8laylock
8laylock
  • Threads: 1
  • Posts: 1
Joined: Jan 21, 2017
January 21st, 2017 at 10:12:10 PM permalink
Hi, I just joined the site, searched for my question and didn't see it so I'm posting it. I just came off a cruise playing more roulette than usual and doing well.
Here was my method:
Spin 1: $5 on 1st 12
Spin 2: $10 on 1st 12
Spin 3: $15 on 1st 12
Spin 4: $20 on 1st 12
Spin 5: $25 on 1st 12
If at any time I won, I started over at Step 1. If I lost, I increased by $5. If I lost 5 in a row, I accepted my $75 loss and started over with $5.

I know that I'm wrong but my rough calculation is I have 4 spins to make money and the 5th to break even on a bet that has odds of 31.57% of hitting. I know I'm missing something but don't know what.

Thank you for your help
ontariodealer
ontariodealer
  • Threads: 7
  • Posts: 999
Joined: Aug 5, 2013
January 21st, 2017 at 11:20:30 PM permalink
it does not matter what you do or how you do it.....eventually on roulette you will lose 5.26% of your bets.
get second you pig
ThatDonGuy
ThatDonGuy
  • Threads: 117
  • Posts: 6219
Joined: Jun 22, 2011
January 22nd, 2017 at 6:52:08 AM permalink
Quote: 8laylock

I know that I'm wrong but my rough calculation is I have 4 spins to make money and the 5th to break even on a bet that has odds of 31.57% of hitting. I know I'm missing something but don't know what.


What you're missing is, when you win, you tend to win small, and when you lose, you lose big.

Assuming a double-zero wheel (since you say you have a 31.57% chance of winning):
6/19 of the time, you win your first bet, and end up +10
13/19 x 6/19 of the time, you lose your first bet and win your second, and end up +15
13/19 x 13/19 x 6/19 of the time, you lose your first two bets and win your third, and end up +15
13/19 x 13/19 x 13/19 x 6/19 of the time, you lose your first three bets and win your fourth, and end up +10
13/19 x 13/19 x 13/19 x 13/19 x 6/19 of the time, you lose your first four bets and win your fifth, and end up even
13/19 x 13/19 x 13/19 x 13/19 x 13/19 of the time, you lose all five bets and end up -75
The expected result is (6/19 x 10) + (13/19 x 6/19 x 15) + (13/19 x 13/19 x 6/19 x 15) + (13/19 x 13/19 x 13/19 x 6/19 x 10) - (13/19 x 13/19 x 13/19 x 13/19 x 13/19 x 75), or about -1.61839 per "run" (until you win, or lose 5 in a row).
charlestfuller
charlestfuller
  • Threads: 3
  • Posts: 43
Joined: Nov 18, 2016
January 22nd, 2017 at 7:00:03 AM permalink
From a distance, when looking at Don's post, you have to win at least 6x more often than you lose. Sounds easy, but it will catch up.
rsactuary
rsactuary
  • Threads: 29
  • Posts: 2315
Joined: Sep 6, 2014
January 22nd, 2017 at 7:00:42 AM permalink
You are doing a version of the Martingale system. And in the long run, it won't work.

There's lots of research on it, on the internet.
DJTeddyBear
DJTeddyBear
  • Threads: 207
  • Posts: 10992
Joined: Nov 2, 2009
January 22nd, 2017 at 11:24:23 AM permalink
It's not much different than any other system,

In short, it looks great on paper.

In fact, it works fine in a casino.

Until the one time that it doesn't. Then it's a complete financial disaster.

---

Here's a homework assignment for you:

Go to a casino and look at the history displays for all the roulette tables there. Check how many times you see a series of six or more numbers that completely miss one of the dozens.

How many spins in a row did it miss that dozen? Seven? Eight? Ten? More?

Can you see how easy it is to lose?
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ ————————————————————————————————————— Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
  • Jump to: