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.

[junkmail3]

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


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I wonder how many Martingale haters will respond before getting to this line.

And don't you need to have a wagering requirement, or a profit goal in order for the RoR to be calculated. Because as it stands, playing this system forever, until you have a heart attack at the table and die, your RoR would be 100% wouldn't it? (I don't know if I'm thinking about it correctly though)

[pzhon]

Your progression doesn't make sense. When you lose on those 7 numbers, you don't just lose one bet. You lose 2. If you want to come out ahead by one unit on your first win, you need to bet 1, 3, 9, 27, 81, 243. If you lose while placing two bets of size 243, you will end up down 728. This happens with probability (7/19)^6 ~ 1/400. So, with probability 399/400, you win 1 unit, and with probability 1/400, you lose 728 units. You'll lose an average of about 1/4 of a unit per spin of the roulette wheel by using this system.

I don't recommend thinking about martingales a lot. There is no way to add up negative numbers to get a positive number. However, many people have deluded themselves into thinking that this is possible by making a system that is too complicated to understand at a glance.

[PokerCat69]

What will likely happen is you will have a few sessions where you don't bust out. You profit $1 on a win so you might be up $50-60 before going home. You'll do this a couple times then one day your progression will bust out and the money you won will not even come close to how much you lost.
Worst case senerio is it loses on your first trip to the casino.

[SheetWise]

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If you lose while placing two bets of size 243, you will end up down 728.

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[img]/images/graemlins/blush.gif[/img]

[pzhon]

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If you lose while placing two bets of size 243, you will end up down 728.

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[img]/images/graemlins/blush.gif[/img]

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At that point, you will have lost 1+1+3+3+9+9+27+27+81+81+243+243=728. <font color="white">3^6-1 = 728 = 222222 base 3. </font>

[LALDAAS]

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If you lose while placing two bets of size 243, you will end up down 728.

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[img]/images/graemlins/blush.gif[/img]

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At that point, you will have lost 1+1+3+3+9+9+27+27+81+81+243+243=728. <font color="white">3^6-1 = 728 = 222222 base 3. </font>

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Yes 728 is correct however when you do win you will be paid 2:1 on 243 which is return $729. The 2:1+your orginal bet.

You were correct my numbers were wrong in my first post.

yours are correct. thus only allowing 5 double ups.