Gambling - Blackjack

Feb 2008
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0
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
 

Laurent

MHF Hall of Honor
Aug 2008
1,174
769
Paris, France
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.
 
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Feb 2008
79
0
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.
 

Laurent

MHF Hall of Honor
Aug 2008
1,174
769
Paris, France
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 \(\displaystyle \frac{1}{2^n}\) (\(\displaystyle n\) losses), and you lose \(\displaystyle 2^n-1\). The expected amount won is therefore \(\displaystyle 1\left(1-\frac{1}{2^n}\right)-(2^n-1)\frac{1}{2^n}=0\). 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.
 
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Apr 2008
1,000
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Teyateyaneng
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?
 

Laurent

MHF Hall of Honor
Aug 2008
1,174
769
Paris, France
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.
 
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CaptainBlack

MHF Hall of Fame
Nov 2005
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someplace
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 \(\displaystyle \frac{1}{2^n}\) (\(\displaystyle n\) losses), and you lose \(\displaystyle 2^n-1\). The expected amount won is therefore \(\displaystyle 1\left(1-\frac{1}{2^n}\right)-(2^n-1)\frac{1}{2^n}=0\). 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
 
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Laurent

MHF Hall of Honor
Aug 2008
1,174
769
Paris, France
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).
 
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Feb 2008
79
0
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?
 
Apr 2008
1,000
100
Teyateyaneng
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 \(\displaystyle \left( \frac{53}{100} \right) ^7\). While losing 5 times in a row is \(\displaystyle \left( \frac{53}{100} \right) ^5\), so you can find by yourself what would be the probability of losing 6 times in a row.