threshold to offer a double is 2/3, not 4/5

As pointed out in the comment above, the analysis of when to offer a double is not correct.

Assume that (1) you double when your probability of winning reaches

d, (2) your opponent follows the same strategy, and (3) all doubles are accepted (which is the favored strategy ifd<4/5). Then it is easy to show (in the Brownian motion model of the authors) that ford<2/3, the probability that your opponent will get an opportunity to redouble (before you win) is >1/2. In this case, an escalating series of doubles is likely, and your expected return is given by an infinite series of growing, alternating-sign terms (which does not converge). On the other hand, ford>2/3, the series converges (to a finite positive value). In this case, offering a double increases your expected return (by a factor of 2), and so is the favored strategy. Thusd=2/3 is the correct threshold for doubling.Mark Srednicki

Dept of Physics

UC Santa Barbara