Martingale in roulette: you play 60 times on black; ball lands red/0 11 consec. times

mathematt

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Dec 18, 2018
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Good afternoon,
I know that in the long run you can never win at roulette. The casino will always win, even if you are using the Martingale technique (that doubles your bet with every loss).
But I am curious about how big the probability is that:
* when you play 60 times on black
* the roulette ball will fall on red or zero 11 times in a row

My quick calculation says is 2,44% ( = ( 2^(60 - 11) * (1 + (60 - 11)) - ( 60 - 1) - (((60 - 2) - (11 + 1))/2 * ((60 - 2) + (11 + 1))) )/ 2 ^ 60), but I am pretty sure it is not totally correct.
Who can correct me please?
 

tkhunny

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Good afternoon,
I know that in the long run you can never win at roulette. The casino will always win, even if you are using the Martingale technique (that doubles your bet with every loss).
But I am curious about how big the probability is that:
* when you play 60 times on black
* the roulette ball will fall on red or zero 11 times in a row

My quick calculation says is 2,44% ( = ( 2^(60 - 11) * (1 + (60 - 11)) - ( 60 - 1) - (((60 - 2) - (11 + 1))/2 * ((60 - 2) + (11 + 1))) )/ 2 ^ 60), but I am pretty sure it is not totally correct.
Who can correct me please?
...or any other game deliberately designed to make money for the house.

The problem with this silly strategy is the exponentially increasing size of the wager upon loss. Too many consecutive losses and anyone runs out of cash.

If you manage only ONE 11-string of losses - rough estimate ...
p(11 losses in a row): (19/37)^11= 0.000654743
Places for your 11-string to start: 50 => 3.7% -- And we haven't even mentioned getting more than one (1) 11-string.
First impression? Your 2.44% is very much too low.

Shall we quit after the #12 win after the first 11-string? That' a different question.

The problems with the whole thing are:
1) The odd premise that one will eventually score a win. It's just not 100%. It's clearly not 100% with...
2) The finite money supply, since "eventually" is now finite.
3) The house odds.

P.S. Full Disclosure: Keep in mind that I am totally biased against gambling so I might make rough estimates of loss probability that are too high and then feel perfectly fine with my error. :)
 
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Jonathan

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De Moivre (1667 - 1754) gave a solution for the probability of a run of successes in a series of trials in his book "The Doctrine of Chances" (problem lxxiv ie 74). It is not easy to find references on the internet - I hope this link works https://arxiv.org/pdf/math/0511652 Not surprisingly the formula is not simple because of the complications tkhunny mentions.
 

mathematt

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De Moivre (1667 - 1754) gave a solution for the probability of a run of successes in a series of trials in his book "The Doctrine of Chances" (problem lxxiv ie 74). It is not easy to find references on the internet - I hope this link works https://arxiv.org/pdf/math/0511652 Not surprisingly the formula is not simple because of the complications tkhunny mentions.
Thank you Tkhunny and Jonathan! Very useful info.
* Tkhunny, how do you get from 0.000654743 to "Places for your 11-string to start: 50 => 3.7%"?
* Jonathan, that's a complex article. I am trying to understand it. Do you understand it? What would be the probability when r = 11 and n = 60?
 

mathematt

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Guys, still there? :)
 

tkhunny

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Thank you Tkhunny and Jonathan! Very useful info.
* Tkhunny, how do you get from 0.000654743 to "Places for your 11-string to start: 50 => 3.7%"?
* Jonathan, that's a complex article. I am trying to understand it. Do you understand it? What would be the probability when r = 11 and n = 60?
One cannot achieve 11 consecutive if you don't start until five before the end of the game.
 

Jonathan

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No, I don't fully understand the article. I posted to back up tkhunny's comments about the complexity of the problem. Others on the forum may be able to help here.
Another approach to determine the probability without complicated calculations would be to run a computer simulation of wheel spins. I'll leave this for you to investigate.
 
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