I've been reading this very fun book and came across a problem that, while not expressly labeled a "paradox" was clearly written for the purpose of frustrating expectations. I understand the explanation, but I still feel in the grip of the paradox, i.e., that I'm missing something essential.
A gambler makes \$1 bets on an American roulette wheel (i.e., 38 numbers, two of which are "house" numbers, payoff on a winning number at 36 to 1). His friend, in an attempt at dissuading the gambler, bets the gambler \$20 at even odds that the gambler will be behind after 36 spins. The question is essentially how this $20 side bet will work out.
The paradox comes from the fact that, intuitively, it seems that since the gambler has a negative expectation on each spin (about -\$0.05 per spin, and -\$1.89 after 36 spins), it must necessarily be more likely than not that he'll be behind after 36 spins. But of course, in order to lose the side bet he must lose all 36 spins, which has probability 0.383, so he'll come out ahead on the side bet (which has expectation \$4.68)
I find it hard to square the fact that the roulette bet has a negative expectation with the conclusion that after the first 36 spins it's more likely than not that the gambler will be even or ahead of the house. The trick seems to be that the problem is "rigged" to require a certain relatively unlikely event (36 consecutive losses), but I can't quite put my finger on it.