Martingale betting system (Roulette)

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Chugga
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Martingale betting system (Roulette)

Postby Chugga » Wed Aug 13, 2008 12:45 am UTC

Excuse me if this isn't the right forum to post this in, but it is concerning probability, so here seemed the greatest fit.
A friend of mine recently came up with a betting system for roulette which he claimed was flawless. Upon doing a bit of research (long live wikipedia!) it turned out to be the martingale* betting system, wherein one doubles one's bet after each loss. The idea is that you are almost definitely going to recover your original bet:
say a starting bet of $5 (which is what he suggested) and 1 to 1 payout;
bet $5 and lose (loss $5) 0.514 prob
bet $10 and lose (loss $15) 0.264 prob
bet $20 and lose (loss $35) 0.136 prob
bet $40 and lose (loss $75) 0.070 prob
bet $80 and win (win $160, overall win $5) 0.964 prob

My maths may be way off (it's not my forte, I'm the first to admit that) but even so, as long as one does not get greedy you could make a reasonable amount of money off this, certainly more than a minimum wage job.
Would somebody care to point out the inevitable flaw in my argument please? I'm not a gambler, so I don't know enough about it apart from what I've researched up on the interwebs.

*http://en.wikipedia.org/wiki/Martingale_(roulette_system)

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Re: Martingale betting system (Roulette)

Postby GreedyAlgorithm » Wed Aug 13, 2008 1:46 am UTC

The wikipedia article sums up why it doesn't work. In your example, for instance, you only have a 0.964 chance of winning $5. That's not enough to make up for the 0.036 chance you LOSE $5+10+20+40+80=$155. And certainly, you can go again, but if you have any maximum amount of money (note: you do) then you're expected to lose in the long run because at some point you can't double your bet again. The expected value is always negative. Think of it like this: you've got a great chance to win $5 but if you lose then you're destitute. It's just like earning minimum wage except after a year you lose all your money and your house too.
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Re: Martingale betting system (Roulette)

Postby JayDee » Wed Aug 13, 2008 3:20 am UTC

GreedyAlgorithm wrote:And certainly, you can go again, but if you have any maximum amount of money (note: you do) then you're expected to lose in the long run because at some point you can't double your bet again.

Not only do you probably have a maximum amount of money, you'd be hard pressed to find a Roulette table without a maximum bet.
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Re: Martingale betting system (Roulette)

Postby phlip » Wed Aug 13, 2008 3:40 am UTC

My favourite type of gambling fallacies are the ones where, if you switch "win" and "lose" in the situation, gamblers would still go for it.

Like, if you win lots, you must be on a streak, and thus more likely to win... but if you lose lots, you must be due, and thus more likely to win.

In this case... they've also invented a gambling system where, most of the time, you lose a couple of dollars, but very rarely, you'll win heaps. It's called the lottery.

Martingale is sort of a reverse lottery... most of the time, you'll win a miniscule amount, but on the rare occasion you lose, it's a huge amount that more than overrides all of those small winnings.

It's a simple enough inductive proof that, for any purely-random game where the payouts are less than fair (eg: roulette... a double-or-nothing payout has less than 50% chance), with any betting strategy that involves placing a finite number of bets (ie: all of them), the expected value is <= 0, with equality only when the number of bets placed is 0.

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Re: Martingale betting system (Roulette)

Postby folkhero » Fri Aug 15, 2008 12:02 am UTC

The other posts have really summed it up pretty well, but I'd like to add the metaphor of picking pennies up off of a traintrack. You make a little bit of money for a while, and then SMACK! WTF? you just got hit by a train.

If you really have enough money to make the probability of losing extremely tiny, then:

1. Why are you willing to risk it (even if you almost certainly won't lose it) in the first place for a pittance as a time reward? (you're already rich, who cares about $5)

2. Why are you wasting your time sitting at a roulette wheel when you could be sailing you yacht to the Mediterranean? (If you actually enjoy the act of gambling then it seems like you would want to gamble for reasonably high stakes to start with.)

