Hey All
I'm sure someone already invented this system but anyway. . . Here is goes
I've spent the last few days playing at BorgartCasino. com
Playing Roulette and betting the columns. . . you need to wait 5 spins before placing a bet. . . the only way you bet is when 1 of the 3 columns hasn't had a hit in 5 spins. . . Bet that column. . . Borgarts min is $10 dollars.
Spin 1 $10
Spin 2 $13
Spin 3 $18
Spin 4 $24
Spin 5 $32
Spin 6 $42
Spin 7 $56
Total 195 dollars
In order to lose 195 dollars you need to lose 12 straight spins. . . . It finally just happened to me after I was up 745 dollars in profit
By no means is it flawless obviously but it's been working well. . reason for the loss was partly because of 0 came up twice in my 7 spins
Thoughts?
Well done! Your right its nothing new, glad your having some success, hope it continues. I've not been on that site, is it a real wheel or RNG?
it's a live real wheel
Does anyone have a better bet progession I could use? I would like to get a couple more spins and still only spend about $200 before calling it quits
Try this
1,1,2,3,4,6,9,14,21,31,47
My min bet is $10 dollars. . . so are u saying. . . 10,10,20,30 . . . . etc then a 470 dollar bet?
There are many ways to play columns, try to create progressions with street for example
12 sleeping dozens happens once for every 100 spins. That's the mathematical arithmetic for it. That's for the double zero wheel too. I need to power test this with a sim. Usually things tend to balance out. This is not a balanced amount. Conditional statistics are not supposed to be this easy. Time for a crash test dummy session.
The odds to lose this bet on the first of five spins is (26 / 38) = .684211.
That times itself 11 more times descends down to = .010
That equals one in every 100 spins.
.684
.467
.320
.218
.149
.102
.070
.047
.032
.022
.015
.010
Does this look correct to everyone?
COOL! I figured out the balance point. It's right in front of me. There are only 15 opportunities to use it for every 100 spins. It's right there in the odds.
So to balance the cost of losing your progression it will need to pay for itself in 15 winning payoffs. I wonder what position in the progression is the average for those 15 attempts?
Playing any dozen on single zero wheels you have a 67.56% chance of losing the first bet.
We all know you play roulette 1 spin at a time. This means you also have a 67.56% chance of losing the second bet.
So too the third bet, and the fourth, and so on.
For the purposes of playing roulette in real time, calculating the odds of future bets beyond the very next bet is of little use.
Quote from: bombus on May 07, 2011, 09:57:31 PM
Playing any dozen on single zero wheels you have a 67.56% chance of losing the first bet.
We all know you play roulette 1 spin at a time. This means you also have a 67.56% chance of losing the second bet.
So too the third bet, and the fourth, and so on.
For the purposes of playing roulette in real time, calculating the odds of future bets beyond the very next bet is of little use.
Boy, that's a huge and glaring mistake in basic binomial distribution odds. You sure you want to be known for saying this?
Here's help: nolinks://stattrek.com/Tables/Binomial.aspx#distribution (nolinks://stattrek.com/Tables/Binomial.aspx#distribution)
I thought binomial test sampling is carried out without replacement?
Quote from: bombus on May 07, 2011, 10:14:42 PM
I thought binomial test sampling is carried out without replacement?
It works for coin toss tests, same coin over and over again.