New Game error? - Need Your help

Without getting too involved in resolving a potential error on a brand new game that was shown at G2E, a VERY mathematically inclined forum member pulled me aside there and said: "There's a math error on this new Roulette game" (pointing to the table) - "right here on the layout - look at the layout!" Hmm.....

I took a look at the layout on the new game (Double Ball Roullete), and said "I know what you're refering to here...."

Double ball Roullete uses two Roullete balls per single spin result, and basically, the payouts are halved but with twice the chance to hit any bet.

However, for the Red-Black type bets, there are THREE results on the layout, as the two balls can land on any number/color each. The three color bet options are:
1. Pure Red.
2. Mixed Red-Black, and;
3. Pure Black
ALL of which pay 3:1. Mip asked, "Dan, where's the problem?" - and I said on the Mixed Red-Black bet paying the same 3:1 as the Pure Red or Black result. To me, it looked like hopping an Easyway, but paying 30:1 just like a hopping Hardways, - instead of 15:1.

I just got a call from an associate on this a just moment ago, and he brought this subject up, - as he had a question on this for research. He said the game was signed by the inventor with full math done, including math approval from a math laboratory - indicating that Mixed Red-Black on this game is supposed to also pay 3:1!

Also know that it is possible for the two balls to land in the same number slot, in a tight fit.

I said "why don't we bounce it by the board to get to the bottom of this one!" He said, "sure, let's get into this, for a resolution by the pros." We need to know...

Here it is, fellas.....

Quote: Paigowdan

The three color bet options are:
1. Pure Red.
2. Mixed Red-Black, and;
3. Pure Black
ALL of which pay 3:1..

So if you always bet $10 on red and $10 on
mixed, you would only lose when it was
black/black. The rest of the time you would
win. Right? Red/red win $20. Mixed win $20.

and if a ZERO hits all three lose correct?

Quote: coilman

and if a ZERO hits all three loose correct?

Zero works the same as in regular roulette - only a bet on zero wins.

Okay here is what I am thinking on the mixed paying the same....it should be lower

Once one ball has landed say in black you no longer have a equal amount of chances of the second ball finding a black spot as there is one less black spot available. Same as if the first ball hit a red spot

Quote: coilman

Okay here is what I am thinking on the mixed paying the same....it should be lower

Once one ball has landed say in black you no longer have a equal amount of chances of the second ball finding a black spot as there is one less black spot available. Same as if the first ball hit a red spot

In this game, you DO - as two balls CAN occupy the same slot. Doesn't happen too often...

http://doubleballroulette.com/ for the layout.

The double ball paying 1:1400 is also interesting... I make that an edge of only 3.1%... which is odd as you'd expect a jackpot to have a higher house edge (though I guess the physics may well mean that a double result is much less likely than 1:38 * 1:38).

But yeah, that Mixed result being 3:1... super, everyone stay very quiet, maybe they won't notice.

I see two problems.

1. Ignoring 0 and 00, there are "two ways" to make red-black, while there is only "one way" to make all-red and only "one way" to make all-black.

2. The "tight fit" for two balls in each slot will mechanically/physically affect the probability when one ball travels into a slot already occupied by the other ball.

In problem "1," ball "A" can land black, while ball "B" lands red, but there is yet another possibility -- ball "A" can land red while ball "B" lands black. So, there are "two ways" to make red-black. I am guessing that payoffs can be changed to remedy this situation.

Problem "2" (mechanical/physical influence) is a little tougher, but, as an inventor, I can think of two obvious solutions. One of the solutions is possibly easy, and the other is more "proper." Of course, the "proper" one would incur more manufacturing costs.

I look at this as a very simple binary problem:

Aside fron Zero/green, which is a rare hit to produce the house edge, it is this:

Two colors from two balls give four possibe outcomes:
Red Red: 25%
Red-Black: 25% \
Black-Red: 25% / - to combine for 50%, minus zero hitting.
Black-Black: 25%

so that the the mixed color result occurs 50% of the time, and "Pure color" result is 25% of the time, aside from zero.
Shouldn't the payouts be:
R-R: 3:1
Mixed: 1:1
B-B: 3:1

e431312-1a

I'm not the greatest at math, but it is fairly straightforward in this case.

