Quote:MattUK...snip...

This is where the problem is - some people doesn't want to know how much they pay for playing. Yet I am sure that a good chunk of gamblers would think otherwise.

I really disagree with your basic premise. People will not pay a fee of any kind to enter a casino for the privilege of losing their money, UNLESS there is no other option. Likewise, people will NOT be happy about paying a percentage simply to place a bet.

Look at all the bitching about OK casinos charging .50/hand for the state. Various casinos have ways around this; the place I played, if you had a players card, the house paid the fee for you. The dealer dropped slugs into the money box every couple of hands to track it. If you didn't use a card, you had to pay the fee every hand as you played.

.50 per person per hand. The casino was so desperate to please their patrons that they ate this fee. If it weren't a major negative to overcome in drawing customers, you think they'd pay it themselves? Or lose hands per hour screwing around with it every hand of every game?

I think your 1% would engender similar resentment. And even if the casino started out with "better" games than those around them, I don't see that lasting any length of time in any competitive casino neighborhood . With what they lose in administering it, they slow the game. It will be the first perq given to hi-limit players, not to charge it.

And patrons simply can't be bothered with math. Either they like the action or the potential odds pay, but most play really, really bad sidebets, slots, and games (including me - I like the action on a couple sidebets despite largish HE). The constant fee = worse losses, less than a full win, greed in perception.

Quote:beachbumbabsI really disagree with your basic premise. People will not pay a fee of any kind to enter a casino for the privilege of losing their money, UNLESS there is no other option. Likewise, people will NOT be happy about paying a percentage simply to place a bet. (...)

And patrons simply can't be bothered with math. Either they like the action or the potential odds pay, but most play really, really bad sidebets, slots, and games (including me - I like the action on a couple sidebets despite largish HE). The constant fee = worse losses, less than a full win, greed in perception.

Just for clarification, this idea is feasible only online. It doesn't make sense in brick-and-mortar casinos.

Your post once again identifies what is the biggest obstacle. Not the odds - they may be even FAR BETTER than in European Roulette. Plus, there are extra benefits not even available in the standard version, like ease of offering happy hours with even better odds. It's the self-deception that gamblers don't pay for playing. This game shatters it, reveals the truth with brutal honesty. "Yes fella, you do pay for playing" - it says and it's enough to despise it. Even through everyone do it with every single casino game! I think it's probably worth a thorough psychological experiment.

The other interesting thing is that you are in dissagreement with your colleague Mission146. She thinks that gambling is all about numbers and finding best odds. This game gives it with ease. For example, 1c charged from every 1$ equals to 98% RTP equivalent which is the level of best slots. But for you it's not all about best odds but about thrill and fun. Perhaps you can have a discussion about it!

Last thing - a thought experiment. You have a choice between standard European Roulette (2.70% HE) and "my" roulette with 1c charged from every 1$ (2% HE). My roulette beats it by a massive 0.70% or 26/37 to be precise. Which one would you play beachbumbabs?

Quote:KevinAAYou can make a happy hour at a brick-and-mortar casino quite easily by increasing the comp rate during that time.

Kevin, we talk about my idea, not happy hours in casinos.

Someone hammered on this math assumption earlier but I don't think the original poster understood: removing the 0 from a roulette wheel doesn't really change the variance very much. The thing that makes roulette variance super high is the 1/36 odds to 36x payout, and the way people mitigate that is by betting on multiple numbers at once. You can't really reduce roulette variance to better than double-or-nothing using standard roulette bets, though. (without making multiple bets at a time, of course)

I think the bad psych assumption being made is that gamblers want to play low-variance games. I don't know of any gambling games that offer lower variance than double or nothing -- maybe slot machines with no bonus round and a very high number of lines? I think the popularity of roulette, craps, and slots-with-bonus-rounds indicates that high variance is fun because for those games you can easily fluke into getting rich for a short time, but you can't fluke into getting poor unless you chase your losses by doubling your bets or something.

(Obligatory note: Zekka is a pretty inexperienced gambler and there are no doubt lots of games he doesn't know about. Don't trust him!)

