The way she describes winning at roulette is bogus.
First of all, if your money is doubled when you hit the right colour, you don't make any profit in the long run. You will stay on the same amount.
Second of all, there is a 0, and often also a 00 on the roulette wheel, which is neither red nor black. This means that the chance you're correct is less than 50%, so you stand to make a loss in the long run.
There is no way to win at roulette, without using a computer to predict where the ball is going to be, based on the movement and spinning of the wheel. And I seriously doubt that even that is possible using just human observation and a trigger in your foot.
I think you're misinterpreting the author's point about roulette. She never claimed that you could profit off of roulette in the long run. You can, however, almost certainly profit off of it in the short run. The strategy she describes has one double one's bet if one loses. If you bring enough money to cover 7 losses (which means you would have to have roughly 100 times your initial bet), then you have a >98% chance of coming out ahead. You won't be ahead by much, and the strategy won't work in the long run (i.e. you can only do it a couple of times--when you inevitably end up with 7 consecutive losses it will more than wipe out any earnings you make), but it is nevertheless a legitimate strategy to "beat" roulette in the short term.
I hear what you're both saying. But I wouldn't call beating roulette in the short term a legitimate strategy exactly because it only works in the short run--strategies are supposed to work for longer than that! The only legitimate value I can see is that if you actually enjoy the experience of playing roulette that it's worth paying for it. Anyone who's willing to buy volatility can win in the short term. Otherwise, stick to a game of skill and get some money on the table while you do it. If you're good you can get some vol. and some positive EV.
A fair coin has a roughly 51% chance of landing with the same side up as it started with. See the paper by Diaconis et al. "Dynamical Bias in the Coin Toss."
A fair coin is one on which p = 0.5.
A coin that has a 51% chance is clearly not a fair coin.
This isn't even what I was talking about, but is still equally easy to turn into a fair coin.
In this case, caller calls in the air after flipper hides the side that starts up. 0.5 * 0.51 + 0.5 * 0.49 = 0.5
For any other case, flip a coin twice. Assign one person HT and the other person TH, reflipping TT and HH. Since p * (1 - p) = (1 - p) * p, the dual flip is a fair flip.
Please don't mention the possibility of sleight of hand next.
-- There are a hundred tickets in a hat. 99 of the tickets win you $1. The last ticket loses $10,000 dollars.
the sentence "it is nevertheless a legitamte strategy to "beat" roulette in the short term" is absolutely false. The expected value of each roll of the dice is negative, summing a bunch of negative expectations leads to an even large negative expectation.
EDIT: Also, some people will talk about infinite bankrolls, there is no such thing! even if there was, infinity + X = inifinity, why are you gambling?
I would pick out of the hat once, maybe even twice. The point is that I'm never going to sum over a bunch of negative expectations. I beat the probability of losing down to something negligible, play once, and then quit forever.
There is a difference between being willing to gamble and beating a game. It is obvious to see that each time you play this game you are losing $9901 dollars (.99x1+.01x-10000). That is the expected outcome each time you play the game.
Probability is often not intuitive. It is easy to think you "beat" the system because you won a couple bucks.
This game may better illustrate my point.
There is a hat with 100 tickets. 99 of the tickets produce a $1 win. The last ticket requires you kill yourself. Would you play this game once or twice?
I'll agree with your semantics. To me, "beating a game" implies that you can consistently come out ahead over a long period of time. Nevertheless, playing an "unbeatable" game over a short period of time is not necessarily irrational. Although your expected value in your first version of the hat game is -$9901, you are not losing $9901 every time you play. If you play long enough it will average out to be this, but every time you play you will either win $1 or lose $10,000. If the chance of a devastating loss is negligible, it's hardly unwise to make a gamble, even if it has a negative expected value--this is why we occasionally do risky things that have a small chance of killing us, like driving or eating sushi. The problem in thinking in terms of an expected value is that an expected value presupposes many trials--given a single experiment, it's a meaningless concept. Indeed, given a single experiment, both the classical and frequency interpretations of probability are meaningless with regards to what the probability of a given outcome really is. In the case of a single experiment, the only interpretation that makes sense is the Bayesian interpretation.
I would not play your second game, but the reason I wouldn't has nothing to do with expected values (notwithstanding the fact that it's technically impossible to calculate an expected value for an unquantifiable outcome like death). In your first game, I feel that I would come out ahead, and in the second game I feel like it's not worth it.
What the author is describing is a 'Martingale' strategy, and it is indeed a surefire way of making a profit (with one critical caveat: you need an unlimited bankroll).
After each win, you always earn a profit of X (where X is your base bet), regardless of how many rounds you have lost before.
Without a sufficiently large bankroll relative to X, however, probability of bankruptcy is high.
No, it isn't. The house can only lose the initial amount.
I bet 50, and lose.
I bet 100, and lose.
I bet 200, and lose.
I bet 400, and lose.
I bet 800, and lose
1600, 3200, 6400, 12800, 25600, 51200, 102400, 204800, 409600, and then...
...I lay 819,200 dollars on the table and WIN! 50 dollars.
The reason they have limits is so that Sheik Abudab can't come in and lay down 500 million dollars and bust them. That is one valid reason to take bad odds: If you can destroy something as a side effect, without really being in danger yourself.
In your Martingale example, the table maximum would most likely be set around $2000. So if you lose 6 times (on your $50, $100, $200, $400, $800, and $1600 bets), you can't double your bet anymore. You'll be stuck with betting $2000 at most, which means you can longer guarantee yourself a profit.
But now that I think about it, you could always just move to another table with a higher limit. Of course, if you're going through all this trouble to make $50, you probably don't have the bankroll to keep going.
Maybe another reason for tables having different limits is that the pit bosses don't trust the dealers (or security?) in the low-roller areas to handle money above a certain amount?
Don't the 'high-roller' tables have a higher minimum bet? Most people don't start out at the $1 table, and then move up to the high-roller table as they keep losing though. Most people just plop down at a table that has a range/minimum that is right up their alley with regards to their average bet. Then they only really move if there is a reason to (like feeling that this table isn't 'lucky' anymore, or the dealer changes, etc).
First of all, if your money is doubled when you hit the right colour, you don't make any profit in the long run. You will stay on the same amount.
Second of all, there is a 0, and often also a 00 on the roulette wheel, which is neither red nor black. This means that the chance you're correct is less than 50%, so you stand to make a loss in the long run.
There is no way to win at roulette, without using a computer to predict where the ball is going to be, based on the movement and spinning of the wheel. And I seriously doubt that even that is possible using just human observation and a trigger in your foot.
[edit: the author is female, sorry about that]