Ignoring any mechanical influence of one ball on another, it seems to me that there are:
- 324 ways to make all-red;
- 324 ways to make all-black;
- 648 ways to make red-black;
- 148 ways to make any combination involving "0" or "00."

The pay-offs should be appropriate for those possible outcomes.

On the other hand, the mechanical influence of one ball on another is much more troubling and complex, and that variable should be eliminated. It seems probable that the tight fit of the slots would yield more red-black combos than mathematically expected (and possibly, more green-red and green-black combos). There are ways to solve this problem.

I somehow missed this game at the show. It should not be confused with Double Action Roulette, by the way.

I agree that the red/black combination would have a 2*(18/38)^2 = 44.88% chance of winning on a double-zero wheel, ignoring any effect that it may be harder for two balls to land in the same hole.

To simplify the question, what is more likely in the flip of two coins?

A. Two heads
B. Two tails
C. One head and one tail
D. All the above equally likely.

The answer is C at 50%. A and B each have a 25% probability.

It is like the easy hop and the hard hop bets using two dice:
if two dice are used, and one die is Blue, and the other die Yellow, then you can have an easy-8 as 5 & 3, with the 5 being blue, and the 3 being yellow, as WELL as the 5 being yellow, and the 3 being Blue - for two results. When accounting for two dice with different results, there are two ways to form it: 53, and 35.
But the hard-8 as 4 & 4 is with the blue die as four ONLY, and the yellow die as 4 ONLY, with NO other possible way to form it, as one result, 4-4 only.
4-4 can only come one way: 4-4, but 5-3 can also come as 3-5.

Now if you had a two different numbered balls in this Roulette game, then Black-Black can only be formed one way:
If ball #1 lands as black and ball #2 lands as black, then you cannot say that if ball #2 is black and ball #1 is black, then that would be a "differing result."
Same with Red-Red.

But you CAN have ball #1 land as red, and ball #2 land as black, as WELL as Ball #1 land as Black, and Ball #2 land as red for another, extra result - for two possible outcomes when mixed red & Black occur.

This must be elusive to some degree...looking at the voting, at 3 yes and 3 no, it is surprising...in math, something apparently simple can be tricky and elusive at times...

Quote: 7craps

I see the two balls for one spin as just 1 ball for 2 spins.

However, from Dan's description, it appears that the interaction of the two balls could skew the results away from the math expectations. As someone who has designed a few mechanical random outcome generators, this possibility bugs me.

Any mechanical bias on the expected results should be eliminated.

In any event, the three events as posted by the game add up to a 75% possibility. So one of them must have an edge... I worry that this could have possibly gone through any gaming math or review....

Quote: tupp

However, from Dan's description, it appears that the interaction of the two balls could skew the results away from the math expectations. As someone who has designed a few mechanical random outcome generators, this possibility bugs me.

Any mechanical bias on the expected results should be eliminated.

In seeing the game demonstrated, the two balls traveling on the same track seldom ever touch or click, and can also land in the same slot. While an occupied slot may slightly cause the other ball to jump out, the game really does operate as "two spins in one."

The balls seldom interact, and if they do, it just may act as randomization anyway.

I was with Miplet when he saw the error. He pointed it out to me. I cocked my head, though a moment, and agreed with him, but with a slight hesitation. I mean, who did the math? Wasn't this certified? What were we missing? We can't really be right, can we?

So, for shits and giggles, we started running a test. In the next 7 spins (albeit a small sample), we had one spin with red/red, one with black/green, and FIVE with red/black. We started to think "Is this installed anywhere?"

So we got the attention of one of the reps, and showed him the problem. I explained, "excluding spins that had a green result, there are FOUR possibilities: red/red, black/black, red/black and ... BLACK/RED." It took a moment, but he got it. We then showed the error to another rep, who also got it after a moment.

Bottom line, they both agreed that it seemed that we were right, but that they'd have to go back to the office to kick it around. After all, it had already gone thru third-party testing and approval.