3. Why not use the money to start your own casino and get the odds on your side? (or invest in a blue chip stock or some sort of mutual fund) Instead of tiny profit with tiny chance of catastrophic loss, you get moderate profit with tiny chance of catastrophic loss. A.K.A. picking up silver dollars off a traintrack.
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Re: Martingale betting system (Roulette)

Postby Charlie! » Fri Aug 15, 2008 3:38 am UTC

It doesn't work, because of house limits on bets. If there were no limit, then it's true that you could just keep betting forever and theoretically make money every time (as long as you had TONS of capital). But with house limits factored in you lose exactly as much as you earn, without any pesky infinities.
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Re: Martingale betting system (Roulette)

Postby JayDee » Fri Aug 15, 2008 4:03 am UTC

Not to mention that you don't get 1-1 pay outs, either. Can't forget those pesky zeros.
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Re: Martingale betting system (Roulette)

Postby crzftx » Fri Aug 15, 2008 6:44 am UTC

The idea is nearly flawless. There is a clearly defined win and loss. Every win gains back enough money to return the losses, and then some. If you had infinite money and there were no house limits, you'd be fine. Even without the limits, you must ask yourself where to stop. You cannot have infinite money. Sure, it's a 1/2^10 chance of losing 10 coin flips. But if you've already lost 7, and you're starting to get low on cash, when do you call it a loss? You will eventually run out of enough money to bet what you need to make up everything.

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Re: Martingale betting system (Roulette)

Postby Blatm » Fri Aug 15, 2008 6:24 pm UTC

Here is a good (but short) analysis of the Martingale system using graphs.

Looks to me like flat betting is linear (obviously), and the Martingale is exponential (can someone figure out the exact function?), which is somewhat counterintuitive, because of the asymptote at 0, and that it will at some point be more profitable than flat betting at some point. Of course, that assumes both infinitely divisible currency and a total lack of variance.

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Re: Martingale betting system (Roulette)

Postby skeptical scientist » Fri Aug 15, 2008 8:45 pm UTC

Blatm wrote:Here is a good (but short) analysis of the Martingale system using graphs.

Looks to me like flat betting is linear (obviously), and the Martingale is exponential (can someone figure out the exact function?), which is somewhat counterintuitive, because of the asymptote at 0, and that it will at some point be more profitable than flat betting at some point. Of course, that assumes both infinitely divisible currency and a total lack of variance.

Both have asymptotes at zero, because as your expected bankroll dips below zero, your probability of having any money left to bet with approaches zero, so your expected bet approaches zero. The slope of the curve at any point is just the house advantage times your expected bet, so the martingale system drops off more steeply at first, since its expected bet is higher (constant betting always bets 10, whereas the martingale system bets 10, 20, 40, 80, or more, depending on the length of its losing streak). However, since the size of a bet is capped at your remaining cash, as the martingale better loses money, the average bet size decreases, so the slope is less negative.
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Re: Martingale betting system (Roulette)

Postby mrbaggins » Sat Aug 16, 2008 10:40 am UTC

As said, it doesn't work, unless you have an infinite bankroll, which you don't.

Even then, in roulette, there is a house cut of a little under 5%. Martingale can NEVER work on a roulette table.

A similar betting method that will let you play around on the table a lot longer (but will still eventually lose you money unless it is an even money on genuine 1:1 odds) is the 'catch-a-streak' method.

1: Put 10 on both red and black
2: Whatever comes up, let it slide (leave the winnings on the one that won) and replace the lost bet.
3: Go to 2, until such times as the winnings pile is sufficient for your greediness.

Worst case scenario is the winning number switching color every spin (Red-Black-Red-Black etc.)

Example of typical set up: XXR means the amount of money on Red, XXB = black
1: +10R +10B (Add 10 to each side) (Total -20)
Black comes up
2: +10R 20B (Replace Red, Let black slide) (Total -30)
Red comes up
3: 20R +10B (Replace black, let red slide (Total -40)
Red
4: 40R +10B (Replace black, slide red (Total -50) [[Take winnings now to be down $10 from 4 bets, or try for 3 reds in a row to get 80, being 30 in front]]
Black
5: +10R 20B (Replace red, slide black (Total -60)
Black
6: +10R 40B (Replace red (Total -70)
Black
Take winnings of 80, you're in front 10.
If you let it slide one more time, and black wins, you'd win 160, having paid 80. A streak of 5 is worth 320.

