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Double or nothing roulette Double or nothing roulette

08-21-2015 , 08:38 PM
Hi guys,

A friend of mine thinks he invented hot water Double or nothing roulette

His logics goes af following
He bets $1 on black
When he looses he bets $2
When he looses again he goes up to $4
And keeps doubling untill he eventually runs black.
This always gives him back $1 when he gets black regardless of the streek because of the following;
First spin win: $1>$2
Second spin: $1+$2 ($3) >$4
3th spin: $1+$2+4 ($7) >$8

To me this looks flawed to me cz he runs 18 against 19
That makes it .053% from a litural flip.

And this is where I'm leveling myself Double or nothing roulette
Give these stats you'll loose a dollar for ever 20 you win?
(1/0.05)

What happend to variance?Double or nothing roulette
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08-21-2015 , 08:55 PM
Your friend's scheme is well known and is called "Martigaling." I'm sure there's a wikipedia page as well as countless other internet pages about it.

Your long term expectation from betting like this is exactly the same as flat betting or any other betting scheme. You will have many many $1 wins, followed by an huge loss when you have no more money to do one more doubling or hit the table limit.
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08-21-2015 , 09:08 PM
Table limit on his table is 25k
He made $40 in the last hour, the chance that he has to double till 25k from $1 is like what?
%0.0000000001?
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08-21-2015 , 09:32 PM
Haha, just kidding, forget my previous post, obviously you're right. He's created a way to mint money, no one's ever thought of it before. My hat is totally of to him.
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08-21-2015 , 09:38 PM
I see your point, but just seems to easy to be treu
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08-21-2015 , 09:42 PM
Nope, it's definitely true. He's done it. Every casino will be bankrupt by the end of the weekend now that the secret is out.
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08-22-2015 , 01:51 AM
Quote:
Originally Posted by Daikie
Table limit on his table is 25k
He made $40 in the last hour, the chance that he has to double till 25k from $1 is like what?
%0.0000000001?
I don't think a table exists with $1 min and $25K limit. If the min is $10 he will only be able to double 11 times to hit a $25K limit. Assuming he can bet 50 times an hour, the chance he goes broke or hits the limit within the first 2 hours is about 4%. And to double 11 times he'll need a bankroll of $41K.

If he did start with a $1 bet it drops to a little under 1%. And obviously this increases with more time played (cumulatively - sessions don't matter).

You are off quite a few orders of magnitude.
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08-22-2015 , 07:13 AM
Also, who is going to risk $25k to win $1?
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08-23-2015 , 07:54 PM
A number of points.

1) probably everyone who has ever thought seriously about winning at a table game has come up with some sort of Martingale scheme, and wondered if it would work. The majority of those people have figured out the math and realized that it won't work; a few have tried and won small amounts; and some have tried and lost large amounts. There are some who have consistently won - they are called casinos.

2) the betting limit at a given table is an irrelevant factor. Once you reach a certain point (or any point, for that matter) the next bet does not need to be on the same table, or even in the same game or the same casino. It just needs to be something close to a 50/50 bet. The game you are playing has no memory, so it doesn't matter what you have done in the prior spins/hands/games... You could lose 8 spins of roulette in a row, then take your money to the sports book and pick a sports bet. If that loses you could double your amount and bet a hand of casino war. If you run into a casino that doesn't want to take your action (and that seems unlikely) - you can leave the casino and go down the street to find another casino that will. Every time you do this you are risking greatly increasing amounts of money on -EV events. You may be "bound to win some time" - but you are still less than 50% to win the next bet. Are you willing to keep making those bets, knowing that if you lose you have to double it again?

3) If Martingaling worked, the entire casino industry would have been wiped out a few hundred years ago. Since it is free money, and everyone knows about it, there would be nothing they could do about it. Rather than just continue to lose money, they would just close up shop and call it a day. Why are casinos still in business? Because the Martingale (and anything similar) doesn't work.
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08-27-2015 , 08:05 AM
Quote:
Originally Posted by VBAces
The majority of those people have figured out the math and realized that it won't work.
Not a chance they've really figured out the math.

