Just getting a few opinions would be great as well. :thumbsup:
We all know that you will get twice as many singles as repeats and so forth....
So my question is the following.
Let's say you are playing a martingale. {1 2 4 8} Are you subject to the same negative expectancy on every bet?
"We all know that you will get twice as many singles as repeats and so forth...."
Bally
So you'll get twice as many
R
B
as you do
R
R
If that's not right, please spell it out for me.
Sam
You will see twice as many singles as doubles.
Twice as many doubles as triples.
Twice as many triples as four timers etc...
A single is 1/1.
A double (parlay) is 3/1.
A triple is 7/1.
A four timer is 15/1.
Bally
Is this what you call a single:
R
R
B
Bear with me......
Sam
A single could be either RED or BLACK.
If I decide to bet BLACK.. the odds are 1/1. Same as if I decide to bet RED.
Going for any double... RED BLACK, BLACK RED, RED RED, or BLACK BLACK is 3/1.
So I am twice as likely to get my single bet up as I am getting the double.
Bally
I have spent hours studying this and even had KonFuSed help me on it. Don't know if you knew of him, but he was quite the math guy! I don't want to appear to be a "math guy" because I'm certainly not. However, I can regurgitate what I learned from him.
If you'd like me to continue, I will. If not, I'll let it go.
And Thanks...
Sam
Quote from: Bally6354 on January 18, 2013, 09:01:33 PM
You will see twice as many singles as doubles.
Twice as many doubles as triples.
Twice as many triples as four timers etc...
A single is 1/1.
A double (parlay) is 3/1.
A triple is 7/1.
A four timer is 15/1.
Bally in theory YES. In real play. it depends when you enter the cycle. You might see 10 doubles and no singles. You might see RRR-BBB-RRR-BBB-RRR And no doubles. Random is awesome it can form all manner of runs and patterns.
Sam
I am just trying to get my head around a few things and would appreciate any input!
Here is the thing.....
Singles are twice as likely as doubles and doubles are twice as likely as triples etc...
That to me also means that we can go in reverse. A second loss is half as likely as an isolated loss and so on....
This is where my negative expectation question comes into things.
The NE supposedly stays the same on every round. HOWEVER my chances of winning are increasing if a second loss is half as likely as the first loss and so on.
Maybe I am completely confusing myself ???
Bally
You can never have a true single. It must be
R
R
B
or
R
B
or
B
R
B
You can say......I just hit B and that was a loss, so my chances of losing twice are less. Only if you made that statement before the first B hit. After that B hit, it's a whole new world.
What are the chances of hitting B B when you first place a bet? 25% VS 75% for red. It appears after the first B, you have a greater chance of getting another B. Not so. Now you chance is 50/50, just as it always is.
That s the way konFuSed explained it--more or less.
Sam
(It may be I don't understand you. The above is pretty elementary. You probably know it already)
You get as many singles as series.
As for the probability it stays the same no matter past spins. 1/2.
Even Chances have a Linear Expectation: the sum of the probabilities of each independent trial.
For two trials the expected success is 1/2 + 1/2 = 1
3 trials = 1/2 + 1/2 + 1/2 = 1.5
4 trials = 1/2 + 1/2 + 1/2 + 1/2 = 2
....
As you see it is linear and so why the amount of Singles = amount of Series
amount of series of two = 2x amount of series of 3 = 2x amount of series of 4 .....
...
So in 2 trials you may expect 1 Win, but at the fourth trial you may expect 2 Wins, the ratio stays the same, it is Linear. =)
The number n of trials for the expected number of Wins to be 1 is
n = 1/p
This is also the expected number of trials before we see the first Win.
Expect a Loss after a Win or Vice versa is a very common pitfall of Gambler's Fallacy.
Hope this helps.
One nice thing you can observe in a binary random distribution is that you can only detect 3 different states.
RR BB Series of Series
RR B R BB Series of Singles
BB R BB Isolated Single / Isolated Series (hovering state)
Cheers
If there was no zero, the probability of getting one or more hits in three trials would be 87,5 %
We can devise a simple flat bet exercise.