As you can see, excluding the first bet, each round only costs 10. But if you can get 3 in a row, you've won 80. Odds of having a streak of 3? 1/8. Convenient, eh?

The flaw is obvious if you consider starting at a certain amount on the table.
#2 is basically just #1, but you could base it on the previous spin. So if the last number is red, you could put 20 on red, and 10 on black. If it comes up black, you're down 10, if it comes up red, you're only up 10. So both #1 and #2 are pointless, you can skip straight to #3.

But:
If the last 2 numbers were black, and you put 40 on black and 10 on red, what are the outcomes? You're up 30 (50 on table, 80 won) or down 30 (50 on table, 20 won). It's ALWAYS an even money bet, regardless of how awesome it looks.

Admittedly its not a gaping flaw, its just even money, but it's important to realise it isn't as powerful as it looks. And that on a roulette table, it's not really even money.
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Re: Martingale betting system (Roulette)

Postby Blatm » Sat Aug 16, 2008 9:15 pm UTC

Of course, the flaw with that, is that roulette isn't 50/50, but about 48/48/4 that it comes up 0.

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Re: Martingale betting system (Roulette)

Postby crzftx » Sat Aug 16, 2008 9:56 pm UTC

mrbaggins wrote:As said, it doesn't work, unless you have an infinite bankroll, which you don't.

Even then, in roulette, there is a house cut of a little under 5%. Martingale can NEVER work on a roulette table.

A similar betting method that will let you play around on the table a lot longer (but will still eventually lose you money unless it is an even money on genuine 1:1 odds) is the 'catch-a-streak' method.

1: Put 10 on both red and black
2: Whatever comes up, let it slide (leave the winnings on the one that won) and replace the lost bet.
3: Go to 2, until such times as the winnings pile is sufficient for your greediness.

Worst case scenario is the winning number switching color every spin (Red-Black-Red-Black etc.)

Example of typical set up: XXR means the amount of money on Red, XXB = black
1: +10R +10B (Add 10 to each side) (Total -20)
Black comes up
2: +10R 20B (Replace Red, Let black slide) (Total -30)
Red comes up
3: 20R +10B (Replace black, let red slide (Total -40)
Red
4: 40R +10B (Replace black, slide red (Total -50) [[Take winnings now to be down $10 from 4 bets, or try for 3 reds in a row to get 80, being 30 in front]]
Black
5: +10R 20B (Replace red, slide black (Total -60)
Black
6: +10R 40B (Replace red (Total -70)
Black
Take winnings of 80, you're in front 10.
If you let it slide one more time, and black wins, you'd win 160, having paid 80. A streak of 5 is worth 320.

As you can see, excluding the first bet, each round only costs 10. But if you can get 3 in a row, you've won 80. Odds of having a streak of 3? 1/8. Convenient, eh?

The flaw is obvious if you consider starting at a certain amount on the table.
#2 is basically just #1, but you could base it on the previous spin. So if the last number is red, you could put 20 on red, and 10 on black. If it comes up black, you're down 10, if it comes up red, you're only up 10. So both #1 and #2 are pointless, you can skip straight to #3.

But:
If the last 2 numbers were black, and you put 40 on black and 10 on red, what are the outcomes? You're up 30 (50 on table, 80 won) or down 30 (50 on table, 20 won). It's ALWAYS an even money bet, regardless of how awesome it looks.

Admittedly its not a gaping flaw, its just even money, but it's important to realise it isn't as powerful as it looks. And that on a roulette table, it's not really even money.



The gaping flaw is Reality.