Quote:ZekkaI don't know of any gambling games that offer lower variance than double or nothing -- maybe slot machines with no bonus round and a very high number of lines?

I think I encountered that machine at The D last year; guaranteed that every spin was a winner. I won something like 10c on a 75c spin until I ran 50$ worth of FP down. Literally every spin paid the exact same amount (regardless of what the actual numbers were, which I've forgotten). Please, I'd like SOME more variance than that.

I'm okay with variance, and I'm okay with edge. I'd be okay with an entry fee if I could get it back somehow at a remotely reasonable rate (20$ to get in and you get a 'free' lunch or something). I don't like commission (I don't consider the vig on the buy commission, as that's there for a different reason), which makes me frown at standard PGP and Bac. And, by extension, that means I'd go to some lengths to avoid this, even if it was at marginally better odds.

Quote:

Happy to help, Tom. You are wrong everywhere. The house advantage is always unpredictable. Out of, say, 100 spins you can have 0,1,2,3,4 or even 5 zeros, losing the bet. Under my idea it would always be a fixed % of the wager. Therefore "theoretical house edge" is not "actual house edge", so to speak. With my idea it's one and the same.

This is not different from getting 0, 1, 2, 3, 4, or even 5 reds when you bet on black, except that this outcome currently has 19/37 odds but after your change it has 18/36 odds. TomG successfully analyzed this about a post after you floated this. It sounded like, from posts like this, your idea was that eliminating the house cut would reduce the odds of getting this kind of bad outcome and I think you frequently used words like "stable" and "predictable."

You're right that the variance depends on the bet made. Betting on one number gets you one payout of 3600% and 36 payouts of 0%, which gets me stddev of ~583%. ~Betting on half the wheel gets you double-or-nothing, which gets you eighteen payouts of 200% and nineteen payouts of 0%, which gets me stddev of ~99.96% (unless I'm all wet.) I guess in relative terms for gambling games, this isn't high, but compared to everything people bet on outside of casinos, I think it's high. Like I said, I guess you can make multiple bets to reduce it even further.

I'm glad you're being explicit that variance isn't important because I read that as implicitly a huge goal of yours and rereading your posts I probably misunderstood them. I'm still very confused by this post:

Quote:

After 100 spins, 2$ per bet, you don't know what will be the true cost of playing the European Roulette. Most likely 6$, maybe 4$ with 2$ or 8$ still sensible. In my game it will be 1$ exactly. And it's not even about the odds.

I still don't see how this is more true for your game than with roulette. The variance of your game is about the same, so I feel like it should be about equally hard to guess how much money you'll end up with.

I'm not sure the commissions version of the game is easier to understand since I'm still likely to end up with sums of money that are really, really different from what I initially put in. I think "true cost" means difference in expected value from what you put in, but I think that for a short series of observations it's hard to sense that. Any short series of roulette outcomes from betting on black on an unaltered wheel would have a reasonably high probability of having happened on a 50-50 wheel, IMHO, and vice versa -- so the game 'feels' the same and you're still about equally likely to walk away massively up as massively down. (Compare two wheels, one that awards exactly 100% and one that awards exactly 95% every round -- the second wheel's outcomes are highly distinguishable from the first wheel's because variance is so low.)

I think most measures of "predictability" either fall down on psychology, stats or both, so maybe there's a psychology pov on predictability that I'm not getting here? I suppose seeing a wheel that's obviously fair and paying in a house commission gives you an exact quantification of the house edge, even though it probably won't be relevant to you if you don't play for very long. Fun stats project: with a fixed bet size, figure out how long it would take on average for a series of red/black outcomes on a MattUK wheel to have less than 5% probability of having happened on a normal roulette wheel. Of course, winnings are going to tend to go to the same place, but I'm using red/black as a proxy for "massively going up each round" vs "massively going down each round" which is probably a really big part of whether Matt's wheel feels different to play.

EDIT: I stated the *interesting stats problem* in a really dumb way previously (the answer to my old formulation was "never") I've changed it.

