If you bet $100, your odds of winning are around 50% and if you lose, bet $200, and so on and so forth until you win, and then cash out with a guaranteed profit. Assuming odds of black are 47.37% as per American roulette, your odds of not winning a single one after 6 tries are 0.02125 (I think) and decrease exponentially after each subsequent try.

In: 93

You’re right, your odds of loosing 6 consecutive plays at roulette playing black are slim. Yes. But to get to that point you’ll have had to lose 5 consecutive plays, which would require you to have bet $100, then $200, then $400, then $800, then $1,600 for a total of $3,100. Then still have $3,200 left over to bet the sixth time.

This strategy is known as the martingale strategy and only works if you have an infinite bankroll, which you don’t.

EDIT: And you’d by stymied by the table limit, too.

EDIT2: Some elaboration on the Martingale Strategy based on some comments.

1. Given an infinite bankroll and no table limit, it *is* a valid strategy that is guaranteed to win. What prevents it from being implemented in real life is the lack of an infinite bankroll and the existence of table limits.

2. It does *not* rely on the gambler’s fallacy. It doesn’t assume that the odds of any given play changes, but rather exploits the fact that any non-zero chance of winning becomes a guarantee on an infinite time line. At some point, you *will* win. And, when you do, your bets are structured as to cover all previous losses plus profit.

3. It doesn’t care about the odds of winning. OP already notes the odds are less than 50/50 (because of the zeros on the wheel). But the odds of winning don’t impact the validity of Martingale, they just alter how long on average you might have to play to realize that guaranteed win. What *does* affect Martingale is the payout structure. For roulette, the odds are less than 50/50 but the payout *is* 50/50 (e.g. you win what you bet). This is why Martingale says to double. If the payout structure was different, you’re subsequent bets would be different (better payouts mean smaller increases, worse payouts mean larger increases).

If the odds of winning were actually 50%, and a lot of people did this, this is what would happen:

-Most would profit $100

-A few people would lose all their money, roughly equal to the total amount everyone else won

The fact that the odds of winning are slightly less than 50% only makes it worse for the gamblers. Basically, you have a high chance of getting a small amount of winnings, and a small chance of losing a huge amount of money. Gaining the small amount won’t affect your life nearly as much as losing the huge amount, so why bother?

The strategy you mention is a known strategy that is mathematically proven to have a guaranteed profit. However this is assuming you have infinite money to bet. The problem in your example of 6 tries is that while you have a 97.87% chance of winning $100 there is a 2.1% chance of losing all of $6400. So if 50 people walk into the casino to try this strategy and there is a $7k limit on the table or something then 49 people will walk out of there $100 richer then they were but one person will walk out having lost $6400. The casino have therefore gained $1500 from these people.

If you lose $100, then $200, $400, $800, $1600, and then win on a $3200 bet, you’ve won $3200 minus $(1600+800+400+200+100) for a total of $100 win.

So you have a 98% chance of winning $100, and a 2% chance of losing $6300.

So the net effect is like the opposite of a lottery. You have a large chance of winning very little money, and a small chance of losing a large amount of money.

If the roulette was 50/50, and you played 64 times, you’d win 63 and lose 1 (on average) which would leave you with $0. And that’s before we factor in the casino’s advantage.

Also, think about it – the odds of getting 10 reds in a row is small but not zero. Say there’s a 1/1000 chance of getting all reds, so you make your bet on black as you’ve said. Every time you lose you have to bet double just to win your initial target back, and each bet is independent so all you need to do is lose 4/5 in a row and you will be betting 10x your initial bet, but the win is still the same. Do you get me? Bet 10, lose, bet 20, lose, bet 40, lose, bet 80, lose, now all of a sudden you are making £160 bet, at 50/50 odds, and all you can win is your money back plus £10.

That bet doesn’t care that you’ve just lost 5 in a row, it’s still a 50/50 bet. Only now you have 50% chance of being up £10, and a 50% chance of being down £310. Ergo, a bad bet. Don’t do it

How had no one said the answer yet? Have you people never been to a casino?

The reason you can’t martingale is because tables have maximum and minimum bets. Lets say it’s a $5 minimum table. The maximum would be $500.

So your martingale goes 5,10,20,40,80,160,320…. Then you can’t double it again.

7 in a row streaks are not as uncommon as you might think. Also in this case you’ve risked $645 to win $5. It’s really not a good play.

Also the people mentioning gamblers fallacy are incorrect here. Each spin is independent but the chance of consecutive occurrences is to be treated as one outcome with a probability 1/2^n

This strategy loses in the long run. If you only try it once, you might eventually get your money back, plus your initial bet. Not great winnings. But if you then keep playing to try to win more–winning just $100 at a time–eventually you will have a bad streak that will overrun your ability to continue the strategy.

This seems counterintuitive, I guess, at least it did to me when I was younger and coded a simulation in Commodore 64 to test exactly this strategy. And it was interesting to watch the money slowly go up for a time and then inevitably bottom out at nothing.

If you want to win in a casino, you should either get really good at poker or counting cards. You can’t out-probability any game.

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