Roulette 20 Reds

Hello Punters ...

Disclaimer

I don't claim anything, up to each individual to judge the content.
For me is just for fun and the challenge to show it can be done.

20 Reds

Turbo Genius use a Marty at the beginning of he's "Against All Odds" parachute.
I copy that part into my solution covering 20 events with 50% probability each.

The binomial basic

EC - 8 attempts ... (Marty)
Dozen - 4 attempts (2 attempts has 50% probability)
Line - 8 attempts (4 attempts has 50% probability)
Corner - 12 attempts (6 attempts has 50% probability)
Street - 16 attempts (8 attempts has 50% probability)
Split - 24 attempts (12 attempts has 50% probability)
Single 50 attempts (25 attempts has 50% probability)

That is 12 attempts with 50% probability ... now you just add 8 Marty for the Even Money.
Now you cover 20 in a row.

That is what I would promote as "Against All Odds"

Cheers

 

There's still that perfect sequence of death. These perfect sequences still happen and take all the progress achieved by these complex progressions.

I have clearly demonstrated that you can have a perfect sequence of first attempted RR bets that fail. Nothing can stop the perfect sequence of death to a gambler. You must have a method that allows for it and deals with it. Only then can you beat it.

 

So how much is this progression going to cost you if it fails? I haven't worked it out.

 

Well, Turbos cost around 1700$ with 1$ Unit size - if I remember correctly.
I assume this one will be more expensive.

Just take your time and think how many times - if ever - you have seen 20 reds or above during your lifetime.
What can happen will happen.

I would like to add a random indication to this, maybe I write about that later.
Sure put more fun to the calculation and more Lotto ...

Cheers

 

Well to keep the cost down - we can use one profit and one break even and continue upwards the ladder.
For example, the Marty can be 1, 1, 3 ,5 ,11, 21, 43, 85,

Cheers