Code: Select all

R  B  (Bank)  R  B  (Bank)  R  B  (Bank)  R  B  (Bank)
0  0  (100)   0  0  (100)   0  0  (100)   0  0  (100)
10 10 (80)    10 10 (80)    10 10 (80)    10 10 (80)
R             R             R             R
20 10 (70)    20 10 (70)    20 10 (70)    20 10 (70)
R             R             B             B
40 10 (60)    40 10 (60)    20 20 (50)    20 20 (50)
B             R             R             B
40 20 (20)    80 10 (50)    40 20 (30)    20 40 (30)
0  0  (80)    0  0  (140)   0  0  (90)    0  0  (90)

You only win 1/4 of the time assuming you pull out after 3, averaging no gain. This is with 50/50 odds, instead of the real ones.

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Re: Martingale betting system (Roulette)

Postby Random832 » Wed Aug 20, 2008 7:08 pm UTC

JayDee wrote:Not to mention that you don't get 1-1 pay outs, either. Can't forget those pesky zeros.


You're mistaken - a bet on red, black, even, or odd _is_ a 1-1 payout, what the zeros mean is that it doesn't have a probability of 50%. Which the OP took into account. The payout (or "odds") is just the ratio between what you win (on top of the amount you bet, if you win) and what you bet; it has nothing to do with probability.

mrbaggins wrote:Even then, in roulette, there is a house cut of a little under 5%. Martingale can NEVER work on a roulette table.


Again, the house cut comes from manipulating the probabilities (1 in 38 instead of 1 in 36), NOT the payout. This system does defeat that aspect by constantly doubling your bet (or multiplying by an appropriate amount for any other payout) - with an infinite bankroll and no bet limit, you could eventually win doing that no matter WHAT the odds of winning were. (well, unless it absolutely literally NEVER lands on your number, which is possible, but highly improbable) - what it runs up against is running out of money (either exhausting your bankroll or reaching the maximum bet; it's the same for the model - you're unable to place the next bet)

Absent that, it would even work if you were betting on just ONE number (2.6% probability) to make only a 1-1 payout:

Bet $5 lose (loss $5 ) .974 probability [n=0]
Bet $10 lose (loss $15 ) .948 probability [n=1]
Bet $20 lose (loss $35 ) .923 probability [n=2]
Bet $40 lose (loss $75 ) .899 probability
Bet $80 lose (loss $155 ) .875 probability
Bet $160 lose (loss $315 ) .852 probability
Bet $320 lose (loss $635 ) .830 probability
Bet $640 lose (loss $1,275 ) .808 probability
Bet $1,280 lose (loss $2,555 ) .787 probability
Bet $2,560 lose (loss $5,115 ) .766 probability
...
Bet $69.017E+69 lose (loss $138.035E+69 ) .00195 probability [n=don't give a shit]
Bet $138.035E+69 lose (loss $276.070E+69 ) .00190 probability
Bet $276.070E+69 lose (loss $552.140E+69 ) .00185 probability
Bet $552.140E+69 lose (loss $1.104E+72 ) .00180 probability
Bet $1.104E+72 lose (loss $2.209E+72 ) .00175 probability
Bet $2.209E+72 lose (loss $4.417E+72 ) .00171 probability
Bet $4.417E+72 lose (loss $8.834E+72 ) .00166 probability
Bet $8.834E+72 lose (loss $17.668E+72 ) .00162 probability
Bet $17.668E+72 lose (loss $35.337E+72 ) .00157 probability
Bet $35.337E+72 lose (loss $70.674E+72 ) .00153 probability
Bet $70.674E+72 win (win $141.348E+72, overall win $5) .99851 probability

...or...
Bet $8.834E+72 lose (loss $17.668E+72 ) .00162 probability
Bet $17.668E+72 lose (loss $35.337E+72 ) .00157 probability
Bet $35.337E+72 lose (loss $70.674E+72 ) .00153 probability
...
Bet [imath]$5 \cdot 2^{n}[/imath] lose (loss [imath]$10\cdot 2^{n}-5)[/imath]) [imath]\frac{37}{38} ^ n[/imath] probability
Bet [imath]$5 \cdot 2^{n+1}[/imath] win (win [imath]$10\cdot 2^{n+1}[/imath], overall win $5)[/imath]) [imath]1-\frac{37}{38}^ {n+1}[/imath] probability
...
[math]\lim_{n \rightarrow \infty} =
\begin{cases}
\text{Bet }\infty\text{ lose (loss $}\infty), & 0\text{ probability} \\
\text{Bet }\infty\text{ win (win $}\infty,\text{ overall win }5), & 1\text{ probability}
\end{cases}[/math]
(but don't you feel silly that you went so high for a measly $5 net payoff?)


