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 02-28-2017, 07:00 PM #1 asymbacguy centurion   Join Date: Jun 2015 Posts: 191 House edge at nozero roulette Hi all. An online casino offers a nozero european roulette (36 numbers) where at the end of any single session the player winnings are decurted by 10%. What's the house edge on such nozero wheel? Obviously every bet is payed in the same way than a regular single zero wheel. Thanks in advance.
 02-28-2017, 10:47 PM #2 RustyBrooks Carpal \'Tunnel     Join Date: Feb 2006 Location: Austin, TX Posts: 23,130 Re: House edge at nozero roulette Normally it doesn't matter what bets you make, because all bets on a normal wheel have the same EV. The bets in this case also have the same EV but because the 10% is deducted from the session winnings, instead of each bet's winnings, it comes out different. Imagine a session composed of 1 bet: Consider if you made a bet on even, which will hit half the time and pays out 2:1 1/2 the time you lose, and lose 1 unit 1/2 the time you win, and win 1 unit, but they deduct 10% so you win .9 EV = (-1+.9)/2 = -0.05 If you make a bet on a single number, then 1/36 times you win 36 but get 10% deducted, netting 32.4 35/36 times you lose 1 EV = (-35+32.4)/36 = 0.07 A strategy that gave you a small chance of a big win would have worse outcomes over time than one that had a larger change of a smaller win. I'm not sure I know how to mathematically arrive at an optimal strategy nor how to figure out it's EV. I could probably simulate it, if I have some time tomorrow.
 03-01-2017, 01:23 AM #3 commercejim banned   Join Date: Oct 2016 Posts: 2 Re: House edge at nozero roulette Math above not entirely correct. Player edge on a single number, single spin session is -9.72% For a given amount wagered per session, greater volatility, i.e. fewer trials and/or longer odds, leads to equal or greater expected loss.
 03-01-2017, 03:05 PM #4 RustyBrooks Carpal \'Tunnel     Join Date: Feb 2006 Location: Austin, TX Posts: 23,130 Re: House edge at nozero roulette Oh, I see, it pays 35:1, not 36:1. Also I say 0.07 when I meant -0.07, but yeah, -0.0972 is correct.
03-01-2017, 04:18 PM   #5
whosnext
Pooh-Bah

Join Date: Mar 2009
Location: California
Posts: 3,908
Re: House edge at nozero roulette

It is easy to see, either conceptually or algebraically, that for betting on a single number and for sessions less than or equal to 36 spins, the house edge is H*(35/36)^S where H is the "haircut" percentage (H was 10% in OP) and S is the number of spins.

At every multiple of 36 for S, there is a "kink" in the house edge formula. Here are a few examples of the house edge in this case:

SpinsHouse_Edge
19.72%
363.63%
722.67%
1082.21%
1441.93%
3601.23%
10000.75%
36000.39%

 03-02-2017, 10:57 AM #6 asymbacguy centurion   Join Date: Jun 2015 Posts: 191 Re: House edge at nozero roulette Thanks! What about house edge on even chances?
 03-02-2017, 11:47 AM #7 Didace Carpal \'Tunnel     Join Date: Nov 2009 Posts: 11,979 Re: House edge at nozero roulette Don't you need to take into account asymmetrical streaks?
03-02-2017, 02:39 PM   #8
whosnext
Pooh-Bah

Join Date: Mar 2009
Location: California
Posts: 3,908
Re: House edge at nozero roulette

For even-money bets, formulas can also be found but, as above, they have frequent "kinks" in them. So it is probably best to just report what the house edge is in various number of even-money spins sessions.

SpinsHouse_Edge
15.00%
22.50%
32.50%
41.88%
51.88%
61.56%
71.56%
81.37%
91.37%
101.23%
111.23%
121.13%
131.13%
141.05%
151.05%
160.98%
170.98%
180.93%
190.93%
200.88%
500.56%
1000.40%
2500.25%
5000.18%
10000.13%

 03-03-2017, 05:17 AM #9 asymbacguy centurion   Join Date: Jun 2015 Posts: 191 Re: House edge at nozero roulette Thanks a lot!
 03-03-2017, 07:39 AM #10 jeccross veteran   Join Date: May 2005 Posts: 2,919 Re: House edge at nozero roulette Doesn't the house edge change according to whether you are up or down though?
03-03-2017, 11:18 AM   #11
RustyBrooks
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Join Date: Feb 2006
Location: Austin, TX
Posts: 23,130
Re: House edge at nozero roulette

Quote:
 Originally Posted by jeccross Doesn't the house edge change according to whether you are up or down though?
We're looking at the game from the position of not having started a session, so no not really.