That night, I read the Wiz' post that Double Action Roulette was starting a field trial at the M the next day. I was about to change all my plans for the remainder of my trip, until I realized that was a different company's two-spin gizmo (with no bet for the red/black combo). (The different company is TCS John Huxley.)


---


On a separate note, the double-hit (two balls in one pocket) pays 1,400. A full parlay on a standard game would pay 1,296. So why pay more? (The TCS version pays 1,200.)

When I asked a rep about it, he said it doesn't happen that often. I countered that mathematically, it should happen once in 38 times. He agreed with the math, but said it just doesn't happen that often. So it seems that the first ball "sometimes" deflects the second ball. That being the case, how can the wheel ever have been certified, since it doesn't follow the associated math?



FYI: The pockets on this wheel are much deeper than on a standard wheel. This provides the room for the two balls to co-exist. However, it reduces the amount of bounce seen by a single ball on a normal wheel. I.E. The ball tends to stay in the first pocket it sees. However, because of centrifugal force, it stays toward the outside, in the ideal position to deflect the second ball.

Quote: Paigowdan

In seeing the game demonstrated, the two balls traveling on the same track seldom ever touch or click, and can also land in the same slot. While an occupied slot may slightly cause the other ball to jump out, the game game really does operate as "two spins in one."

The balls seldom interact, and if they do, it just may act as randomization anyway.

I agree that touching/clicking adds to the randomization, but I saw them touch often, as well as often appear to be running for extended periods locked together.

Quote: thecesspit

In any event, the three events as posted by the game add up to a 75% possibility. So one of them must have an edge... I worry that this could have possibly gone through any gaming math or review....

It had, but was caught, that's to this forum's activity and interest, and concern by some.

Firstly, it is a brand new game awaiting its first install, and spotting a final glitch may have helped it.

There should be no Schadenfreude over a math error that slipped through on it, but was later caught.
It was discussed here civilly and as just a math problem. Some here initially voted that it all seemed correct. Some seemingly simple math issues are a tad slippery, and may miss an editor's eye at times. It happens.

Quote: DJTeddyBear

On a separate note, the double-hit (two balls in one pocket) pays 1,400. [snip] When I asked a rep about it, he said it doesn't happen that often. I countered that mathematically, it should happen once in 38 times. He agreed with the math, but said it just doesn't happen that often. So it seems that the first ball "sometimes" deflects the second ball. That being the case, how can the wheel ever have been certified, since it doesn't follow the associated math?

This is exactly what was troubling me, just from reading Dan's brief description.

If I were the manufacturer, I would be very concerned.

Quote: DJTeddyBear

FYI: The pockets on this wheel are much deeper than on a standard wheel. This provides the room for the two balls to co-exist. However, it reduces the amount of bounce seen by a single ball on a normal wheel. I.E. The ball tends to stay in the first pocket it sees. However, because of centrifugal force, it stays toward the outside, in the ideal position to deflect the second ball.

It sounds like they considered the problem, but they failed to eliminate it. I hope that the retool isn't too hefty.

Quote: Paigowdan

It had, but was caught, that's to this forum's activity and interest, and concern by some.

Firstly, it is a brand new game awaiting its first install, and spotting a final glitch may have helped it.

There should be no Schadenfreude over a math error that slipped through on it, but was later caught.
It was discussed here civilly and as just a math problem. Some here initially voted that it all seemed correct. Some seemingly simple math issues are a tad slippery, and miss an editor's eye at times. It happens.

Shrug... if I paid for my car to be serviced and then discovered it had no brakes as I drove it off the lot, I'd be asking the mechanic for a refund pretty damn quickly.

Guys, the 1400:1 payout is NOT on two balls in the same slot.

It is on two balls in the same slot number, AND you had bet that exact slot number to hit with a double ball result.

Quote: thecesspit

Quote: Paigowdan

It had, but was caught, that's to this forum's activity and interest, and concern by some.

Firstly, it is a brand new game awaiting its first install, and spotting a final glitch may have helped it.

There should be no Schadenfreude over a math error that slipped through on it, but was later caught.
It was discussed here civilly and as just a math problem. Some here initially voted that it all seemed correct. Some seemingly simple math issues are a tad slippery, and miss an editor's eye at times. It happens.