Most stop at the "well if you had infinite money and no table limits, it would work"... which is also fundamentally incorrect.
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08-27-2015 , 03:54 PM
my friend did this in vegas...

....the dealer laughed at him when he finally busted.

Just remember the more you bet in a negative -EV game (i.e. roulette) the more you will lose. In order to make casino games +EV you need to order free cocktails while betting the minimum.
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08-31-2015 , 08:26 AM
If you want to increase your chance of winning to almost 100%, do the patented "David Lyons System" for single-zero roulette.

Bet every number except zero and 29 with a single dollar. If you win you profit a dollar (yay!); if you lose you do the same except with 36 dollars on every number. If you win you win your losses from the first spin back, plus a dollar (yay!). If you lose those two spins in a row (only 0.3% chance of happening!) then simply keep multiplying the stakes by 36 until you win.

This is virtually guaranteed to turn a profit of a dollar... I mean losing three spins in a row has a 0.016% probability (that's 6331 to 1) so you are 99.94% to win your dollar.

Easy game.

PS Better yet, play the system at the same table as me; I always cover 29 and zero and if you lose three in a row I will toss you $50 to take the sting out of your losses.
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08-31-2015 , 01:37 PM
Quote:
Originally Posted by David Lyons
Not a chance they've really figured out the math.
This.

Quote:
Most stop at the "well if you had infinite money and no table limits, it would work"... which is also fundamentally incorrect.
It would work if you define success as "bankrupt the casino" (but better yet would be to bet the casino's bankroll on every spin, because that would bankrupt them as quickly as possible). It wouldn't work if you define success as profit since you can't profit when you already have infinity dollars.

Edit: struck that part out because that would be martingaling. Your first bet would be the casino's bankroll, then if you lost, their bankroll would double, so you'd double your bet. And so on.

Last edited by heehaww; 08-31-2015 at 01:47 PM.
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08-31-2015 , 03:31 PM
Quote:
Originally Posted by heehaww
It would work if you define success as "bankrupt the casino"
You don't need infinite money for this, and it can happen in the real world. Whereas infinite anything can never happen in the real world.
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08-31-2015 , 04:07 PM
Quote:
Originally Posted by NewOldGuy
You don't need infinite money for this, and it can happen in the real world.
You need infinite money and limits to turn a +EV game (for the casino) into a -EV game. You don't need infinite money to make the game -EG for them or at least risky (high-RoR), but you'd need larger limits than the ones offered. And of course you don't need infinite money or limits to get lucky or cheat, or for the casino to go bankrupt for reasons unrelated to its Roulette revenues. I don't know what you're getting at. When a system is said to "work" or "not work", it's implied that you mean in the long run. You'd need infinite limits to "beat" Roulette in the long run.

My point was, "with infinite bankroll and limits, Martingale would work" is basically true. Only in an impossible scenario would Martingale work, which is another way of saying Martingale doesn't work. And even in the impossible scenario, it wouldn't work because of martingale's awesomeness, it would work because of the weirdness of infinity and measure theory.

In real life, Martingale is so bad that if Roulette were +EV for the player, you'd still go broke by martingaling! (Unless you're sufficiently rich compared to the table limits.)

Last edited by heehaww; 08-31-2015 at 04:13 PM.
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09-01-2015 , 05:59 AM
Quote:
Originally Posted by heehaww
In real life, Martingale is so bad that if Roulette were +EV for the player, you'd still go broke by martingaling! (Unless you're sufficiently rich compared to the table limits.)
When it's +EV I like to reverse-labouchere.
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09-01-2015 , 08:33 PM
To illustrate the problem of martingale imagine you have a game that is so ridiculously positive as to have 60% chance to win every time. This is a paradise of an edge actually.

Now say you have 1000 units of capital and want to reach 10000 to give up and leave as a winner.

You play 1 then 2 then 4 then 8 etc if they fail until you run out of money.

In order to make it to 10000 you will need to win 9000 times, i mean 9000 such martingales because in each one of them that works you go up +1.

ie +1 win 1, -1 then +2 win 1, -1,-2,+4 you win 1 etc

Now ask yourself what is the chance to reach 9000 profit that way?

