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!
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
"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
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
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.
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
QuoteAnother 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.
QuoteThe 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
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
You are mixing up the laws of statistics and the house edge
winkel
I probably am. Thanks for your answer.
Sam
Edit: If anyone read this post before I got back to it, forget about it.
I found the answer.
Sam
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
;D ;D ;D
all the best berlinerbruce
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
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....)
:P
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
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/ (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
CD...
I appreciate your raising this issue. It's an answer I needed. AND...I found the book by KFS which I plan to print. No, I had not seen it.
KFS
Thanks for your explanation. You are one of the few who will take a person's question and truly answer it. Are you a teacher in real life? I did figure out the paradox from what winkel said. (I was up most of the night with this one.) But your post gave me a whole new way to look at things.
And thanks for that e-book! Man that is a work!
Thanks to all who answered!
Chickendinner, for this I would buy you a...um...er.....chicken dinner!
Samster
KFS
In your example you seem to be saying this:
Even though there is a push when the first dozen OR the large numbers hit, there is still an edge to be paid to the house on all those bets.
I am thinking this is the case and when the zero or the other uncovered number hits, you pay the edge in a lump sum. Then I get to thinking, isn't that true with all bets? You win for a while and then you lose. When you lose you pay the edge.
This raises a totally new and different--dare I say exciting--issue for me, but I must dwell on in for days lest I make myself look dim.(er)
Pass on this whole thing if you're done with it and I will, too!
Sam
Thanks for the dinner Sam, but unfortunately it seems, to me at least (hang on, another post - you as well), that my original question remains in dispute.
Does excluding the zero in a betting system give the house a futher 2.7% advantage (or edge)?
Let's forget outside bets and imagine a hypothetical roulette table that pays odds 36-1 on the inside numbers. Great thinks the hypothetical player - no house edge! Well, not quite. The only rule is you can't bet zero. So therefore the house has a 2.7% advantage over the long-term: a 2.7% edge. Agreed?
So on a real European Roulette table that pays untrue odds of 35/1 (giving them a 2.7% edge), excluding the zero (in a bet selction) will always give the casino a further 2.7% edge.
Yes there are will always be 37 or 38 slots in which the ball can slot, we can't change that. But by always excluding 1 number (eg, the zero) in a bet selection, there are only 36 or 37 slots in which you can win.
If I'm wrong, I am more than happy to admit so. And for those of you who think I am wrong, could you please explain why in very simple terms. Maybe I should have tried a bit harder at Maths when I was at school...
CD
CD
Does excluding the zero in a betting system give the house a futher 2.7% advantage (or edge)? I'm sure KFS can answer that.
Let's forget outside bets and imagine a hypothetical roulette table that pays odds 36-1 on the inside numbers. Great thinks the hypothetical player - no house edge! Well, not quite. The only rule is you can't bet zero. So therefore the house has a 2.7% advantage over the long-term: a 2.7% edge. Agreed? Yes
Both winkel and Monte Carlo have said it is possible to increase the house edge.
Bloomone and I studied his idea of reducing the "volatility" and we both agreed with him that it is possible to increase the edge. Now, after reading KFS, I simply don't know.
About the time you "know" something, you get shot down!! Ain't it fun, though?
Sam
Hi,
CD, you are confusing "will not allow you to bet Zero" and "will not pay you for a hit on Zero"
The latter equals to keeping an extra bet - stealing.
But if you are not allowed to bet the Zero, there is no difference at all...
Suppose you bet 1u on every number 1 - 36 for 37 times = 37 x 36u = 1332u
You will hit the winning number 36 times winning 36u = 36 x 36 = 1296.
Giving you a loss of 36 u and -36 / 1332 = -2.7%
Now, if the casino refuses to pay for a bet on Zero:
For 37 times you bet 37u (1u on every number) = 1369u
You will win 36 times 36u = 1296u...
Giving you a loss of 73u and -73 / 1369 = 5.33%
Regards,
KFS
Sam,
Regarding your "pay-the-edge" question...
In my mind we all pay all of the time... You have to bet (pay) to win.
I think about it like this:
There are 37 persons at the table. They all bet one $10-chip each on one number each so all of the numbers are bet.
Zero hits. (or whichever)
The dealer puts the marker at the lay-out and collects all the money.
Then he counts 36 chips and hand them to the winner.
The dealer keeps 1 chip.
Who's chip?
No; not a trick question... ;)
I like to think that we all payed 1/37 of that chip...
Have a nice week-end!
KFS
KFS
I wrote: This raises a totally new and different--dare I say exciting--issue for me, but I must dwell on in for days lest I make myself look dim.(er)
You wrote: I like to think that we all payed 1/37 of that chip...
You read my mind!
Thanks for your tutorial and...
Good day!
Sam
Thanks Kon-Fu-Sed - I understand what you are saying.
Perhaps my example was wrong, though. I guess what it comes down to is that 99% of systems use statistics as a method for bet selection. So if the player ignores one number, he or she ignores a possible outcome and is therefore using only 97.3% of the available numbers as a guide to select their bets. I'm undoubtably mixing statistics and house edge here. But I think the bottom line is that a system that only uses 97.3% of the numbers has a 2.7% disadvantage over a system that uses 100% of the numbers. The house edge itself will not change; it will always by 2.7% or 5.26%. But in order to make optimal use of sequential and binomial probability theory (and standard deviation), 100% of the numbers should be tracked and played. I know that the absolute probability for each spin will always be 1/37 or 1/38, but if mathematical estimates are all we have to go on when placing a bet, all the numbers must be included in whatever bet selection method is used.
Thanks for your help anyway mate, and hopefully I am not talking too much bollocks in this post ;)
Cheers
CD
Good morning CD, an All,
Yes: The way I see it, you are absolutely correct when you describe it like that.
/KFS
Thanks KFS. I'm glad to hear that I'm on the right tracks. Now I just need to find that optimal system...!
Cheers
CD
Winner Winner Chicken Dinner