Shrug... if I paid for my car to be serviced and then discovered it had no brakes as I drove it off the lot, I'd be asking the mechanic for a refund pretty damn quickly.

Yeah, if you had gone out on the road. Granted, it's a tad late, but it'll be repaired before its first install.

Quote: Paigowdan

Guys, the 1400:1 payout is NOT on two balls in the same slot.

It is on two balls in the same slot number, AND you had bet that exact slot number to hit with a double ball result.

Some were asking why this would pay more than a parlay but it makes perfect sense that it would. If it paid less than a full parlay it would be a sucker bet to the point that no self respecting player could play it. If it paid the same it would be pointless. If they give you a little extra payout for making a huge longshot that is a great way to invite action.

Quote: Paigowdan

Guys, the 1400:1 payout is NOT on two balls in the same slot. It is on two balls in the same slot number, AND you had bet that exact slot number to hit with a double ball result.

Thanks for the clarification. However, the apparent discrepancy between the math and the results is still worrisome.

The Double Ball promo video shows that the "Any Double Ball" bet pays 50-to-1, yet the expected frequency of hits is one out of every 38 spins. What is limiting the frequency of "Any Double Ball" hits so that the house has an edge at 50-to-1? How is that edge calculated?

Yes, this came up in another thread. With that version, I already poked some holes in the math. Shame on the mathematician and on the testing lab.

Quote: tupp

Thanks for the clarification. However, the apparent discrepancy between the math and the results is still worrisome.

The Double Ball promo video shows that the "Any Double Ball" bet pays 50-to-1, yet the expected frequency of hits is one out of every 38 spins. What is limiting the frequency of "Any Double Ball" hits so that the house has an edge at 50-to-1? How is that edge calculated?

I bet the odds of occupying the same spot is less than 1 in 50 because there is less space for the second ball. This would also be affected by the size of the ball and the wheel. To determine the actual frequency, it would take many thousands of spins to have a decent confidence. But, I suspect that the house edge is high enough on that one that it won't make much of a difference between one thousand or ten thousand samples.

This also wasn't noticed until the last 2 hours of the show. I'm curious to know how many people looked at the layout and didn't notice. Say this goes unnoticed and gets installed. How long untill the APs see it? Would your average player notice and say something?

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e431312

I think they should have left the 50-1 payout on any number hit twice, although they could have set it to 35-1. The 1400-1 makes it impossible to have an advantage since the best possible odds of winning are 1 in 38² or 1 in 1444.

I still think there is an advantage on the 4 numbers bet, which I discussed on the previous thread because I'm still betting that the odds of getting the two balls on the same slot is worse than 1 in 50.

Of course, who needs that when you can bet on red/black at 3-1.

Quote: Paigowdan

Guys, the 1400:1 payout is NOT on two balls in the same slot.

It is on two balls in the same slot number, AND you had bet that exact slot number to hit with a double ball result.

Dan, we know that.

But the fact remains that there should be some double hit, on average, 1 in 38 times. It's just that there is also only a 1 in 38 chance that it occurs on the particular number you selected.

Here's my problem with it: The TCS Double Action game gets two results from a single ball. It doesn't have any of the potential 'deflection' problems that we are discussing with this Double Ball system. It has a bet for the double result that pays 1,200 to one. Why does the same type of bet on the Double Ball system pay 1,400 to 1?

Quote: bigfoot66

Some were asking why this would pay more than a parlay but it makes perfect sense that it would. If it paid less than a full parlay it would be a sucker bet to the point that no self respecting player could play it. If it paid the same it would be pointless. If they give you a little extra payout for making a huge longshot that is a great way to invite action.

Don't underestimate the stupidity of a payer, or overestimate his ability to calculate the odds. The long-shot opportunity is all he sees.

Quote: tupp

Thanks for the clarification. However, the apparent discrepancy between the math and the results is still worrisome.

The Double Ball promo video shows that the "Any Double Ball" bet pays 50-to-1, yet the expected frequency of hits is one out of every 38 spins. What is limiting the frequency of "Any Double Ball" hits so that the house has an edge at 50-to-1? How is that edge calculated?

