Mr. Brown always bets a dollar on the number 13 at roulette against the advice of Kind Friend. To help cure Mr. Brown of playing roulette, Kind Friend always bets Brown $20 at even money that Brown will be behind at the end of 36 plays. How is the cure working?

(Most American roulette wheels have 38 equally likely numbers. If the player’s number comes up, he is paid 35 times his stake and gets his original stake back; otherwise, he loses his stake)

Solution:

If Mr. Brown wins once in 36 turns, he is even with the casino. His probability of losing all 36 times is \displaystyle \left( \frac{37}{38} \right)^{36} \approx 0.383 . In a single turn, his expectation is

35\left( \frac{1}{38} \right)-1\left( \frac{37}{38} \right) = - \frac{2}{38}\ \text{dollars}

and in 36 turns

-\frac{2}{38}\left( 36 \right) = -1.89 \ \text{dollars}

Against Kind Friend, Mr. Brown has an expectation of

+20\left( 0.617 \right)-20\left( 0.383 \right)\approx +4.68 \ \text{dollars}

And so, all told, Mr. Brown gains +4.68 – 1.89 = + 2.79 dollars per 36 trials; he is finally making money at roulette. Possibly Kind Friend will be cured first. Of course, when Brown loses all 36, he is out $56, which may jolt him a bit.