Joined: Apr 21, 2015
  • Threads: 19
  • Posts: 53
November 25th, 2015 at 4:44:41 AM permalink
When constructing a betting system we need to base it around the variances and statistics we know for the bets we are making.

What if we were to be betting on 35 numbers in european roulette and removing bet from each number after it hits.

Most people would either flat bet or increase their bets on numbers that have not yet hit.

I know that on average around 23 different numbers will hit over 37 spins, with the rest being repeats, and around 18 different numbers will hit over 24 spins. (From memory..)

I know that i will hit more numbers at the start of betting while i have 35 bets on the table than i will at the 24th spin when i only have around 18 bets/ numbers remaining.

Knowing that I am more likely to hit my bets at the beginning of cycle while i have 35 bets on the table, it would not make sense to have my low bets on now, only to increase later when i have less chance of hitting my numbers.

In this scenario wouldn't i be better to bet higher at the beginning and gradually drop my bets down as my chance of hitting my remaining numbers declines?

For example:
Bet $5 on 35 numbers, (removing bet as the hit) for 12 spins
Bet $4 on remaining numbers after 12th spin, (removing bets as they hit)for a remaing 12 spins
The 19 (on average) numbers that dont hit after the 24th spin will have cost me $108 each.

The numbers that do hit will have won me between $175- $115 if hit in first 12 spins
And $80-$32 if hit in 12-24th bet.

From memory the average win if you add them all up is around $102, but because you are more likely to win in the first 12 bets than the second 12 i believe the average win would be higher.

What are your thoughts on this?
Joined: Jul 22, 2014
  • Threads: 28
  • Posts: 5552
November 30th, 2015 at 9:52:28 AM permalink
First, without a serious statistical sampling size (tens of thousands of spins) stating "on average" 23 different numbers will hit over 37 spins is a bit more voodoo than you think. Each spin is an independent trial. It goes back to the old "If black hits 19 times in a row, why wouldn't you bet red the next 19 spins??" The ball/wheel don't care at all if black was the last 19 in a row, the next spin has a "50/50" (yes I quoted it because it's not really a 50/50 with the zeros) of hitting.

So let's actually do some math with your system:

On your first spin, you're betting a total of $175 ($5 on 35 numbers) with a 35/37 percent chance of 'winning' $5 (lose $170 and 'win' 175) and a 2/37 chance of losing $175.

EV(1st spin) = P(W) + P(L) = (35/37)(5) + (2/37)(-175) = 4.7297 - 9.4595 = -4.7298

In optimal scenario, say you hit one of your numbers making $5 profit and thus removing one of your bets from the next spin, then what is your EV of the next spin? Well, your total bet is now $170 ($5 on 34 numbers) with a 34/37 percent chance of winning $10 (lose the other $165 to 'win' $175) and a 3/37 chance of losing $170.

EV(2nd spin) = P(W) + P(L) = (34/37)(10) + (3/37)(-170) = 9.1892 - 13.7838 = -4.5946

...and so on:

EV(3rd spin) = P(W) + P(L) = (33/37)(15) + (4/37)(-165) = 13.3784 - 17.8378 = -4.4594


I hope you can see a pattern here. You're EV is negative regardless of your best case scenario. Not only that, as expected, the less you bet the less your negative EV becomes. You can apply the same logic/math to the rest of your system to see that at no point in time by placing negative expectation bets (no matter the combination of them) will you ever have a positive expectation.
Playing it correctly means you've already won.
Joined: Mar 15, 2013
  • Threads: 2
  • Posts: 219
November 30th, 2015 at 12:35:37 PM permalink
Quote: Romes

First, without a serious statistical sampling size (tens of thousands of spins) stating "on average" 23 different numbers will hit over 37 spins is a bit more voodoo than you think.

IIRC this average number of unique numbers to total numbers ratio is indeed the case and it is a direct consequence from the nature of the Normal distribution.

  • Jump to: