reverse progression in a single number bet

[toby]

I need to foolproof this:

a)1 number

b)5 units until a hit, 4 units after a hit, 3 units 2 and 1. We need 5 hit within 179 or less spins to stop.

c)in any profit we reset.

d)after 179 spins without profit we start the second one with 10 units in the number, 8, 6, 4, and 2

e)the 3rd cicle is 15, 12, 9, 6, 3 etc...

The odds for a 5 or more hits within 179 spins is +53.32%.

20 or more of the 37 numbers will get 5 or more hits in 179 spins.

[winkel]

There´s only one problem:

How to get one number out of the 20 that hit +5times

br
winkel

[toby]

There´s only one problem:

How to get one number out of the 20 that hit +5times

br
winkel

You only pick one, it has a 53.32% of being the one which hits 5 times.

Pick 4 numbers at the same time in the same table.

Reset in any profit.

Another data: more than 29 numbers of the 37 will get profit(+1 or more) if you play it for 179 spins or until a +1.

[Jish]

I think 53.32% is way too low, as we all know with martingale there will always be that streak of X numbers that will make your bets so high you lose your etire bankroll, but on the other hand single numbers pay 35-1 where EC pay 1-1, so what im saying is if you dont get your numbers in 179 spins ths could happen 5,6,7,8+ times in a row, it dosent happen very often but it does happen

[toby]

You lose playing even money because you have 48.5% chance of winning, letalone diffeencies between BR´s player and the bank.

When you have 53% chance of being in profit you pick the best chance to get a good luck strike.

I guess you have more than 53% when you play several numbers as diferent players.

Even though, this system sinks in the end because of the house edge plus bankrupt risk.

You need to pay attention to the numbers to play.

You need numbers that break even.

Math doesn´t work alone.

BR

[toby]

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