Roulette paradox 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.
 A: The issue is that you are thinking of things in binary terms: ahead or not ahead, when really it matters by how much the gambler is ahead or not ahead.
Take a very simplistic game: You win $1$ dollar on a die roll of anything except a $6$, but if a $6$ is rolled, then you lose a million dollars. You're more likely than not to come out ahead after a single roll, but your expected value is definitely negative, and it's intuitively clear that this would not be a fun game for your wallet to play.
A: Peter gave a good example as to how you can have a negative expected value while playing a game you're likely to win (short term). Just to expand on that, the reason why the gambler comes out ahead, not ahead of the house (unless I am misunderstanding what you said, but I think you just misspoke), but ahead overall, is because the bet sizing isn't consistent. Same idea as Peter's example except the bet sizing/payout is inconsistent in a way that favors the bettor rather than the house.
After $36$ spins you have a negative expected value of ~$1.89$. Using what you provided, the chance of losing all $36$ spins being $.383$, means $1-.383=.617$, or, $61.7$% of the time the player wins $20$ dollars. This gives the side bet a positive expectation of $(.617)20=12.34$. Which, will overall gives $+10.45$.
We can use this idea to see what the minimum bet must be to net a positive expected value overall. Obviously, if you bet more than 20 dollars on the side bet it will increase your expected value further. We know that we win the side bet $61.7$% of the time. So, we need to find:
$$(.617)x \gt 1.89$$
$$x\gt ~3.063$$
So, as long as the side bet is for $\ge 3.07$ (we'll round up to the next cent as it's hard to pay/bet someone fractions of a cent), the gambler will will also have a positive expected value overall.
I'm currently taking my first Prob & Stats course so if I made a mistake be gentle and I shall correct it/delete it if necessary. Nevertheless, I hope this is helpful.
