Forums | MacLife

Martingale system

I think you mean 30 rolls in a row.

And check that link. It has the equations.

You're having a mathematic misconception about probabilities.

The outcome of one roll does not affect the potential outcomes of the next.  You have a 50% chance of winning each time.  In the same way you could flip a coin 10 times and get all heads, just because you lost the first roll doesn't mean you have increased odds of winning the second.

Roulett?
Considering that the diameter of the wheel and the ball are usually the same, if you time how long it takes for the ball to revolve around the table, your chances increase and you can win most of the time

Daniel wrote:

You're having a mathematic misconception about probabilities.

The outcome of one roll does not affect the potential outcomes of the next.  You have a 50% chance of winning each time.  In the same way you could flip a coin 10 times and get all heads, just because you lost the first roll doesn't mean you have increased odds of winning the second.

I specifically worded it to say that the probability of losing is 1/(2^n).  The probability of losing N times in a row gets smaller and smaller with each new roll.

I think I answered my own question though in the original post.

I wrote:

The odds of losing X times in a row get exponentially higher as I keep going....  Is it because at the same time, the amount of money I invest needs to get exponentially higher also?

Last edited by Fried Chicken (2008-02-18 8:45 am)

Now Blackjack, that's where the real money's at. 

Fried Chicken wrote:

I specifically worded it to say that the probability of losing is 1/(2^n).  The probability of losing N times in a row gets smaller and smaller with each new roll.

The probability of losing stays constant.

And each subsequent bet would have to be sum(1,infinity,n) (which is like n^2+n/2 or something) for it to be a guaranteed net positive gain overtime, which means pretty freaking quickly, you'll run out of money.

edit: actually, i don't think this would result in a guaranteed gain. I don't think that's possible with roulette...

Last edited by mo' ron (2008-02-18 4:42 pm)

Aqua OS X wrote:

Daniel wrote:

You're having a mathematic misconception about probabilities.

The outcome of one roll does not affect the potential outcomes of the next.  You have a 50% chance of winning each time.  In the same way you could flip a coin 10 times and get all heads, just because you lost the first roll doesn't mean you have increased odds of winning the second.

Yes, but as your coin toss data set increases in size you are more likely to record data that's closer to 50/50.

That still doesn't alter the probability of each individual toss.

One conceivable strategy would be to bet on the ball landing in a red space for a certain number of spins, for example, 38.
There are 18 red spaces on a roulette table with 38 total spaces. Dividing 18 by 38 yields a probability of landing on red of 47.37%. This probability can be used in a binomial distribution and made into an approximate standard normal distribution.
Doing so indicates that, if one were to spin the wheel 38 times, there is a 99% probability that the ball would land on red at least 10 times. There is an 83% probability that in 38 spins, the ball will land on red at least 15 times. Out of 38 spins, there's a 50% chance that 18 will be red.
However, the break-even point is 19 spins, since the bet on red is 1:1, and the probability of 19 red spins in 38 is only 37%. This indicates the difficulty of winning by only betting on red.
The results occur because, as indicated by the 18 divided by 38 equals 47.37% figure, the ball will land on red less than half the time. This percentage applied in the binomial and standard normal distributions creates the vast divide in probability from 18 red spins to 19 red spins out of 38 spins. It is very unlikely for anyone to spin much more than 18 red spins out of 38 spins.

D'Eyncourt wrote:

Fried Chicken wrote:

Daniel wrote:

You're having a mathematic misconception about probabilities.

The outcome of one roll does not affect the potential outcomes of the next.  You have a 50% chance of winning each time.  In the same way you could flip a coin 10 times and get all heads, just because you lost the first roll doesn't mean you have increased odds of winning the second.

I specifically worded it to say that the probability of losing is 1/(2^n).  The probability of losing N times in a row gets smaller and smaller with each new roll.
[snip]

This is the gambler's fallacy. Assuming a fair roulette wheel, the probabilty that you will lose is ALWAYS THE SAME with each spin: 20/39 or 21/40 (depending on whether or not the wheel has only a single green 0 or an additional green 00). Get that into your head: in a game of pure chance, the history of outcomes does not matter. They only come into play if you want to predict the outcome over the next n spins or you are analyzing the behavior over the past n spins, but for THIS NEXT SPIN the odds are the same.

