1. ## Gambling - Blackjack

I have a question about blackjack (or Spanish 21, which is what I would play, but it's pretty much the same thing).

What's to stop someone from doing the following:

Bet $10; if you lose, Bet$20; if you lose,
Bet $40; if you lose, Bet$80; if you lose,
Bet $160; if you lose, Bet$320; if you lose,
Bet $640; if you lose, you're pretty screwed since the max. bet is$1000...

I am assuming this is why they have a max. bet. But, what are the chances of losing 7 times in a row... pretty slim, no? Anyone got any insight into this? I am going to try do this at a casino but I'd like some thoughts !

Could also maybe do this for roulette or craps..

Or go up in increments of $50 (more risky, as you'd have to only lose 5 times before getting to the max bet). EDIT: Or what about doing: Bet$5 (if you win, your profit is $5) If lose, bet$15 (if you win, your profit is 15-5= $10) If lose, bet$35 (if you win, your profit is 35-20= $15) If lose, bet$75 (if you win, your profit is 75-55= $20) If lose, bet$155 (if you win, your profit is 155-130= $25) If win, start your bet with$5 again

2. Originally Posted by DiscreteW
What's to stop someone from doing the following:
To prevent such martingales (this is the name (in common language, not maths) of this kind of gambling strategy) from working, casinos always impose maximums on bets, so that the strategy fails if the amount you bet gets too large.

Intuitively, one is inclined to think these strategies would anyway work in most cases (what are the odds of failing ten times in a row??). This is however an illusion. Casinos win, in average. You can do the computation for one of your strategies. In fact, there is a mathematical theory, called "martingale theory", which proves that there is no strategy at all (no matter how complicated) that allows the player to win in average. Too bad! Then what's the point of doing maths, you may ask... Well, the longer you keep away from playing, the more you save.

Laurent.

3. Originally Posted by Laurent
To prevent such martingales (this is the name (in common language, not maths) of this kind of gambling strategy) from working, casinos always impose maximums on bets, so that the strategy fails if the amount you bet gets too large.

Intuitively, one is inclined to think these strategies would anyway work in most cases (what are the odds of failing ten times in a row??). This is however an illusion. Casinos win, in average. You can do the computation for one of your strategies. In fact, there is a mathematical theory, called "martingale theory", which proves that there is no strategy at all (no matter how complicated) that allows the player to win in average. Too bad! Then what's the point of doing maths, you may ask... Well, the longer you keep away from playing, the more you save.

Laurent.
I can understand how over a long period of time the casino would come out top. But, trying this once or twice.. I think its very unlikely for someone to lose 7+ times in a row... in roulette for example, if you bet on red all the time, the chances of it landing on black 7+ times in a row seems scarce to me. Perhaps I am missing something.

4. Let's let the numbers talk by themselves...

Suppose you use the following stategy:
Bet 1. If you win, you win 1 and you stop. If you lose, you lose 1 and then you bet 2.
If you win, you win 2-1=1, and you stop. If you lose, you lose 2+1=3 and then you bet 4.
If you win, you win 4-3=1, and you stop. If you lose, you lose 4+3=7 and then you bet 8...

Doing this, you always win 1, except if your next bet exceeds the maximum bet allowed in the casino ...in which case you lose much money. This happens with probability $\frac{1}{2^n}$ ( $n$ losses), and you lose $2^n-1$. The expected amount won is therefore $1\left(1-\frac{1}{2^n}\right)-(2^n-1)\frac{1}{2^n}=0$. This is not surprising, because the game is fair: at each bet, your expected win is 0. With whatever strategy (with bounded amounts), it would have been the same. Fair games are not a good deal for casinos, and I suspect you pay a tax when you buy token or when you convert them into dollars back again. Anyway, what is for sure is that you can't have positive (>0) averaged winnings...

You can be "almost" sure to win something. But in the other case, however rare it is, you lose so much that it compensates for your winnings.

Laurent.

5. You can be "almost" sure to win something. But in the other case, however rare it is, you lose so much that it compensates for your winnings.
I don't want to be off topic, but this seems an interesting problem. How do you know that in average the total of (small) amounts you win compensate a big loss of money?
Say each time you are winning money you stop to play and restart once again. And that the max bet of the casino is 1000 dollars as said DiscreteW... In average does a big loss compensate exactly the "little" but numerous winnings?