Well if your capital is K at the time you have to at least survive a martingale that is no more than n=IntegerValue(LogK/Log2)+1 in length. Eg if you have 3455 you cannot experience log3455/log2->12 losses back to back , you will be destroyed. In fact even if you experienced 11 you are only left with 1407 to bet and will not recover everything if you win that bet. I doubt many people that started from 1000 and went up all the way to 2455 that way would be happy to risk that last 1407 by the way having been taught a lesson in bad luck and reckless gambling. In any case the chance to go bust is a bit different but it will be at least the chance to lose 12 in a row here (the chance to lose 11 and then win plus lose later etc but this will turn out to be near the chance to lose 12 often).

Ok so you are looking to at least survive this product;

P((1-0.4^IntegerValue(LogK/Log2+1)), from K=1000 to 9999)

That comes out as 89.4% chance to survive it. 10000 Simulations show 89.2+0.3%.

If you wanted to reach instead 1000000 ( not only it would take years of betting ) but it would leave you about 20% risk of ruin.

Think about it. For such edge to have 20% chance to not reach your goal of huge wealth and to take essentially hundreds of thousands of bets is a ridiculous result. This is true because there are ways to get there with close to 0 risk of ruin.

If you bet instead only 1 unit at a time the risk of ruin would become practically 0 (0.4/0.6)^1000 lol. It would also take forever to reach target.

But if you tried kelly you would be betting 2*0.6-1=0.2 fraction of your capital every bet and you would get there real fast with risk of ruin (in principle 0 but because you would hit the table/market minimum eventually sometimes and then die) less than 0.2%. But you would require only ~350 steps on avg.
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09-07-2015 , 03:33 PM
Fun exercise;

Reckless player with edge win probability p>50% in a 1-1 payout game declares with audacity i will play martingale and i dare the universe to stop me. He starts with capital N units. What is the chance he never busts ?

You know that this is possible (why? consider extreme cases where he grows eventually big enough for the probability of ruin per martingale trial to never be able to catch up with that growth because within the typical time needed to experience a ruin of that capital the capital has already grown to much larger levels from all the successful trials requiring even smaller chance to be ruined than the original capital on the first trial). Investigate what role N,p play in the answer if you cannot obtain a closed form result.

Last edited by masque de Z; 09-07-2015 at 03:39 PM.
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09-09-2015 , 03:05 AM
Quote:
Originally Posted by masque de Z
Fun exercise;

Reckless player with edge win probability p>50% in a 1-1 payout game declares with audacity i will play martingale and i dare the universe to stop me. He starts with capital N units. What is the chance he never busts ?

You know that this is possible (why? consider extreme cases where he grows eventually big enough for the probability of ruin per martingale trial to never be able to catch up with that growth because within the typical time needed to experience a ruin of that capital the capital has already grown to much larger levels from all the successful trials requiring even smaller chance to be ruined than the original capital on the first trial). Investigate what role N,p play in the answer if you cannot obtain a closed form result.
To simplify the problem slightly (cheating?) I assume that if you can't make a full bet, you bust (e.g with a tank of 5, i bust if i lose two bets because I would only have 2 remaining, not the 4 needed to get back ahead).

For any tank N = 2^n + x - 1, where x \in (0, 1, ..., 2^n - 1), you will bust if you lose n spins. So for a full run through of the sequence gives:
bust with probability q^n
+1 to tank with probability of 1-q^n

Therefore, to go from 2^n + x - 1 to 2^(n+1) - 1, you need to 'not bust' 2^n - x times. So you reach tank 2^(n+1) - 1 with a probability of
[1-q^n]^[2^n - x]

So the probability that you never bust is:

[1-q^n]^[2^n - x] * PROD(i=1 to inf) [1-q^(n+i)]^[2^(n+i)]

I can't really prove that that infinite product doesn't go to zero but I can see that it mightn't.

EDIT:

[1-q^(n+1)]^2 = 1 - 2*q^(n+1) + q^(2n+2) > (1-q^n) for large enough n and q > 1/2

shows the ROR is > 0 I think.