I suspect that the 50 to 1 refers to the actual chance of two balls landing in the same pocket. I assume that this is based upon testing. However, I would expect that if that were the case, then the payout would be a fairer 1,750 to 1. (I.E. If the odds of any double are 50 to 1, and 38 to 1 that it be the number you select = 50 * 38 = 1,900 to 1. So a 1.750 to 1 payout seems more reasonable.)

e431312

Quote: PaiGowDan

Guys, the 1400:1 payout is NOT on two balls in the same slot.
It is on two balls in the same slot number, AND you had bet that exact slot number to hit with a double ball result.

Quote: DJTeddyBear

Dan, we know that.

No. At the time (earlier in the thread), not everyone got that aspect or condition then. Some thought the payout was 1400:1 on ANY double ball single number result.

Quote: DJTeddyBear

But the fact remains that there should be some double hit, on average, 1 in 38 times. It's just that there is also only a 1 in 38 chance that it occurs on the particular number you selected.

If ANY double ball same-slot result paid out 50:1 with a concurrent 38:1 true odds of hitting, it would have one hell of a player advantage. This must have been in an earlier, "pre-math review" version (and yes, such thing do existing as earlier, not-fully-debugged game versions, not uncommon). The 50:1 player-payout on ANY 38:1 true odds may have simply been wrong - an error of an earlier, undebugged version. That is, unless there were some specific mechanical guidelines or specifications that mechanically discouraged, by some percentage, the chance of two balls landing or residing in the same slot or pocket. This is a bit of a stretch, as it would require a new gaming "mechanical specification" to Roulette wheels using this product in terms of exact slot depth and length - with is unfathomable. Since the game is perhaps attempting to emulate a true "two results in one spin," it is assumed that the "two-balls-in-one slot" effect is negligible in kicking out the second ball, should it also try to land in the same slot.

One might assume that the game inventor and his mathematician had both considered and analyzed the mechanical effect of this aspect in the game's operational design. Smaller Roulette balls might help. Big balls would hurt, as it's tougher to pack two big balls into a tight little slot, where they'd be more likely to be kicked out.

Quote: DJTeddyBear

Here's my problem with it: The TCS Double Action game gets two results from a single ball. It doesn't have any of the potential 'deflection' problems that we are discussing with this Double Ball system. It has a bet for the double result that pays 1,200 to one. Why does the same type of bet on the Double Ball system pay 1,400 to 1?

Because the 1,200:1 version has a 14% house edge on this high payout bet, while the Double ball Roulette at 1400:1 has about a 3% house edge on this high-payout bet.

Quote: tupp

The Double Ball promo video shows that the "Any Double Ball" bet pays 50-to-1, yet the expected frequency of hits is one out of every 38 spins.

The expected frequency should be 1 in 38 spins for an "ANY Double ball single number result," assuming no mechanical wheel alterations to pop out landing in the same slot. It should pay 35 or 36 to 1, keeping in accordance with the "direct number" 35:1 payout.

Quote: tupp

What is limiting the frequency of "Any Double Ball" hits so that the house has an edge at 50-to-1?

It cannot have a house edge at 50:1, period. It must be at 34:1 to 37:1 player playout, - but it should be at 35:1, matching the standard payouts of the standard Roulette game.

Quote: DJTeddyBear

]I suspect that the 50 to 1 refers to the actual chance of two balls landing in the same pocket. I assume that this is based upon testing. However, I would expect that if that were the case, then the payout would be a fairer 1,750 to 1.

No, two balls landing in the same pocket should be about 38:1 also, unless particularly shallow and narrow slots on certain Roulette wheels would prevent two balls from occupying the same slot.

Quote: DJTeddyBear

(I.E. If the odds of any double [ball number] are 50 to 1, and 38 to 1 that it be the number you select = 50 * 38 = 1,900 to 1. So a 1.750 to 1 payout seems more reasonable.)

The odds should be 38:1 or close to 38:1, and 35:1 on player payout, assuming standard wheels that can accomodate two balls landing in a slot.
With an ANY double-ball same slot result, it should pay 35:1 or less.
With a specific double ball same slot result, it should be in the 1,200:1 range.

Dan -

We're in agreement on most points.

