Mathematicians.......please take a look at this!

[TwoCatSam]

The question is not whether this is the way to bet or not but rather...is my math right?

Would placing the bets just as I have indicated give the house a 50% edge?

Anyone?

From Lohnro's Thread:



OK, here goes.

Trying to reduce the wheel to 8 numbers, 7 European.

Place $10 on the first dozen and $15 on the large numbers.  Seven numbers are uncovered.  13 thru 18 and 0.  Now place $1 on any five of those numbers.

Iteration.......

1st dozen hits, you break even.
Large number hits, you break even.
One of your $1 bets hits, you win $6.  (36 minus a total of 30 bet)
One of your uncovered numbers hits, you lose $30.

We will not concern ourselves with the break-evens.

Winners = 5/37 = 13.5% win rate at $6
Losers  =  2/37 = 5.4% loss rate at $30

Over one thousand spins.....

Loses

1,000 X 5.4% = 54 losses X $30 = $1,620

Wins

1,000 X 13.5% = 135 wins X $6 = $810

Seems to me you are flat-betting and loosing 50% of your bankroll.

I could use some help!  Or confirmation!

Sam
Taking of the math professor's hat!

[Herb]

Assuming that the wheel still has 37 or 38 pockets on it, then your edge is still -2.7% and -5.26%, regardless of how many numbers you chose to bet.  Even if you bet all of the numbers, the house edge is the same.

In short, you were just unlucky when you played.

-Herb

[ChickenDinner]

"Assuming that the wheel still has 37 or 38 pockets on it, then your edge is still -2.7% and -5.26%, regardless of how many numbers you chose to bet."

Hmmm...I don't want to question a math guy. However, I don't understand how this can be right.

This system always excludes 2 numbers so it only gives you a 94.6% coverage.

Is it not correct that if a system excludes 1 number, e.g., the zero, or any amount of numbers for that matter, the house edge is increased because your chances of winning are reduced?

For example, excluding 1 number, 36/37, gives a system only 97.3% coverage on odds of 35-1 so the house gains another 2.7% on top of the 2.7% that already exists.  Exclude 2 number, 35/37, gives you 94.6% coverage & the house now has a 8.1% edge, exclude 3 gives you 91.8% coveage and the house now has a 10.9% edge, etc.

Am I right......?

Can anyone back me up here........?

Winkel......?

Sam's mate Ken......?



Cheers
CD

[TwoCatSam]

I agree that on a 37 numbers wheel, when two are uncovered, the house rake is 2.7%   Normally........

Here or over there, Monte Carlo posted his system and myself and someone else discovered he actually raised the house rake from 2.7%.  Another trusted member once stated he knew of a way to reduce the house rake to 1%.

I suppose the question is this:  Is the house edge always 2.7% on European?  Is there any way it can change?

Will someone look at my examples and show me the error in my math?

Must I go to Gambler's Glen and post this as a system so someone will harshly point out to me what I have written?

Sam

[Herb]

It doesn't matter whether you bet every single number or just one number, the house edge remains unchanged.

If you don't like the answer, then ask a different question.

[TwoCatSam]

Another example of changing the house edge.

You bet a unit on red, a unit on black and a unit on zero.  Here are the iterations:

Red = lose a unit to zero
Black = lose a unit to zero.
Zero = win a net of 33u.  (35u won on the zero minus the two bed on red and black.)

Here's how is shakes out.....

2.7% of the time you will hit zero.  That is 27 times per thousand spins in a perfect world.

Wins....

1,000 X 2.7% = 27 hits at 33u = 891. 

Losses...

973 losses at 1u each for a loss of 973u.

So you won 891 but lost 973 for a net loss of 82 units.  So what was the loss percentage?

82/1000 = 8.2%  1,000 X 8.2%  = 82. 

So we can conclude that betting a unit on red, black and zero simultaneously results in a net loss (edge for the house) of 8.2%.

Why is this important?

The universal truth has been the house edge in European roulette is 2.7% and cannot be changed.  winkel says it can. 

Using deductive reasoning and logic we may reasonably ask this question:

If one universal truth is proven wrong, might others also be wrong?

Sam

[Herb]

Another example of changing the house edge.

You bet a unit on red, a unit on black and a unit on zero.  Here are the iterations:

Red = lose a unit to zero
Black = lose a unit to zero.
Zero = win a net of 33u.  (35u won on the zero minus the two bed on red and black.)

Nope, the house edge still hasn't changed.  It's still -2.7% and -5.26% on the double zero wheel.

The universal truth has been the house edge in European roulette is 2.7% and cannot be changed.  winkel says it can. 

With regards to the random game of roulette, Winkel's wrong.


-Herb

[winkel]

One big mistake in this discussion

You are mixing up the laws of statistics and the house edge

if the casino would offer a fair game it would pay on numbers 36 for 1 and not 35 for 1 at european roulette

so that your return is one less than it shouldt. it is low -2.7%
This is just the same figure as the normal probability of 2.7% to hit a certain number.

So if you cover all numbers you will always loose the house edge, that is simple.
All chances are reduced by this house edge.

but it is another thing to create a strategy that hits more often than it loses.

How often does it have to win
in 37 spins it has to hit one more time, than it should.

winkel

[TwoCatSam]

You are mixing up the laws of statistics and the house edge

winkel

I probably am.  Thanks for your answer.

Sam

[TwoCatSam]

Edit:  If anyone read this post before I got back to it, forget about it.

I found the answer.

Sam

[berlinerbruce]

hi lads, hey SAM can you look at my last post to you under brainstorming, you will See that the house edge is scratched and dented at times



all the best berlinerbruce

[TwoCatSam]

Bruce

Is a "scratch and dent" sale in order?  Can we pick up some bargains.

winkel

That single sentence set me on the right course and I found the solution to my puzzle!  As I said, I counts on me toes!!

Sam

[ChickenDinner]

I understand that "true odds" would be 36-1, but if 2.7% of the board (the zero) will never be included in a bet selection, I cannot understand why this is not another 2.7% long term advantage (edge) to the casino.

Don't statistics and the house edge share the same bed in the long term?

CD (begining to regret I ever mentioned this subject....)

[insidebet]

Chicken,

Single 0.  house edge is 1/36=2,70%

Double0.  house edge is 2/38=5,26%

I don't think that your bet selection changes any of that.  There are still 37 or 38 slots for the ball to fall in, whether you bet 0 or not.

Insider

[Kon-Fu-Sed]

Hello All,

Sorry for my late reply but I wasn't here yesterday...

Sam's questions are exactly in line with the "Misdirected Intuition - a classic" chapter in my "Roulette Probability Made Easier" nolinks://vlsroulette.com/reference-area/roulette-probability-made-easier-t2193/.

You mix bets and losses in an in-correct way and it's easy to show:

37 times you bet 3u = 111u bet in total.
Once you hit Zero and get 36 units back.
36 times you hit Black or Red and get 2u back = 72u.

36u + 72u = 108u and you're 3u down.

-3 / 111 = -2.7%


And the same for the question in the top post:

10u on Doz1
15u on High
5 x 1u on Singles

Total bet = 30u x 37 = 1110u

Doz1 hits = 30u back 12 times = 360u
High hits = 30u back 18 times = 540u
A Single hits = 36u back 5 times = 180u

Total back = 1080u = a loss of 30u

-30 / 1110 = -2.7%


Remember this: You PAY and you GET BACK.
Never mix "pay" and "lose" because your loss is already in the "pay".


Regards,
KFS

[Kon-Fu-Sed]

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