Can you beat Roulette using maths?

[robert Ihnot]

One method of cheating is called, "post posting." The player waits until the ball is about to land and then place his bet, preferably when the dealer is watching the ball land.

Today in Vegas, they are very careful about all these things--having learned, probably, the hard way. The roulette table is alway manned by two people. After a certain time, the dealer waves his hand over the machine and everyone is expected to keep his hands completely out of that area.

One writer claimed to have "won"/collected a lot by conceling "chocolate chips," worth $1000 each in his hand. If his bet was successful, when he went to collect his winning he would slip the chocolate chips under his, usually irregular pile, and then turn to the dealer and claim to have been underpaid! The dealer's eyes would roll in amazement when he now realized that a pile of $1 white chips, also contained thousand dollar chips underneath.

Of course today, the cameras watch everyting and 20 or so minutes would be spend unstairs going over that play!

 

[matticus]

say you go to the casino every week with $127 and play a game that pays 1:1. You start by betting a dollar, double your bet with every loss, and start back at a dollar with every win. So, you would need to win 127 times without hitting a streak of seven losses. What would your probability of winning on any given round have to be to make this a profitable strategy? I have taken several higher math classes, but have stayed away from probability and statistics courses, and I have no idea how to solve such a problem.

 

[Mensanator]

say you go to the casino every week with $127 and play a game that pays 1:1. You start by betting a dollar, double your bet with every loss, and start back at a dollar with every win. So, you would need to win 127 times without hitting a streak of seven losses. What would your probability of winning on any given round have to be to make this a profitable strategy? I have taken several higher math classes, but have stayed away from probability and statistics courses, and I have no idea how to solve such a problem.

Bets that pay 1:1 aren't 1:1.

 

[robert Ihnot]

say you go to the casino every week with $127 and play a game that pays 1:1. You start by betting a dollar, double your bet with every loss, and start back at a dollar with every win. So, you would need to win 127 times without hitting a streak of seven losses. What would your probability of winning on any given round have to be to make this a profitable strategy? I have taken several higher math classes, but have stayed away from probability and statistics courses, and I have no idea how to solve such a problem.

I think this might have an easy answer. 127 wins and 6 losses = 133 plays. This to win you want the probability of p = 127/133, (or better). If you play 133 games, using binominal simulation, you will win 127 times.

 

[skeptic2]

About six years ago I read the book 'The Eudaemonic Pie' by Thomas A. Bass. It's premise, as I recall, is that the roulette tables are not perfectly level and the ball will tend to fall when it is on the high side of the wheel. A team of college students built small computers to help them calculate the initial speed and position of the ball and indicate the likely number it will fall into.

This book is very engaging and although the author claimed the method actually works, it seems to me it has a fatal flaw. The error in determining the position and speed of the ball over a short interval at the beginning of the roll, is magnified by the ratio of the total roll time divided by that short measurement interval. The error always works out to be so great as to make the initial measurement useless.

 

[dilletante]

This book is very engaging and although the author claimed the method actually works, it seems to me it has a fatal flaw. The error in determining the position and speed of the ball over a short interval at the beginning of the roll, is magnified by the ratio of the total roll time divided by that short measurement interval. The error always works out to be so great as to make the initial measurement useless.

That would be true if they only clocked one revolution of the ball, but errors are minimized by clocking every revolution as it slows down and betting as late as possible. Some computer groups made substantial money on roulette, but Farmer wasn't one of them.

 

[skeptic2]

I don't have a copy of the book and my memory is a little vague after so much time but I believe they did collect data on how fast the speed of the ball decayed. The error was in pushing a button at the exact instance the ball passed each of two points on the table. Even an error of a few hundredths of a second, after being magnified by the total roll time, is enough to prevent an accurate determination of exactly when or where the ball is going to fall.

 

[dilletante]

I am not quite sure what your point is. If your point is that Farmer's group had problems and did not make money, I have already stated that. If your point is that no one could successfully use such a system to gain an advantage at roulette, you are misinformed. I personally know people who played around the world and were quite successful.

 

[peppies]

There are a number of different betting systems that you can use with any type of gambling game: http://www.lolblackjack.com/blackjack/betting-systems/

Although, all of them are futile as I have learned. They only work best with games that pay out even odds (1:1) but the actual odds will be less (because the casino needs to make a profit). The only way you can really win is through questionable techniques like card counting for blackjack and dice setting for craps. Those are the only two ways I know of where you can gain a skill and win (except for poker of course).