Last edited by Banzai-; 09-09-2015 at 03:18 AM.
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09-13-2015 , 01:04 AM
Quote:
Originally Posted by Daikie
Table limit on his table is 25k
He made $40 in the last hour, the chance that he has to double till 25k from $1 is like what?
%0.0000000001?
A $1 to $25k table is quite weird, but let's assume it's right.

If he constantly loses, his bets have to go :
1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, and if he loses that last one, it's over, he can't double that bet ( would be over the $25k limit ).

So, he can bet 15 times before busting.
The odds of losing a single roll is 19/37 (18-18 colors, and the 0 = loss), so a bit more than 50%.

The odds of losing 15 rolls in a row are :
1 / ((19/37) ^15 )
= 1 : 21964.5265662

If you do lose, you'll lose $32767 (sum of all the bets)
On average, by the time you lose that $32767, you will win about 21964 cycles, so $21964.
You lost $10803 overall.

You say your friend won $40 so far with that strat? Big surprise.
This is the only thing this strat can do well. Win people small amount of money with relatively little risk.

If somehow you have $32767 in your pockets and some gangster threatens to break your legs unless you can pay back a debt of $32777 you owe him, this is a decent strategy if you have no other way to find 10 bucks somewhere else.
Your odds of losing are ridiculously low.

But the investment is massive for very little profit.
If he follows the strat correctly, he needs to have over $32k with him, and then he'll play for $1 pots that takes him 1.5 minute each ( 40 cycles an hour )? Seems like a horrible way to make money starting from $32k... even WITHOUT the risk of losing.

So let's play a bit more with the stats.
He plays 40 cycles, i.e. approximately 1 hour.

His odds of winning $40 are (1-(1/21964.5265662))^40, or 99.81804976%
Seems quite acceptable!
His odds of losing are 0.181950239%, but these are the odds to lose $32767.
On average, he loses $59.6196349362
Yikes, he loses $19 more than he wins on average! But so far so good, it's still just a very small chance of losing, so the 'average' isn't relevant I guess?

So he decides to try it for a full week!
He does the same thing for 40 hours.
1600 cycles.

His odds of winning $1600 are (1-(1/21964.5265662))^1600, or 92.974363887%
Still seems good, but the risk of losing it all are now more present.
His odds of losing are 7.025636112%, but these are the odds to lose $32767.
On average, he loses $2302.09018482

But hey, it's just 7%!
So the guy decides to make it a full time job. This is obviously a gold mine, how come no one ever thought about this before?

So he does it for a full year. 40 hours a week, 50 weeks, 2000 hours, so 80000 cycles.

His odds of winning $80000 are (1-(1/21964.5265662))^80000, or 2.619152489%
He is almost guaranteed to lose. Note that this is the odds of losing AT LEAST once, not just once.

On average, he will lose 3.64223648341 times over the 80000.
3.64223648341 * $32767 = 119345.162852
And he will win 79997 * $1, or $79997
For an overall loss of (119345.162852 - 79997) = $39348.162852 for a good year of work.

So, how come so many people think this is a good idea? How come people come up with these stories about how they won this much or that much? Well, scroll back up. It's easy to win small amounts with very little risk with that method. But if the odds are against you, and the longer you play the more obvious it becomes.

That guy who plays a week just has 7% to lose it all, and after a full week of grinding, he'll think that his method is a sure case... Then he'll lose $32k.

And the money he won before that lost won't nearly make up for the loss.

This is kind of the same result of, say, if a lottery company would give 2x more money than they get, on a big prize pool with 1 out of 100m odds to get.

They'd sell tickets after tickets and just look WOAH, it's basically free money, we've made 50m already!
But then someone wins the jackpot and they lose $100m.

This is more or less the same logic, only, your earnings are not even that great for the investment.

Even if there was no risk of losing ( and over the long run, again, it's more than a risk, it's basically a certainty )... You'd go to the casino with over $32,000 then come out 2 hours later with $32,080?
BEFORE expenses?

Seems to me that if you have $32k to gamble, you should not do it for pennies.

Especially not if, as demonstrated in this post and this thread and millions of times everywhere on the internet, it's not even a winning strategy anyway.

