folditnow
Joined: Sep 25, 2017
• Posts: 7
September 25th, 2017 at 5:00:02 PM permalink
I'm trying to wrap my head around this. I understand that Roulette spins are disconnected events and one spin does not affect the others. If you discount the Zero/Double-Zero so that we are only talking about the 36 possible single numbers and I cover 10 streets on every spin (30) leaving the house 2 (6 numbers - regardless what they are) My odds of hitting one of my numbers is 30/36 (no zeros) or 5:1 in my favor. My question is, of those six numbers, what is the probability that the house will hit one of those six numbers four times in succession before one of my numbers hits?
EdCollins
Joined: Oct 21, 2011
• Posts: 1270
September 25th, 2017 at 5:37:48 PM permalink
Hi folditnow,

Welcome to the forums.

However, the payoff for a winning number on a roulette wheel is, no matter what the bet, always LESS than what the true odds say the payoff should be. This is true whether you are betting on particular color, or on a range of numbers, or on odd/even, or on a specific number, or whatever. And that payoff that is less than what it should be is what gives the house its edge.

There is no Martingale System or method of betting that can overcome this edge.

Your "odds of hitting a number" (notice the quotes) might be "in your favor," (again, in quotes) depending upon how you bet. (30/38 in your example. I don't know why you discount zero and double zero), but each time you win you are winning LESS than what you should be winning, if there were no house edge at all.

Eventually this house edge will make itself felt. You can't get around it or ignore it.
odiousgambit
Joined: Nov 9, 2009
• Posts: 8407
September 25th, 2017 at 5:54:14 PM permalink
Usually this kind of question gets asked for one of two reasons,

* the OP has taken a beating from a losing streak

or

* the OP is plotting a martingale attempt

8/38 times the house wins the bet, or 0.2105263157894737, which is one in 4.7499999999999996438. 4 times in a row is one in about 509 ...

I am not the math-go-to-guy but I think that is correct as 5 to the 4th power is 625, for a check

So if it is case #1, you were modestly unlucky
If it is case #2, you are probably tempted to press your luck and martingale. That brilliant idea is centuries old and has been resoundingly discredited as a way to gamble but has been perfect for driving gamblers crazy.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell! She is, after all, stone deaf. ... Arnold Snyder
DJTeddyBear
Joined: Nov 2, 2009
• Posts: 10300
September 25th, 2017 at 6:45:09 PM permalink
Quote: folditnow

... If you discount the Zero/Double-Zero so that we are only talking about the 36 possible single numbers ...

People always want to ignore those green spots. But it's those green spots that put holes into whatever plans you're cooking up.

Bottom line: you'll give yourself a headache working on the math for a system that just won't work.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/  Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
DRich
Joined: Jul 6, 2012
• Posts: 6841
September 25th, 2017 at 7:10:08 PM permalink
Quote: folditnow

what is the probability that the house will hit one of those six numbers four times in succession before one of my numbers hits?

If you are discounting the green numbers completely then I think it is 1 in 1296.
Living longer does not always infer +EV
folditnow
Joined: Sep 25, 2017
• Posts: 7
September 25th, 2017 at 7:48:34 PM permalink
Thanks for the responses guys!

I realize the payoffs are skewed in favor of the house. Casinos were not built off of winners and I'm not trying to suggest it's possible to beat the long term odds because I know better. I discounted the zeros for clarity, removing the decimal so it's easier to calculate, because lets face it, how often am I going to sit and play long enough for the decimal to get me? Nobody sits at a table of any kind long enough for the Casino's edge to reach equilibrium, so there are streaks, highs and lows that can be plotted on a graph. If you look at these over a long period of time it will approach a waveform that alternately dips and rises over a baseline, eventually approaching the center. We call this entropy. Everything in nature operates in this manner. In order to have highs, you must have lows. Everything is cyclicle in nature, approaching a mean over time.

So this is not a fools errand. I realize that I cannot beat the game long term. If I sat there continuously playing I will eventually lose. But good gambling is taking the best bet with the best odds. Yes you lose sometimes. But if there is any possible advantage you can acheive, you had better grab it.

Now, the Martingale. Yes it has it's weakness, in the fact that eventually you will hit the table maximum or run out of money. Martingales are normally run on outside bets, which are all 1:1 even money bets. What I am suggesting is (like bookmakers do) spreading that action over several or more seperate bets so that your exposure to risk is divided by your bet. Simply put, I can make a 10 dollar bet on black and cover 18 numbers, but the house also has 18 numbers, plus the zeros, for either 19 or 20 chances to win against my 18. Obviously in the long run I am screwed.

If instead I placed 10 one unit street bets, i'm dividing my risk over 30 numbers (as opposed to one of two colors), increasing my chance to hit. Black/Red is 18 to 19/20, My way is 30/37 (Euro) . I will eventually get beat, but, If I Martingale just these 10 units individually one time, my relative chance to hit with the martingale is much higher than the 1:1 chance of an outside bet (which is not actually a true 1:1), but i will only be able to do this a certain amount of times due to the street bet maximums (which should also tell you that this idea is sound. They only put maximums on things that could hurt them.

I've been testing this very successfully, I just cant figure the multiple odds of hitting on successive spins. Can't figure the formula, not a mathemetician.
pacomartin
Joined: Jan 14, 2010
• Posts: 7895
September 25th, 2017 at 8:47:27 PM permalink
Quote: folditnow

Now, the Martingale. Yes it has it's weakness, in the fact that eventually you will hit the table maximum or run out of money.

It is a myth that table maximums were instituted to stop Martingale players. In fact, casinos love Martingale players, just as they love any gambler with a system. Gamblers with any kind of system will always play for longer periods of time.

Table maximums are there for one reason only. A casino, like any business must worry about cash flow. If individual bets are too big, the gambler might take home too much money on any given day.
RS
Joined: Feb 11, 2014
• Posts: 8623
September 25th, 2017 at 8:51:42 PM permalink
Quote: folditnow

I'm trying to wrap my head around this. I understand that Roulette spins are disconnected events and one spin does not affect the others. If you discount the Zero/Double-Zero so that we are only talking about the 36 possible single numbers and I cover 10 streets on every spin (30) leaving the house 2 (6 numbers - regardless what they are) My odds of hitting one of my numbers is 30/36 (no zeros) or 5:1 in my favor. My question is, of those six numbers, what is the probability that the house will hit one of those six numbers four times in succession before one of my numbers hits?

Probably of hitting a number is 30/36 = 5/6.
Probability of not hitting a number is 6/36 = 1/6.

Probability of not hitting a number 4 times in a row is 1/6^4 = 1/1296 (what Dr. Itch wrote).
folditnow
Joined: Sep 25, 2017