 

[Hurkyl]

I once read an article on betting systems, where:

• You start with X money
• You are trying to get Y money

and the problem is to maximize the likelihood of reaching your target.

IIRC, the best method turns to be to make the maximum bet possible until you either reach your target or you go broke.

(Where the game is a typical game where the odds slightly favor the casino)

 

[DaveC426913]

I once read an article on betting systems, where:

• You start with X money
• You are trying to get Y money

and the problem is to maximize the likelihood of reaching your target.

IIRC, the best method turns to be to make the maximum bet possible until you either reach your target or you go broke.

(Where the game is a typical game where the odds slightly favor the casino)

This occurred to me independently last time I was in a Casino.

It seems to me, the way to win at Blackjack is, not to bet like Scrooge and ultimately fritter away your money, but to go to the table and put all the money you planned to spend that evening on your first hand. You win or lose. Your chances of winning this one hand are better than your chances of winning over the next 20 or 50 subsequent deals.

 

[mgb_phys]

With maths - no
with physics - yes http://en.wikipedia.org/wiki/The_Eudaemonic_Pie
with statistics - yes http://en.wikipedia.org/wiki/Joseph_Jagger

 

[crimsonaaron]

Correct me if I'm wrong but there are 37 spots in a roulette wheel with equal chances for the ball to land in each one. The payout for a single number is 35:1. This means that overall you would be losing money. If you use progression of betting an additional dollar every time ($1 the first time $2 the second $3 the third and so on) then the payout for any given spin can be modeled by y=35x while the amount that you have paid could be modeled by y=x(x+1)/2. So the amount of net gain would be modeled by y=35x-x(x+1)/2. The x-intercepts for this equation(where your net gain would be 0) are 0 and 69. So if the first time you won was on your 69th spin then you made make back all that you spent. Considering the odds are 1 in 37 for any given number, you should easily be able to do this unless you are extremely unlucky. Anything before that and you would be winning money. Once you win then reset the cycle back with just $1. It is slow but I think it works. If the payout is too small then you could up it by doing higher intervals(for example 5, 10, 15...) but not starting from higher spot. I have seen people say that progression does not work which I assumed was this, but if so why doesn't this work?

 

[SteveL27]

This is a pretty old thread ... I seem to remember posting this once before, but evidently not in this thread.

Some physics grad students in the 1970's built a system for beating roulette. It worked. It's based on the idea that the wheel is a mechanical system, subject to biases. They carefully tracked the behavior of the wheel, and when they spotted a bias, they made money off it.

They used primitive wearable computers to calculate everything. Very interesting story. They had to shut the experiment down when one of the students got badly burned by a short in the wearable computer.

http://en.wikipedia.org/wiki/Eudaemons

 

[scepticus]

The answer to the question as put is YES.
We"Can" but that does not mean that we will ,only that it is possible
The House has an Edge ( in 37 numbers ) of 2.75 but that's it !It is the gambler's aim to beat that edge.By using maths we can, but this is all theoretical as each spin is Random and therefore unknowable until "after the Event ".It is nonsense to claim either that we are certain to win or certain to lose- unless you claim to be a clairvoyant !
For example.
Combining two ideas we can claim a mathematical advantage in roulette but an advantage does not mean certainty only that we are more likely to win than lose .
If we choose to bet only RED/ODD numbers plus BLACK/EVEN numbers we bet 20 numbers which means that we should win 20 times in every 37 spins.
If we chose to bet one dozen and one column we are betting twenty numbers so we should win 20 times over 37 spins.
Combining these two- and betting only those which they have in common- we have two sets of 20 over 37 which , multiplied together gives 400 over 1369 which gives the bettor an advantage if betting 10 or fewer numbers- and this does occur in some pairings.
The key to winning at betting is Bet Selection .All gambling involves uncertainty so why this bias against roulette ? Gambling is gambling is gambling .

 

[Mentallic]

Sorry scepticus, but you're way off base. The contents of this thread was mainly about discussing if the Roulette wheel has any uneven bias that would cause the randomness of a number to appear to not be completely random afterall - thus allowing someone to beat Roulette.

I'm feeling optimistic that the users on this Math forum can understand the definition randomness and its relationship to gambling.