Remember, you can't lose more money on one bet than you put down, so the payout doesn't matter until you win, though a lower payout does put a limit on how much you can increase your bet by each time and actually make it all up when you win. The probability doesn't matter at all (it just changes how fast the cumulative probability changes) The flaw - and the ONLY flaw - is that you can't actually keep arbitrarily increasing your bets.
Last edited by Random832 on Wed Aug 20, 2008 8:37 pm UTC, edited 9 times in total.

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Re: Martingale betting system (Roulette)

Postby Matterwave1 » Wed Aug 20, 2008 8:13 pm UTC

All betting systems will fail because in the end, there is house advantage, there are table limits, and they have a bigger bank than you (so even if there were no house advantage, they'd still most likely win). If you had infinite money, you could break the house...(just keep betting 1 trillion dollars per bet, eventually you'll win and they'll go bankrupt) but why would you be gambling?

Just think of betting like playing war (you know with cards, and if you get a higher card, you capture your opponent's card), except you get 20 cards and the house gets 50 million. Eventually, you'll go broke before the house does. (barring some extreme amount of luck I suppose)

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Re: Martingale betting system (Roulette)

Postby Random832 » Wed Aug 20, 2008 8:43 pm UTC

Matterwave1 wrote:If you had infinite money, you could break the house...(just keep betting 1 trillion dollars per bet, eventually you'll win and they'll go bankrupt) but why would you be gambling?


The funny thing about this is, the size of the house's bank is practically irrelevant here anyway - they're never at risk for more than your original $5 bet; the rest of the money you get from the payoff at the end all comes from the bets you lost leading up to it

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Re: Martingale betting system (Roulette)

Postby phlip » Wed Aug 20, 2008 10:29 pm UTC

Random832 wrote:
Matterwave1 wrote:If you had infinite money, you could break the house...(just keep betting 1 trillion dollars per bet, eventually you'll win and they'll go bankrupt) but why would you be gambling?


The funny thing about this is, the size of the house's bank is practically irrelevant here anyway - they're never at risk for more than your original $5 bet; the rest of the money you get from the payoff at the end all comes from the bets you lost leading up to it

I think he was suggesting more that the first bet you make be bigger than the house's coffers, and start doubling from there.

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Re: Martingale betting system (Roulette)

Postby Matterwave1 » Thu Aug 21, 2008 6:16 am UTC

Yea, my point about breaking the house doesn't work if you start at $5, only if you start at a bet that's more than the house has.

But, I was just saying, you can't really break the house short of rigging the games (or possibly counting cards in blackjack which, according to the movie 21, you'll get your ass kicked over)

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Re: Martingale betting system (Roulette)

Postby intimidat0r » Mon Aug 25, 2008 8:29 am UTC

I too thought of this a few months ago.

I wonder why people keep returning to this idea.
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Re: Martingale betting system (Roulette)

Postby codyhotel » Mon Aug 25, 2008 3:42 pm UTC

I would just like to add to this conversation my thoughts.

You are all wasting your time discussing this, because if any of you EVER try this in a Casino you will be ejected after your third bet. Casino's know every betting tactic known to man, you can't use simple ones like this.
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Re: Martingale betting system (Roulette)

Postby Random832 » Mon Aug 25, 2008 3:46 pm UTC

codyhotel wrote:I would just like to add to this conversation my thoughts.

You are all wasting your time discussing this, because if any of you EVER try this in a Casino you will be ejected after your third bet.


Doubtful - more likely they'd happily let you keep betting until you hit the table limit, then they get to keep your money.

Casino's know every betting tactic known to man, you can't use simple ones like this.