6. Originally Posted by arbolis
In average does a big loss compensate exactly the "little" but numerous winnings?
Yes, provided you make a (very natural) boundedness hypothesis on how much you can lose (in total). Otherwise you can always win, but after a possibly very large time (with infinite expectation) and after periods of possibly very large losses. As soon as I have some spare time, I'll post a more detailed answer with definitions and proofs.

7. Originally Posted by Laurent
Let's let the numbers talk by themselves...

Suppose you use the following stategy:
Bet 1. If you win, you win 1 and you stop. If you lose, you lose 1 and then you bet 2.
If you win, you win 2-1=1, and you stop. If you lose, you lose 2+1=3 and then you bet 4.
If you win, you win 4-3=1, and you stop. If you lose, you lose 4+3=7 and then you bet 8...

Doing this, you always win 1, except if your next bet exceeds the maximum bet allowed in the casino ...in which case you lose much money. This happens with probability $\frac{1}{2^n}$ ( $n$ losses), and you lose $2^n-1$. The expected amount won is therefore $1\left(1-\frac{1}{2^n}\right)-(2^n-1)\frac{1}{2^n}=0$. This is not surprising, because the game is fair: at each bet, your expected win is 0. With whatever strategy (with bounded amounts), it would have been the same. Fair games are not a good deal for casinos, and I suspect you pay a tax when you buy token or when you convert them into dollars back again. Anyway, what is for sure is that you can't have positive (>0) averaged winnings...

You can be "almost" sure to win something. But in the other case, however rare it is, you lose so much that it compensates for your winnings.

Laurent.
Betting on red/black are not fair bets as you still lose some or all of your stake on a 0 or 00.

RonL

8. Originally Posted by CaptainBlack
Betting on red/black are not fair bets as you still lose some or all of your stake on a 0 or 00.
Oh yes, I forgot about the zero and was surprised the game could be fair... What I wrote holds for any (hypothetical) game where with probability 1/2 you win twice what you bet, and otherwise you lose. Like a roulette without the 0 (and 00).

9. Okay, lets assume the house (casino) has a 3% advantage over the player. So, the casino will win 53% of the time and the player will win 47% of the time.

Can someone calculate for me the chances of losing (that is, a player) 7 times in a row?

If it's not too much work, what about 5 and 6 times?

10. Originally Posted by DiscreteW
Okay, lets assume the house (casino) has a 3% advantage over the player. So, the casino will win 53% of the time and the player will win 47% of the time.

Can someone calculate for me the chances of losing (that is, a player) 7 times in a row?

If it's not too much work, what about 5 and 6 times?
I think the answer of losing 7 times in a row is simply $\left( \frac{53}{100} \right) ^7$. While losing 5 times in a row is $\left( \frac{53}{100} \right) ^5$, so you can find by yourself what would be the probability of losing 6 times in a row.

11. ive actually thought about this many time before - this method of gambling. at first i thought making money would be almost a certainty. but, as someone else said, the casino always wins on average. if you start with a $20 bet and doble it every time u lose, when you eventually win (if you do) you get all your money back plus a profit of your original bet, ie. in this example$20. BUT if you dont win before reaching the maximum bet:
bet $20 - lose bet$40 - lose
bet $80 - lose bet$160 - lose
bet $320 - lose bet$640 - lose
(say the maximum bet is $1000) so you have the chance to win at least one out of 6 hands. however if you by chance do not win any of these 6 hands, you lose a total of$1260. this means to make up this loss, you would have to win 63 times gambling this way with a \$20 starting bet. im thinking its going to be very unlikely to win at least 1/6 hands 63 times in a row.

12. Originally Posted by Laurent
Yes, provided you make a (very natural) boundedness hypothesis on how much you can lose (in total). Otherwise you can always win, but after a possibly very large time (with infinite expectation) and after periods of possibly very large losses. As soon as I have some spare time, I'll post a more detailed answer with definitions and proofs.
If you still have the time, I would love to see the proofs.

13. There is a simple way of proving to yourself that these strategies don't win in the log run, and that is to learn some programming get yourself a good pseudo-random number geberator and simulate the run of play. You will have to run the simulation for a large number of plays to even out the fluctuations to see what is really happening (10 million is not unheard off - meaning I have done this number or more on a number of occaisions).

The good point about this is that it will keep you away from the gaming tables, and it will have taught you a valuable and marketable skill (programming and Monte-Carlo simulation)

CB