Having a Series of 3 or more as trigger to bet on the opposite outcome, we stop on at +1, 0, -1 or -3
Seems a disavantage...
Below attached are sessions from Random.org (1000 outcomes), one a day.
Surprised on how well it goes...
Example of LW registry:
Random.org 17/01
WLWLWWWLLWLWLWWWLWWLWWWLLWWLWWWWLLWWLWWLLWLLWLLWWLLL
WWWWWWWLLWLWWWLLWWWLLWLLWWLWLWWWLLL
LWLWLLWWLLWLWWWLWLWLWLWLLL
WWLWLWWWLWLLL
LLWLWLLL
LLWWLWLLL
LLL
WWWWWWLWWWWWWWLLL
LWWLWLWWLWWWLWWLWWWWWWWLWLWLWWLLWLLL
LWLLWLLWWWLLWWWLWWWLWLWWWWLLL
LLWLLLWWLLLWWLWWLLWLLL
WLLL
W
+16 UNITS
15.01 -10
16.01 +7
17.01 +16
18.01 +6
19.01 +18
Quote from: Bally6354 on January 18, 2013, 09:44:59 PM
Sam
I am just trying to get my head around a few things and would appreciate any input!
Here is the thing.....
Singles are twice as likely as doubles and doubles are twice as likely as triples etc...
That to me also means that we can go in reverse. A second loss is half as likely as an isolated loss and so on....
This is where my negative expectation question comes into things.
The NE supposedly stays the same on every round. HOWEVER my chances of winning are increasing if a second loss is half as likely as the first loss and so on.
Maybe I am completely confusing myself ???
Chance to get same EC in next bet and to change with opposite is equal. So, RB is as likely to appear as RR is or BB is. Gambling on roulette or any similar gambling runs on the principal of chaos or randomness and not strictly on probability. Probability theories hold good in long run only. For instance, chance to get 7 reds in a row is very remote in a 185 spins session but you may get that not just once but many times in 185 spins. However, if you see the same thing with its mathematical probability in say 1 million spins, it would be very close to that.
Those who are too much obsessed with probability or mathematics starts playing a sleeper EC after certain losses and end up one's bankroll with martingale.
Quote from: albalaha on January 19, 2013, 08:33:46 AM
Those who are too much obsessed with probability or mathematics starts playing a sleeper EC after certain losses and end up one's bankroll with martingale.
Those who are too much obsessed with probability or mathematics
WILL NEVER USE A MARTINGALE PROGRESSION!
MG, I agree. If you're "obsessed" with probability or mathematics (and understand it) it means you WON'T use a Martingale. ;)
I wonder but I always read someone suggesting playing martingale on a sleeper EC based on its probability to win.
Quote from: Bally6354 on January 18, 2013, 09:44:59 PM
Here is the thing.....
Singles are twice as likely as doubles and doubles are twice as likely as triples etc...
That to me also means that we can go in reverse. A second loss is half as likely as an isolated loss and so on....
This is where my negative expectation question comes into things.
The NE supposedly stays the same on every round. HOWEVER my chances of winning are increasing if a second loss is half as likely as the first loss and so on.
Bally, you're right, but with an important caveat. Although the odds don't change, in any
predefined sequence of bets, the most likely outcome is a win on the first bet, which is twice as likely as a win on the second bet, which is twice as likely as a win on the third, etc. Note this is valid only for ECs.
Thus,
for ECs, the probability of a win on the first bet is, 0.5, the probability of a win on the 2nd is 0.5 × 0.5 = 0.25 etc.
