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Posted: 31 October 2012 12:48 AM   [ Ignore ]
Total Posts:  1
Joined  2012-10-31

Hey, been thinking about roulette for a while now. I’ve been looking at the progressive dozen betting as being fairly safe.



I just wanted to know the probability of certain things

You have a 2/3 chance of missing, what’s the chance of that happening 13 times?

Is that (2/3)*(2/3)*(2/3) 13 times?

I realise the whole gamblers fallacy deal but obviously there is a number of how often that would happen. Like 0.0009 chance or whatever.

Posted: 31 October 2012 01:08 AM   [ Ignore ]   [ # 1 ]
The Midnight Skulker
Sr. Member
Total Posts:  1398
Joined  2010-01-28

What you are looking for is called joint probability, the probability of multiple events all occurring.  Joint probability is simply the product of the probabilities of the individual events, so in your dozens problem, the probability of losing 13 bets in a row is, as you stated, (2/3)^13.  I would note, however, that this applies only to a no-zero wheel.  For a European wheel the probability is (25/37)^13, and for an American wheel it is (26/38)^13.


My name is Skulker Luis de Midnight.
You killed my bankroll.
Prepare to die.

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