If you're allowed to stop a session whenever you want and start a new one then there will be some strategies to limit house edge.

 03-03-2017, 12:06 PM #12 nickthegeek centurion   Join Date: Sep 2011 Posts: 189 Re: House edge at nozero roulette It's very simple to program this problem. All you need is the cumulative binomial distribution function. In R, this function is called pbinom. If N is the number of spins in your session, p the probability of hitting and H the house cut, you can define the following R function: Code: ```houseEdge<-function(N,p,H=0.1) { odds<-(1-p)/p q<-odds+1 breakEvenPoint<-trunc(N*p) -1 + H*pbinom(breakEvenPoint,N,p,lower.tail=FALSE) + q*p - H*q*p*pbinom(breakEvenPoint-1,N-1,p,lower.tail=FALSE) }``` It's fully vectorized. If you try: Code: `houseEdge(1:20,0.5,0.1)` you reproduce whosenext's numbers (post #8). You can also see how the edge changes depending on p. For instance: Code: ```> houseEdge(100,c(1,2,4,9,18)/36,0.1) #[1] -0.023672795 -0.016589894 -0.011253887 -0.006884977 -0.003979462``` tells you the edge on a 100 hand session depending on how many numbers you bet on. It's interesting to notice that the edge decreases when p approaches at 0.5.
 03-04-2017, 06:22 AM #13 asymbacguy centurion   Join Date: Jun 2015 Posts: 191 Re: House edge at nozero roulette That's interesting. So the theorical best approach should be to bet even chances (or a cluster of 18 numbers whatever taken) for "long" time? What about a p>50%, as betting two dozens or 27 numbers over 36, for example?
03-04-2017, 07:17 AM   #14
nickthegeek
centurion

Join Date: Sep 2011
Posts: 189
Re: House edge at nozero roulette

Quote:
 Originally Posted by asymbacguy That's interesting. So the theorical best approach should be to bet even chances (or a cluster of 18 numbers whatever taken) for "long" time? What about a p>50%, as betting two dozens or 27 numbers over 36, for example?
House edge still decreases for p approaching to 1. This can be easily understood. Consider the trivial case in which you bet on all the 36 numbers (p=1). You just get your initial bet back. You are not cut, since you are never winning (nor losing). Basically you are not even playing.

On the other hand, when p is small you are rarely winning, but when you do, you likely win huge and are cut huge, increasing the house edge.

03-04-2017, 10:22 AM   #15
Didace
Carpal \'Tunnel

Join Date: Nov 2009
Posts: 11,979
Re: House edge at nozero roulette

Quote:
 Originally Posted by asymbacguy So the theorical best approach should be to bet even chances (or a cluster of 18 numbers whatever taken) for "long" time?
Depends what you mean by "best". What is your goal?

03-05-2017, 05:04 PM   #16
asymbacguy
centurion

Join Date: Jun 2015
Posts: 191
Re: House edge at nozero roulette

Quote:
 Originally Posted by nickthegeek House edge still decreases for p approaching to 1. This can be easily understood. Consider the trivial case in which you bet on all the 36 numbers (p=1). You just get your initial bet back. You are not cut, since you are never winning (nor losing). Basically you are not even playing. On the other hand, when p is small you are rarely winning, but when you do, you likely win huge and are cut huge, increasing the house edge.
Yep, Thanks nick!

03-05-2017, 06:16 PM   #17
asymbacguy
centurion

Join Date: Jun 2015
Posts: 191
Re: House edge at nozero roulette

Quote:
 Originally Posted by Didace Depends what you mean by "best". What is your goal?
Good question.

I'm not really interested in roulette but this nozero roulette might improve some strategies based on progressions.
I mean that the probability to be ahead (and then restarting the process) is improved as any bet will have a fair expectancy.
As nick stated, a low p tends to increase the house edge along with a high volatility tipical of a low p.
On the other hand, frequent and more "controllable" winnings tend to produce more small "cuts", so enlarging in frequency the money deducted from the wins.

So what could be a "best" approach, meaning for it a less disadvantaged approach, considering both the probability to win (and its volatility) for any bet and the subsequent global cut effect?

I'd think that a p slightly over of 50% might have the best results in terms of volatility and in terms of cut effect.
Roulette is a perfect symmetrical and independent game, especially on this type of wheel, but certain opposite situations studied in form of multiple distributions tend to produce low volatility values.
This fact should increase the likelihood to be ahead several times by the use of multilayered progressions.

An example might be to get two fictional fake "even" chances as dinamically choosing 19-20 numbers on one side vs the remaining 17-16 numbers on the other one.
We could be wrong about our dynamic selection, still our bets won't be EV- before we'll decide to quit a winning session.

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