Roulette yet another post

[LALDAAS]

I of course understand no matter what betting strategy you can come up with this is still a -EV game and will never over come the HA

However It is still a fun game. With all the martingale posts I have been thinking about it allot.

I have a system I like and was wondering if a math wiz would like to crunch some numbers on this one for me.

This is a cover bet Martingale+1 system.

Starting off with $1 bets or what ever the table min.

Lets assume the min is $1 and max is $500

Lets also assume this is a single Zero table.

Place 1 unit on 1 to 18 (1:1)
also place 1 unit on 3rd 12 (2:1) on the same bet.

What happens here is a cover bet.

On a bet on the [1 to 18] if hit it is a break even bet. You are paid 1:1 on a 1 unit bet. So you gain your loss on the [3rd 12] bet.

On a bet on the [3rd 12] if you hit you are paid 2:1. So you gain your loss on [1 to 18] plus 1 unit.

With this bet you are covering 30 of the 37 numbers. (Again we are playing the single zero table.)

So here we are breaking even on 18 #'s, winning on 12 #'s and losing on 7#'s

When we do lose on these 7#'s then apply the martingale system to our loses. Doubling up our bet until a win occurs returning to our original betting unit.

Also to add for a faster winning gain adding 1 betting unit per double up of the martingale system to make of for the lose of a unit per a winning spin winning spin.(what I mean by this is gain a unit rather just breaking even plus 1 unit.)

Example:

1,1
2,2 +1+1=6
6,6 +2+2=14
14,14 +3+3=34
34,34 +4+4=76
76,76 +5+5=162
162,162 +6+6=334

the +1 only allows 7 consecutive loses. Before reaching the table max.

Now I hope I explained this well enough of you guys to understand.

So if you are looking for that one more opportunity to double a lose forget about the +1. Leaving the betting structure for a lose as:

1,1
2,2
4,4
8,8
16,16
32,32
64,64
128,128
with the table max you cannot double up to the full amount however if you choose to go 250,250 and do hit the return is 250 losing 5 units.

Any consecutive loses more the 8 is well a bust.

Mathematically challenged individual (such as my self) were to look at this:

breaking even on 18 #'s, winning on 12 #'s and losing on 7#'s

Ones got to think in the short run you just might be able to run away with a few bucks.

So what’s the risk of ruin here, I am just curious.

I would like it to be known I have never tried this aside from a play money table. I have just been screwing around bored at work and thought I would share.