 

[Hurkyl]

There's an easier way to win most hands: simply place a single-number bet on every number from 1 through 34. 34 out of every 37 spins you come out ahead.

Exercise: explain why the house still has the edge when you adopt this strategy.

(P.S. in the U.S., roulette wheels have 38 numbers)

 

[AlephZero]

The key to winning at betting is Bet Selection.

The key to winning at roulette is to own the casino. When was the last time you saw a casino that had to close down because it ran out of money?

 

[scepticus]

Thanks Guys
But the Heading was" can you beat roulette with maths" and I took it at face value.
I was hoping someone would show me where my maths was wrong and not just post soundbites.

 

[Hurkyl]

Thanks Guys
But the Heading was" can you beat roulette with maths" and I took it at face value.
I was hoping someone would show me where my maths was wrong and not just post soundbites.

Try my exercise. Really, the best way to understand is to do the computation for yourself, and compute your expected winnings. Do you know how to do that?

Fortunately, the computation isn't that hard.



It's important to actually consider all of the cases. One of the classic traps is the following strategy:

• Bet 1 dollar.
• If you win, stop. Otherwise bet 2 dollars.
• If you win, stop. Otherwise bet 4 dollars.
• If you win, stop. Otherwise bet 8 dollars.
• Et cetera

At 1:1 payoff it looks like this strategy guarantees that you'll come out a dollar ahead, right? The problem is that there is a tiny problems

There's an upper limit on how much money you have available to spend gambling​

Sometimes, there's also a limit on how much the casino will allow you to bet.

Your odds of coming out ahead are around 85724 in 85725. That's very likely, right? Surely this is a good strategy to beat the house!

It's easy to lull yourself into that false sense of security -- the problem is that one time in 85725, you lose all of your money. You are risking your entire million dollars for a chance at coming out a single dollar ahead. The house expects to come out over 11 dollars ahead each time a millionaire tries this strategy.

 

[scepticus]

Thanks Hurkyl
What you describe is known as the Martingale system
I am talking about level stakes.
You don't really answer my question - tell me where my maths is wrong.

 

[PlayingMonk]

I came up with a system that I believe allows one to win at roulette in the long run but I'm not 100% sure that it's valid (I haven't tested it in practice) and I want to share the basic idea here and see if any of you guys can tell me why it is wrong or if there is anything to it.

To increase the odds we'll assume we're using a European roulette wheel. This is basically a variation on the famous martingale system. The algorithm I would use would be something like this:

Step 1 - Bet $1 on black
Step 2 - If you win go back to step 1, if you lose bet $2 on black
Step 3 - If you win go back to step 1, if you lose bet $4 on black
Step 4 - If you win go back to step 1, if you lose bet $8 on black
Step 5 - If you win go back to step 1, if you lose bet $16 on black
Step 6 - If you win go back to step 1, if you lose bet $32 on black
Step 7 - If you win go back to step 1, if you lose bet $64 on black
Step 8 - Go back to step 1

So, the main difference between this and the martingale system is that here you set a cap for yourself that you will not double and will just accept the loss at the point instead of continuing on until you go broke. In this example you will lose $64 every time you lose money.

Now, to do the math. There are 37 spaces on a European roulette wheel and the chances of getting black and winning on any of these spins is 18/37. The chances of losing on a given spin are 19/37. The chances of losing seven times in a row are (19/37)^7 which is roughly .0094159282. The odds of going through the algorithm and not losing seven times in a row should then be .9905840718. Therefore, on any given run through of the algorithm there is about a 99.06% chance of winning $1 and about a .94% chance of losing $64.

So, (1)(.9905840718) + (-64)(.0094159282) = .9905840718 - .6026194048 = .387964667

Because we get a positive number we should win money in the long run.

(I did the math for an american wheel and got a value of about .27, so it should still work on an american table but not quite as much as quickly).

So, would this work or am I messing something up?

 

[Hurkyl]

In this example you will lose $64 every time you lose money.

You lose all 7 bets, not just the last one. That totals to $127.

 

[Hurkyl]

Thanks Hurkyl
What you describe is known as the Martingale system
I am talking about level stakes.
You don't really answer my question - tell me where my maths is wrong.

I can't tell you where your math is wrong unless you actually do the math and show your work.

