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[Albalaha]

I wonder but I always read someone suggesting playing martingale on a sleeper EC based on its probability to win.

[Bayes]

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.

[Bayes]

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.

[Bally6354]

Thank you very much for explaining that Bayes. 

[Ralph]

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.

[Gizmotron]

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.

[Ralph]

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.

[Gizmotron]

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.

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?