Casinos actually benefit from people thinking that stupid betting tactics work, because when they fail the casino gets to keep the money.

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Re: Martingale betting system (Roulette)

Postby phlip » Mon Aug 25, 2008 11:37 pm UTC

intimidat0r wrote:I wonder why people keep returning to this idea.

Beacause humans suck at comparing a likely low payoff with a rare high payoff (or, in this case, a rare large loss)... so with no solid reason to think either is bigger, they just guess (hope?) that the one they want to be better, in fact is. It's the same reason that lotteries are popular... there, there's a likely small loss, and a rare high payoff.

Random832 wrote:Casinos actually benefit from people thinking that stupid betting tactics work, because when they fail the casino gets to keep the money.

Indeed... casinos love a gambler with a system, 'cause they'll make more bets, more regularly, than a more unstructured punter.

Casinos have enough cash to take the long-term view, and Martingale (and, indeed, every other betting strategy ever) is in the Casino's favour, long-term.

It's only the tactics that actually skew the odds in your favour (measuring for any biases in a roulette wheel that might skew the distribution, counting cards in blackjack, etc) that are called "cheating" and will have you thrown out.

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Re: Martingale betting system (Roulette)

Postby Random832 » Tue Aug 26, 2008 1:29 am UTC

I ran a simulation, with the following parameters
Infinite bankroll (only net winnings/loss recorded)
Table limit is $10,000. Once an attempt is made to place a higher bet, the gambler learns about the table limit, realizes the system can't possibly work (since he can't do what it's telling him to do right now), and walks away.
Starting bet is $1.
Once you win (having made a profit of $1), you start over with a bet of $1.
Two zeros on the wheel.

I did get the occasional run where the gambler made enough $1 profits to be up by the time he finally hit the limit on one attempt, but typical attempts ended up down by between 110% to 150% of the table limit

For one run I gave the win exactly a 50% chance (i.e. no zeros), there was an overall win about 40% of the time, and the wins were more impressive than the losses (average loss 10200, average win 18400). With one zero, there were 75 losses to 25 wins, average loss 10500, average win 9800

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Re: Martingale betting system (Roulette)

Postby zealo » Fri Oct 31, 2008 3:53 pm UTC

what happens in you simulation where the player has a 'target'? say, he walks away when he is $5 up, $10 up... whatever?
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Re: Martingale betting system (Roulette)

Postby Random832 » Sat Nov 01, 2008 6:27 am UTC

zealo wrote:what happens in you simulation where the player has a 'target'? say, he walks away when he is $5 up, $10 up... whatever?


Had to go looking for the code; wasn't sure I'd saved it.

With a target of $500, an initial bet of $1, and a table limit of $10,000; what we have is a large number of cases where the player makes the target (winning $500) and a small number of cases where the player busts the table limit having lost a total of about $16,000 - the average (an approximation of the expected value) seems to hover between a loss of $100 and $600 over multiple sets of runs.

So you have a high chance of winning a small amount, and a small chance of losing a large amount. Kind of like a reverse lottery.

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Re: Martingale betting system (Roulette)

Postby phlip » Sat Nov 01, 2008 7:14 am UTC

Like all other aspects of life, I take all my gambling advice from Wikipedia.

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enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
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Re: Martingale betting system (Roulette)

Postby Macbi » Sat Nov 01, 2008 11:10 am UTC

The table limit doesn't actually matter, since if you hit it you can just bet the max twice.
i.e: You've kept doubling to the table limit of $1,000,000 so now if you lose you bet $1,000,000 on black twice:
If you win both you've made you're $5 and you can start again.
If you lose one win one, you pretend it didn't happen and bet twice again.
If you lose both then you have to bet the max 4 times.

