a very solid progression

[beretta28]

I illustrate  below a very solid progression that is giving me good results so far.

1 2 4 8 10 12 14 16 18 20 Stop: 105 units . 
Of course up and down depending on L or W. 
Reset each time if + 1 unit!

The first part is a Marty,the second is  a D'Alembert max 20 units

Target:+ 10 units per session.

Bet selection: switch from R to B and viceversa at every spin(not so important as progression)

Could anyone calculate the risk of loss? It's 6,96% for the first four spins,but how to calculate the probabilities of losing all the bankroll in 10 spins(4 Marty + 6 D'Alembert)

Comments appreciated

[Jean-Claud]

This isn t deferent than the 1,2,3,4,5,6,7,8 etc.

[beretta28]

With the progression 1,2,3,4,5,6,7,8. . .  you don't win if you hit the 3th or 4th spin.
With my progression YES! It's much better.

[mistarlupo]

Could anyone calculate the risk of loss?

(18 / 37) ^ 10 = ~0.074%

[Number Six]

And those 0.074's added together eventually equal 1.00 

[beretta28]

Thanks for your answer,but I'm afraid that you miscalculate.

If the risk of loss  was 0,074%,I'd be very rich!

Once again for the first four spins the risk of loss is 6,95%(very easy to calculate,it' a classical Marty and you win 1 unit and

you restart). ,but if you lose 4 spins in a row, for the six other spins it's a D'alembert(+2 units if loss,less 2 units if win until you get + 1 unit).

I'm afraid that the risk of loss is close to 11,5% for the whole progression of ten terms,but I'm not so sure.

[Bayes]

The chance to lose one bet is 1 - 18/37 = 19/37, and so the chance of losing 10 in a row is (19/37)¹⁰ = 0.001275 = 0.1275%

But read this article.

[beretta28]

Once again your calculation is not correct.

The probability of ten losses in a row is what you have written,but I Knew this easy calculation.

My question (or my progression) is more complicated

Four spins:OK it's a martingale and the risk of loss of four spins in a row is 6,95%.

But after that it becomes a D'Alembert.

So,for example you lose 10,you lose 12,you win 14,you go down to 12 and you win,then you lose 10, you go up to 12 and you win  again and so on. Restart if + 1 unit.

What is the probability of losing 6 terms of a D'Alembert,NOT IN A ROW,but after x spins with several up and down.

Thanks again

[kav]

Beretta is correct. The above calculations are for Martingale. This is different.

The risk of ruin of the second part  10....20 is roughly 15%. (you need to win 16 units by risking 90 units)
But it must be applied to the first part, so we have 15% of 6,95%= 1,044% That's roughly the total risk of ruin.

[NS]

No money management will make you win if you don't have a positive EV in the first place. That's what you should know.

Martingale or not the martingale, you'll lose your bankroll sooner or later if you continue to play a negative EV game.

[Bayes]

Beretta,

Initially you were asking for the chance of losing 10 bets in a row, and the staking is irrelevant. The problem you want solved is vague - what do you mean by "several up and down". Unless you can say exactly what you're looking for, in terms of numbers (not "x" and "several") then you won't get any help.

It sounds like a computer simulation would more easily give you the answers.

[beretta28]

Thank you KAV:I think you had the good math approach(20 bankroll  units, for winning 16 units) and your % is  probably correct.


NS: what you state has known for centuries.
It's obvious that we must looking for a system, winning with flat bet
It doesn't exist.
The goal is to find something with a very very low risk of loss and to  hope that loss occurs as late as possible,later than our death!

As a famous french mathematician said: "IT'S IMPOSSIBLE THAT THE IMPROBABLE DOESN'T HAPPEN"

[beretta28]

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