I understand that, but with this in mind, if the odds are always (near) 1/2, then the probability of losing 4 turns in a row is:
1  *  1  *  1  *  1
2      2      2      2
Which is equal to 1/(2^4)...?
I'm not trying to be stubborn here, I'm just trying to understand this thoroughly in my head.

Fried Chicken wrote:

D'Eyncourt wrote:

Fried Chicken wrote:

I specifically worded it to say that the probability of losing is 1/(2^n).  The probability of losing N times in a row gets smaller and smaller with each new roll.
[snip]

This is the gambler's fallacy. Assuming a fair roulette wheel, the probabilty that you will lose is ALWAYS THE SAME with each spin: 20/39 or 21/40 (depending on whether or not the wheel has only a single green 0 or an additional green 00). Get that into your head: in a game of pure chance, the history of outcomes does not matter. They only come into play if you want to predict the outcome over the next n spins or you are analyzing the behavior over the past n spins, but for THIS NEXT SPIN the odds are the same.

I understand that, but with this in mind, if the odds are always (near) 1/2, then the probability of losing 4 turns in a row is:
1  *  1  *  1  *  1
2      2      2      2
Which is equal to 1/(2^4)...?
I'm not trying to be stubborn here, I'm just trying to understand this thoroughly in my head.

Read the wiki article to understand the math.  The probability of losing four times in a row is indeed 1/16.  However, that probability cannot be used to predict the outcome of an individual roll.  In other words, the 1/16th probability of losing four times in a row is valid only when you haven't yet rolled at all.  You cannot use those probabilities once you've actually started playing, because the probabilities of each roll are their own closed system and cannot be affected by any other rolls.  Consequently, if the probability of losing the first roll is 50%, then regardless of any other outcomes of any other rolls, the probability of losing the fourth roll is 50%, just as if the fourth roll had been the first.

Where you're falling down is taking the probability of the whole set (four losing rolls which has a 1 in 16 chance of happening) and trying to incorrectly apply that math to predict the outcome of an individual event in that set.  You can't do that.  If you could, your original equation would look like 1/2 * 1/4 * 1/8 * 1/16, which is pretty obviously wrong.

Fried Chicken wrote:

I've been wondering something about Casino games, which I just cannot shake off.
Let's take Roulette as a very easy example, which has an almost 50/50 chance of winning.

Let's say I place $1 on black.  I spin and lose.  The next turn, I place 2 dollars on black.  1 dollar to get ahead, and $1 to make up for the previously lost $1.  The odds of me losing twice in a row are (very close to) 1/4.  But I lose again.  Now I have spent $3 total on losing bets.  So the next round, I place $4.  1 dollar to make money, and the 3 other dollars to make up for my losses.  But alas, I lose again.  The odds of this happening are (very nearly) 1/8.  So I now proceed to place $8 on the board.  7 dollars to make up for my losses, and $1 to make money.  I win, and get back 16 Dollars.  Of those, $15 are for making up for bets/lost money, and $1 is profit.

Tell me, why does this system not work?  The odds of losing X times in a row get exponentially higher as I keep going....  Is it because at the same time, the amount of money I invest needs to get exponentially higher also?  What's up?

You don't win enough to make it worth your while.

If you win the first time - you win a dollar, but you have to have a LOT of money to keep betting should you lose that dollar and wish to make it back.

The amount of capital you have to have available to give yourself 99.9% odds of coming out $1.00 ahead if you stop as soon as you are a dollar ahead is actually pretty large compared to the $1.00 you win.

Now if you want to come out $100 ahead - you multiply that initial capital by $100 - and it's a lot of money to earn less than a days wages at a standard working class job.

Furthermore - when you lose, you lose big. And yes, it will happen. The chances are low on any given day, but if you try to make a living off it - it will happen and when it happens it will happen big.