Note: This is entirely equivalent to just betting the max til you're positive again.
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Re: Martingale betting system (Roulette)

Postby phlip » Sat Nov 01, 2008 11:37 am UTC

OK, simulating that:
  • Roulette wheel with 2 zeros (ie 18/38 chance of winning each round)
  • Start with $100,000 in the coffers
  • Table limit of $10,000
  • Each round, bet the smallest of: what it'd take to get to $1 above my current maximum, or the table limit, or how much money we actually have left (ie, start with $1, double after each loss until we hit one of the other limits, then just bet the limit until we're back to normal)
  • Leave only when we reach a target of $500 profit, or go broke

After 10000 simulations, that got 9874 wins of $500 each, and 126 losses of $100k, for an average loss of $12,106.30.

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enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
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Re: Martingale betting system (Roulette)

Postby Random832 » Sat Nov 01, 2008 7:37 pm UTC

Macbi wrote:The table limit doesn't actually matter, since if you hit it you can just bet the max twice.
i.e: You've kept doubling to the table limit of $1,000,000 so now if you lose you bet $1,000,000 on black twice:
If you win both you've made you're $5 and you can start again.
If you lose one win one, you pretend it didn't happen and bet twice again.
If you lose both then you have to bet the max 4 times.

Note: This is entirely equivalent to just betting the max til you're positive again.


I can't simulate this with an infinite bankroll because it can take arbitrarily long to get back to positive and the script will freeze. I will note that the typical/average amount of bankroll size required to win any given target is several orders of magnitude larger than the target itself

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Re: Martingale betting system (Roulette)

Postby Xanthir » Sat Nov 01, 2008 11:44 pm UTC

phlip wrote:After 10000 simulations, that got 9874 wins of $500 each, and 126 losses of $100k, for an average loss of $12,106.30.

This is the part that gets me every time. Any variant on Martingale always shows these absolutely ridiculous numbers. Win $500, or lose $100k. That's... not a well-tuned utility function.
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Re: Martingale betting system (Roulette)

Postby phlip » Sun Nov 02, 2008 12:51 am UTC

Xanthir wrote:This is the part that gets me every time. Any variant on Martingale always shows these absolutely ridiculous numbers. Win $500, or lose $100k. That's... not a well-tuned utility function.

Sorry, but to avoid miscommunication... are you saying my simulation was bad, or are you saying that Martingale is bad?

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enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
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Re: Martingale betting system (Roulette)

Postby Xanthir » Sun Nov 02, 2008 1:49 am UTC

phlip wrote:
Xanthir wrote:This is the part that gets me every time. Any variant on Martingale always shows these absolutely ridiculous numbers. Win $500, or lose $100k. That's... not a well-tuned utility function.

Sorry, but to avoid miscommunication... are you saying my simulation was bad, or are you saying that Martingale is bad?

Martingale is bad. Sorry about that.
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Re: Martingale betting system (Roulette)

Postby qinwamascot » Sun Nov 02, 2008 7:56 am UTC

Just to clarify, although I believe this was already pointed out, Martingale actually has a lower expected value than just betting the minimum amount over and over. Well, not really, because they're both 0 in the end. But from a calculus perspective Martingale is worse.
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Re: Martingale betting system (Roulette)

Postby GreedyAlgorithm » Sun Nov 02, 2008 8:57 am UTC

qinwamascot wrote:Just to clarify, although I believe this was already pointed out, Martingale actually has a lower expected value than just betting the minimum amount over and over. Well, not really, because they're both 0 in the end. But from a calculus perspective Martingale is worse.

.........what

Their expected value is 0. Both of them. That means Martingale does not "actually have a lower expected value". What do you mean, "from a calculus perspective" it's worse? I think you are confused about something.
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Re: Martingale betting system (Roulette)

Postby qinwamascot » Sun Nov 02, 2008 10:02 am UTC

GreedyAlgorithm wrote:
qinwamascot wrote:Just to clarify, although I believe this was already pointed out, Martingale actually has a lower expected value than just betting the minimum amount over and over. Well, not really, because they're both 0 in the end. But from a calculus perspective Martingale is worse.

.........what

Their expected value is 0. Both of them. That means Martingale does not "actually have a lower expected value". What do you mean, "from a calculus perspective" it's worse? I think you are confused about something.