Casinos should probably give a couple millions to the first guy who came up with that idea. He certainly made them a fortune by having countless people going to the casino and losing that $32k... But at the same time, 100 other players went in and won $80 each, so there's just 1 guy who'll trash the strategy and a hundred who'll praise it and say it works, so it just keeps going, making 1 big loser, a few very small winners who'll end up going into the "big loser" category if they keep going, but most of all, one HUGE winner, the casino.

It doesn't matter how you bet and what strat you have. There's no way to beat the casino at that game unless you bet so much they can't afford the variance, but unless you're Bill Gates or something, it's not happening because we're talking 8-9-10 digits $ a roll.
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09-13-2015 , 08:47 PM
Quote:
Originally Posted by bobtrom
So, how come so many people think this is a good idea? How come people come up with these stories about how they won this much or that much? Well, scroll back up. It's easy to win small amounts with very little risk with that method. But if the odds are against you, and the longer you play the more obvious it becomes.

That guy who plays a week just has 7% to lose it all, and after a full week of grinding, he'll think that his method is a sure case... Then he'll lose $32k.

And the money he won before that lost won't nearly make up for the loss.
There is one small clarification that makes systems like this even worse that it sounds above. The big loss can even be first, and won't always be after a bunch of little wins.
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09-14-2015 , 01:52 AM
If you have to play roulette the best way to play it is by setting a goal to reach and then minimizing the amount of betting that gets you there. Alternatively you may increase the probability to get there under some EV constraints or not.

So this is the only legitimate way to play roulette if forced to participate.

Ask yourself what is the chance to show a profit of 100 if you have a capital of 32768.

The answer in this is to start betting as much as needed to reach the goal in one move using the best EV bet for this.

(US roulette assumed here)

Is it to bet 100 on red? Or is it better to bet (assuming you cant bet fractional amounts) $3 in eg 17 and then next time, 3 again and then upgrade to 4 to cover the previous losses then 4,4 then 5 ,5, 5 etc until you get there or lose everything. Always bet in a way that i you win you get there in one move for the position you find yourself (bold betting theory). Or is it better to bet in a combination of bets in numbers or groups of numbers that gets as close to the target as possible (to be determined) since you cant have factions.


Find how much betting on avg does it take to reach +100 playing martingale and you will see its a ton of activity compared to the avg amount of betting it takes to reach 100 the other way!!!
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09-14-2015 , 10:25 AM
Quote:
Originally Posted by masque de Z
Always bet in a way that i you win you get there in one move for the position you find yourself (bold betting theory).
I think this theory / fact is best expressed by "find the system that hits the target by exposing the least amount of your roll to edge".

Roulette is successful because the thing that produces the most endorphines (winning frequently) costs the most in edge. Witness the number of degens covering 80-90% of the table each spin.
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09-14-2015 , 10:32 AM
Quote:
Originally Posted by bobtrom
The odds of losing 15 rolls in a row are :
1 / ((19/37) ^15 )
= 1 : 21964.5265662
...
He plays 40 cycles, i.e. approximately 1 hour.
...
His odds of winning $40 are (1-(1/21964.5265662))^40, or 99.81804976%
His odds of losing are 0.181950239%
Not quite. The series of 15 are not independent (but even if they were, you'd wanna raise to the power of 26, not 40).

Using the streak formula given here, I get that the chance of at least one losing streak of 15 within 40 spins is about 1 in 1669. About half the probability predicted by the independence estimate (when raising to the power of 26).

Quote:
On average, by the time you lose that $32767, you will win about 21964 cycles, so $21964.
For the same reason, this too must be wrong, but I don't have a nifty formula for the average time until a streak.

Edit: here is what might be the correct formula (thanks to Aaron Brown) for the expected number of martingale crashes (a little different than streaks) within N spins.

Let r = streak length required to cause a crash
Let q = probability of individual loss (in this case 19/37)

E(#crashes in N trials) = [(N-r)(1-q) + 1] / (q^-r + 2r - N - 1)

According to that, we'd expect 1/1668 crashes in 40 spins of Euro roulette. I find that to be a believable answer, given that P(at least one streak) = 1/1669.

Last edited by heehaww; 09-14-2015 at 10:49 AM.
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