My point about the 50 to 1 thing was, it was either an error/typo that never got corrected before the video was produced (fix that!), or, because of mechanical limitations or deflection, the results of extensive testing show that you get a double only 1 in 50 times.

Frankly, I'm inclined to believe the latter.




FWIW: I have a solution to the problem.

Design the pockets so that they are steeply angled towards the center. This will counter the centrifugal force, taking the first ball away from a position where it has a chance to deflect the second ball.

Quote: 7craps

DJTB,

Did you try to place both balls in the same pocket while you were there...
Or ask the Dealer to do it?

I did it myself.

I had a ball in a pocket, with the wheel not moving, and dropped the second ball straight down the edge of the wheel towards that pocket.

It was difficult to get my aim right, but on the couple times the second ball hit the first ball, there definately was deflection towards a neighboring pocket.

Of course this was a totallly unscientific test. A ball traveling in an arc around the wheel, along with the wheel moving, may produce completely different results.

I quickly reviewed points made here.
1.) Tupp's early fractioning of RR/BB/RB/zero-other 324/324/648/148 seems spot-on. 4 FOR 1, 4 FOR 1, 2 FOR 1, 9 FOR 1 puts the H.E. at 10.25% for the first three, and 7.76% for the zero-other.
2.) The only way these numbers will hold up is by using a PRNG/Video approach. The mechanical ball-slot operation has some H.E. issues with the 2nd ball bouncing out of an occupied number. This should favor the House's already enormous 10% adv. A suggestion to deeply dimple each and every slot twice (for each ball) would more-closely maintain the H.E. for the mechanical game.
3.) I like 35 FOR 1 on the "Any Double"... H.E. of 7.89%. For the Jackpot of Pick the EXACT Double... meh, 1300 FOR 1.

IMHO video game ONLY, mechanical issues are frought with problems/expense/engineering. Suggested payoffs 4.2 FOR 1, 4.2 FOR 1, 2 FOR 1, and 9.2 FOR 1.

Separately, yes, and its very complex... involving among other things, velocity of the ball, velocity of the wheel and approaching angle, considering ball to ball contact or ball-to-wall-to-ball contact, yotta, yotta, yotta.

On the house edge:
I would assume that distributor in charge of this new game project would or had already addressed the house edge issue to reduce it into normal Roulette ranges - that is, - make the "Zero" action occur when the other ball had hit only one of the other two colors;

For example, if one ball is zero and the other red, THEN the zero is a zero action; however, if one ball is a zero and the other ball on a black colored number, then the zero ball is dead, - no zero. This would affect the percentages H.E. on some bets, such as the "Pure Black," "Mixed color", and "Pure Red" bets, etc.

One two-ball mechanics: Roulette wheels cannot be easily adjusted or replaced. The two-ball system must work with the standard Roulette wheels, and pretty much do.

Quote: Paigowdan

On the house edge:
I would assume that distributor in charge of this new game project would or had already addressed the house edge issue to reduce it into normal Roulette ranges - that is, - make the "Zero" action occur when the other ball had hit only one of the other two colors;

For example, if one ball is zero and the other red, THEN the zero is a zero action; however, if one ball is a zero and the other ball on a black colored number, then the zero ball is dead, - no zero. This would affect the percentages H.E. on some bets, such as the "Pure Black," "Mixed color", and "Pure Red" bets, etc.

One two-ball mechanics: Roulette wheels cannot be easily adjusted or replaced. The two-ball system must work with the standard Roulette wheels, and pretty much do.

We would hope, but with the arguments of the 50 to 1 payout on any double in this thread probably makes some of us skeptical. And as at least one other poster in this thread has mentioned it's closer to 50 to 1 because it's much more difficult to get two balls in the same spot (and is also game equipment dependent, ball size vs. slot size, etc).

Another way to take care of zeros for outside bets is to color one ball green and only allow this ball to enforce zeros on any outside bet. I have a feeling they did not do that though, because on the website there sample table layout says: "BOTH BALLS MUST LAND IN THESE OUTSIDE BETS". So except for mixed red/black goofup, the return on the outside bets (4 for 1) are roughly 4*(18/38)^2 - 1 ~= -10.25%.