No. I'm not. The average total winnings on a Martingale-style betting system are lower than on my proposed system (although in both systems the losses outpace the winnings). I don't feel like writing down the formulation I'm using for the calculus, so I'll just say that to find lim(0/0), we use L'Hopital's rule to get the fraction out of indeterminate form, and differentiate the numerator and denominator. The function I'm using is the money earned relative to the money gambled, so differentiating gives the rate of money earned relative to the rate of money gambled. It is lower in the Martingale than in a minimum-betting scenario.

Alternatively, if you have some goal value to reach, call it M, and some initial investment, call it I, then for all M>I the chance of reaching it before bankrupting is lower in Martingale than in a minimum-betting style. This isn't what I was saying before, but you can see why one is better than the other, although they're both bad.
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Re: Martingale betting system (Roulette)

Postby Random832 » Mon Nov 03, 2008 5:14 am UTC

GreedyAlgorithm wrote:
qinwamascot wrote:Just to clarify, although I believe this was already pointed out, Martingale actually has a lower expected value than just betting the minimum amount over and over. Well, not really, because they're both 0 in the end. But from a calculus perspective Martingale is worse.

.........what

Their expected value is 0. Both of them. That means Martingale does not "actually have a lower expected value". What do you mean, "from a calculus perspective" it's worse? I think you are confused about something.


Um, no. Their expected value is _negative_. Did you guys both somehow forget about the existence of negative numbers? Martingale's is indeed lower (i.e. greater magnitude negative) because the expected value of each bet (being, remember, negative) is multiplied by the amount that is bet.

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Re: Martingale betting system (Roulette)

Postby jestingrabbit » Mon Nov 03, 2008 7:59 am UTC

Random832 wrote:
GreedyAlgorithm wrote:
qinwamascot wrote:Just to clarify, although I believe this was already pointed out, Martingale actually has a lower expected value than just betting the minimum amount over and over. Well, not really, because they're both 0 in the end. But from a calculus perspective Martingale is worse.

.........what

Their expected value is 0. Both of them. That means Martingale does not "actually have a lower expected value". What do you mean, "from a calculus perspective" it's worse? I think you are confused about something.


Um, no. Their expected value is _negative_. Did you guys both somehow forget about the existence of negative numbers? Martingale's is indeed lower (i.e. greater magnitude negative) because the expected value of each bet (being, remember, negative) is multiplied by the amount that is bet.


If its a fair game its 0 by definition. If the house takes a cut, as in roulette, its negative, but there was talk about a fair game.
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Re: Martingale betting system (Roulette)

Postby Nebulae » Mon Nov 03, 2008 9:50 am UTC

With an infinite bankroll, as long as you could keep playing, you always win, no matter what the Win/Loss % is.

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Re: Martingale betting system (Roulette)

Postby jestingrabbit » Mon Nov 03, 2008 10:55 am UTC

That's not true. There is at least one event where you lose every time. Sure, this is a zero probability event, and so will be the the union of the "lose every time" events, but if we're going to assume that you have an infinite amount of money and still care about gambling then technicalities are clearly the order of the day.
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Re: Martingale betting system (Roulette)

Postby mrbaggins » Mon Nov 03, 2008 12:38 pm UTC

jestingrabbit wrote:
Random832 wrote:
GreedyAlgorithm wrote:
qinwamascot wrote:Just to clarify, although I believe this was already pointed out, Martingale actually has a lower expected value than just betting the minimum amount over and over. Well, not really, because they're both 0 in the end. But from a calculus perspective Martingale is worse.

.........what

Their expected value is 0. Both of them. That means Martingale does not "actually have a lower expected value". What do you mean, "from a calculus perspective" it's worse? I think you are confused about something.


Um, no. Their expected value is _negative_. Did you guys both somehow forget about the existence of negative numbers? Martingale's is indeed lower (i.e. greater magnitude negative) because the expected value of each bet (being, remember, negative) is multiplied by the amount that is bet.


If its a fair game its 0 by definition. If the house takes a cut, as in roulette, its negative, but there was talk about a fair game.


Just because the game is fair doesn't mean you can't come up with a system doomed to fail.
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