For any other bets, the
pattern is the same, but the probabilities differ. The "random variable" which gives the distribution of wins on the 1st, 2nd, 3rd etc bets is described by the geometric distribution. So for example, taking a double-dozen bet, the probability of a win is 2/3 or around 67% (ignoring the zero), so your chance is 67% that you will win this bet on the
first spin. If you don't win until the 2nd bet, it means you must have lost on the first, the probability of which is 1 − 2/3 = 1/3, so the chance of winning
first on the 2nd bet is 1/3 × 2/3 = 2/9 = 22.2%. Again, if you don't win until the 3rd bet it means you've lost the first 2, so the probability of winning
first on the 3rd bet is 1/3 × 1/3× 2/3 = 2/27 = 7.4%.
You can see this pattern looking at my simulation of JL's "7-on-1" system which bets on two dozens:
Wins on step 3 = 4731
Wins on step 4 = 1611 <--- this is roughly one third of the number of bets which won on step 3
Wins on step 5 = 545 <--- this is roughly one third of the number of bets which won on step 4
Wins on step 6 = 173 <--- this is roughly one third of the number of bets which won on step 5
Wins on step 7 = 56 etc....
Wins on step 8 = 32
Wins on step 9 = 8
Wins on step 10 = 2
Wins on step 11 = 1
Wins on step 12 = 1
In general, if the probability of winning a bet is
p and the probability of losing it is
1 − p, then the chance you will win your first bet on the x
th spin is
:
(1 − p)x−1p
So there is nothing special about the so-called "law of series", it's just a special case of the above formula when p = 0.5.
Quote from: albalaha on January 20, 2013, 03:25:11 PM
I wonder but I always read someone suggesting playing martingale on a sleeper EC based on its probability to win.
There's no fallacy involved in using a martingale as long as you don't think your chance has improved by waiting for virtual losses.
Thank you very much for explaining that Bayes. :thumbsup:
I have (in fun) tested JL 2/3 play using progressions. I did not do it because I thought it would win, as 2/3 bet I have always thought is the worse bets. It will have a minor chance to stand 100 trials in a row. The probability of have 1000 in a row (and I mean spins with bet) must be way out of common experience. What is the odds, one or max two in 10 millions here are yearly hit by thunder lightens, so rare things can happen.
You should also be aware that the odds for losses in a row are calculated differently.
The odds to lose is 33%
The odds to lose twice is .33 x .33 = .10
The odds to lose three times is .33 x .33 x .33 = .033
You have a 97% of winning all three step, double dozen Marti's.
In other words, you will lose three times for every 100 attempts.
Quote from: Gizmotron on January 20, 2013, 06:42:25 PM
You should also be aware that the odds for losses in a row are calculated differently.
The odds to lose is 33%
The odds to lose twice is .33 x .33 = .10
The odds to lose three times is .33 x .33 x .33 = .033
You have a 97% of winning all three step, double dozen Marti's.
In other words, you will lose three times for every 100 attempts.
Yes but we must look at the probability including some reasonable devination in our favour, we can not say the outcome must the same every 1000 times we try. If somebody claims an outcome we can not say it is not possible by just look att the 2/3 times trials.
Quote from: Ralph on January 20, 2013, 07:04:03 PMYes but we must look at the probability including some reasonable devination in our favour, we can not say the outcome must the same every 1000 times we try. If somebody claims an outcome we can not say it is not possible by just look att the 2/3 times trials.
I don't count on odds. I observe what is currently happening and how effective I am at synchronizing with effective bets. Odds and distributions have nothing to do with winning sessions for me. If you wait for it, the casino will pull its pants down like a virgin on prom night. Every once in a while randomness lines up for you like ducks in an arcade. You can shoot them down with ease. Normally, randomness offers you a steady opportunity, a chance to work out an aggregate win. And all along the way probability keeps churning out irrelevant numbers. If more people did what I do, there would no longer be a house advantage. People don't lose because it's mathematically divined. They lose because they plan in advance to lose and then execute their plan. For most people, there are stages of learning they must earn the hard way, before they can advance. That applies to many here. Just by being here, they admit to being on a path of learning. I'm so thankful that I was never stuck like JL is. I found the tools to move on. I actually hoped I could really beat this thing. But who knows? Maybe the illusion of winning is good enough for some people. Maybe it's good enough for what you want from an online forum?