(note: zekka is not a real statistician and may have garbled things badly. hopefully he didn't)

--- Revisited later ---

First of all, my formulation of variance is a little particular. If you constrain a bet to a binomial distribution (failure if you lose money, success if you win money) then a 50/50 bet is about as high-variance as you can get -- even though AFAICT a 50/50 bet is lower variance than a 1-out-of-36 bet in roulette because the payout is also reduced. This post works with the binomial distribution version because I think that formulation corresponds to how it will feel to play your game. (especially since the difference between how much you get when you win and how much you get when you lose is always large, at least as large as your bet.) It also makes the math easy.

For two bet types, I ran 10,000 tests of this experiment:

- I'm a gambler and I place bets on your wheel, writing them down.

- I run a Fisher Exact Probability test every time to see if my outcomes could feasibly have come from a normal roulette wheel.

- If the p-value that these outcomes came from a normal roulette wheel is ever < 0.05, I stop.

- stop after 10,000 games no matter what the distribution was:

Here were my results:

red-black bets: on average 318 plays, stddev = 1199 plays

single number bets: on average 788 plays, stddev = 3250 plays

(Note: I could have gotten the math wrong. I really hope a real statistician will run this experiment to confirm the results)

Here's my code:

#include <math.h>

#include <stdlib.h>

#include <stdio.h>

#define N_TESTS 10000

// red black

// #define NUMERATOR_MATT 18

// #define NUMERATOR_ROULETTE 18

//

// #define DENOMINATOR_MATT 36

// #define DENOMINATOR_ROULETTE 37

//

// single number

#define NUMERATOR_MATT 1

#define NUMERATOR_ROULETTE 1

#define DENOMINATOR_MATT 36

#define DENOMINATOR_ROULETTE 37

#define P_ROULETTE (((double) NUMERATOR_ROULETTE)/DENOMINATOR_ROULETTE)

double mean(int n, double arr[]) {

// TODO

double sum = 0;

for (int i = 0; i < n; i++) {

sum += arr;

}

return sum/n;

}

double stddev(int n, double arr[]) {

// TODO

double m = mean(n, arr);

double sumdiff = 0.0;

for (int i = 0; i < n; i++) {

double k = arr - m;

sumdiff += k * k;

}

return sqrt(sumdiff / n);

}

double z(int obs_n, double obs_p, double pop_p) {

double p1 = obs_p;

double p2 = pop_p;

double p = pop_p;

return (p1 - p2)/((p * (1 - p) * sqrt(1/((double)obs_n))));

}

double significant(double z) {

// one sided, 95%

return z > 1.645;

}

int test_once() {

int tries = 0;

int successes = 0;

while (tries < 10000) {

if (tries > 1) {

int n = tries;

double p = successes/((double) tries);

if (significant(z(n, p, P_ROULETTE))) { return tries; }

}

tries++;

if (rand() % DENOMINATOR_MATT < NUMERATOR_MATT) {

successes++;

}

}

}

int main() {

double test_results[N_TESTS];

for (int i = 0; i < N_TESTS; i++) {

test_results = test_once();

}

printf("%d tests; mean steps: %f; stddev steps: %f\n", N_TESTS, mean(N_TESTS, test_results), stddev(N_TESTS, test_results));

}

So I think the odds of walking away with more or less will feel much the same for a long time. I think most gamblers would feel like they were winning and losing about as much as on a normal wheel, except they would resent having to pay commissions.

EDIT: one paragraph SUCKED and included a major typo that says the opposite of the truth, fixed both

Quote:ZekkaThis is not different from getting 0, 1, 2, 3, 4, or even 5 reds when you bet on black (...).

I still don't see how this is more true for your game than with roulette.

I don't care about red and black because they are part of the game and they will not change. The only thing that changes is green 0 - it's being replaced by a FIXED FEE BEFORE SPIN. And that is why my idea revels the true cost of gambling. It replaces unlikely (1/37 = 2.7%) loss of entire bet by a small fixed fee (therefore: "guaranteed loss") of 1% (which correspondents to 2% house edge because it's charged before spin). That way the player always knows what he or she has paid for playing. What is truly revolutionary is that the cost of playing is radically separated from the (edgeless) game. I think it's the current system, which blends them together, which is wrong and unjust.