 

[scepticus]

Hi Hurkyl
Possible scenario
We choose the 3rd Dozen and the 1st Column
So, to begin with ,we choose the numbers
1,,4,7,10,
13,16,19,22
25,26 27 28, 29 30 31 32,33,34,35,36
a total of 20 numbers an advantage of 20/37
We also consider all the Red / Odd and Black / Even numbers
These are ;
1,2,3,4,5,6,7,8,9,10
19,20,21,22,23,24,25,26,27,28 - again 20 numbers an advantage
of 20 / 37
The chances of both winning on the same spin are 20 / 37 multiplied by 20 / 37 which is 400 / 1369.
Choosing only those numbers which they have in common we now have 1,4,7,10,19,22,25,26,27,28 a total of 10 numbers
Each bet costs 10 units and for each win we receive 36 units
So over an average 1369 spins we spend 13690 units and receive 400 times 36 which is 14400 a surplus of 710 units
So we can use maths to win at roulette unless my maths is wrong which I am asking people to show.
BUT this can only be theoretical as each spin of the wheel gives a random number so no one can claim, with certainty ,that we MUST lose or win at roulette, Where uncertainty exists certainty can be claimed by clairvoyants but not by mathematicians.

 

[turbo]

Unless the wheel is rigged and you find out why and how it is rigged, you will always lose at roulette. There is no mathematical program that will beat that wheel. You will lose.

 

[Millennial]

Martingal can beat the wheel in theory, but not really in practice (they have rules to prevent it, and it consumes your money fast.)

 

[turbo]

Martingal can beat the wheel in theory, but not really in practice (they have rules to prevent it, and it consumes your money fast.)

"Betting strategies" cannot overcome the house advantage in roulette. It is not possible, if the wheel is honest. The house wins. Why is this thread still alive?

 

[Number Nine]

Martingal can beat the wheel in theory

No, not with a finite amount of money. "In theory", with a finite amount of money, you will lose.

 

[Hurkyl]

The chances of both winning on the same spin are 20 / 37 multiplied by 20 / 37 which is 400 / 1369.
Choosing only those numbers which they have in common we now have 1,4,7,10,19,22,25,26,27,28 a total of 10 numbers

Why would that be the chances? Presumably, you are applying the theorem that, for independent events A and B (where A and B are each of your 20/37 events):

P(A and B) = P(A) P(B)​

but why would you think A and B are independent? They are clearly related in a non-trivial fashion depending on the layout of the roulette wheel, so it should be at least somewhat surprising if they turned out to be independent.

But we don't need to speculate: we can directly compute the odds of one of those 10 out of 37 numbers coming up in a spin of the wheel and see that the odds aren't 400/1369.

 

[daveb]

Assuming the wheel is fair, you can minimize your chance of losing by starting a bet at 1 dollar on some number. If you lose, increase the bet by 1 dollar but bet the same number. Up to the 70th bet, assuming you win, you come out ahead. The 71st bet means you come out even (at 2556 dollars), and beyond that you are behind. Once you win, go back to 1 dollar with a different number (if you can remember all numbers rolled up to that point your loss is minimized further by picking a number that still hasn't been rolled).

Now, the chance of a specific number not coming up in 71 consecutive tries is (37/38)⁷¹, or about 15%, so 15% of the time, this strategy doesn't work

 

[skeptic2]

How much money have you won with this strategy?

 

[scepticus]

Thanks hurkyl for taking the trouble to reply
I am not a mathematician but I understood them to be independent since they do not depend on each other.
so if we ASSUME they ARE independent does my maths stack up ?
There is no real need for a real roulette wheel as we could use a RNG
Thanks

 

[scepticus]

Hi Skeptic 2
I do not use this method I use another but using maths.
All I am doing here is trying to prove that ,yes,we CAN beat roulette by using maths
but can does not mean certainty.

 

[Hurkyl]

Thanks hurkyl for taking the trouble to reply
I am not a mathematician but I understood them to be independent since they do not depend on each other.

They do depend on each other, because they are both interdependent upon what numbers are available and how they're laid out and what colors are assigned to them.

so if we ASSUME they ARE independent does my maths stack up ?

If they were independent, then they would intersect in about 10.8 numbers. Obviously in the real world you can't have a fractional number of numbers, but it's no problem to consider it in the hypothetical. So, each bet would cost about 10.8 chips. Now, repeat your calculation on how much you spend and receive.