jwblue
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April 11th, 2013 at 9:49:03 PM permalink
I put this thread on this board since it is more active than Slots.

Someone won the progressive jackpot a few days ago only five months after the last progressive win. It makes me believe they might decrease the odds for the next progressive win.

Is it possible for the Megabucks slot machines to be changed to different odds for the progressive to be won?

Someone estimated the odds of winning the Megabucks progressive at 1 in 49,000,000.

Quote:

... What if each reel has 368 stops and there is one Megabucks symbol on each reel? You would have one chance in 49,836,032 to hit the Megabucks, and these are the number on the Megabucks par sheet I have.




I do not know if my understanding of stops is correct. Is it only possible for a stop to have a maximum of 368-1 odds? Why couldn't a stop have lower odds?

If there are 368 virtual stops, is it possible to program all the Megabucks slot machines to have more stops?

What if there is an RNG that is not for each reel individually, but for all three reels? (i.e. The random number generated 1,000,345 will show the combination, "Cherry, Bar, Megabucks symbol.") Is it possible to program a slot machine that way? If so, there could be 100,000,000 combinations. Like the previous example, more than one random number could be for the same combination.
tringlomane
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April 11th, 2013 at 10:13:20 PM permalink
They could change the probability since Nevada offers server based gaming now. Would they? Unlikely since any rarer jackpot is illegal unless the probability is specifically stated on the pay glass.

For the latter part of your question. Two possibilities: Less coin-in and/or slower progressive meter increase. I wouldn't be shocked if both of these factors were the case. Less coin-in is a near certainty given the recession though. I highly doubt they decided to up the payouts of the smaller prizes.
onenickelmiracle
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April 11th, 2013 at 10:55:13 PM permalink
I agree with Tring, but think changing the PB would be legal with SBG. They don't promise you jack anymore compared to the days where much information was available in the help screen. I also think this should be a regulation of SBG, but I have NEVER seen it mentioned in pop slot literature or any reading.

The coin-in meter incrementation has lowered and most of these are now 86% compared to 89% the way they used to be. You will never see record JPs again ever.
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jwblue
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April 11th, 2013 at 10:55:41 PM permalink
Quote: tringlomane

They could change the probability since Nevada offers server based gaming now. Would they? Unlikely since any rarer jackpot is illegal unless the probability is specifically stated on the pay glass.




They don't state the probability.

What does "rarer jackpot is illegal" mean?

IGT has never stated anywhere what the odds are of hitting the progressive is. It is a closely guarded secret. A representative from IGT told me so. I am not sure how they can get away with not disclosing the odds. Where is the Nevada Gaming Commission? Every state lottery discloses the odds of winning their lottery. By not disclosing the odds, it gives them carte blanche to change the odds whenever they want which is absurd.

The progressive hit yesterday after 5 months. I am guessing short time period between jackpots means less profit. This inclines IGT to reduce the odds of the next progressive hitting if it is legal.

The question is whether it is legal.
onenickelmiracle
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April 11th, 2013 at 11:10:05 PM permalink
Quote: jwblue

They don't state the probability.

What does "rarer jackpot is illegal" mean?


Maximum odds of a winning combination in most states are 50,000,000 to 1 and 100,000,000 to 1 in NJ. This can be gotten around I think as long as it isn't a winning combination as I understand it. Example: games which don't multiply top symbol in bonus such as Cleopatra II.
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tringlomane
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April 11th, 2013 at 11:16:57 PM permalink
Quote: onenickelmiracle

Maximum odds of a winning combination in most states are 50,000,000 to 1 and 100,000,000 to 1 in NJ. This can be gotten around I think as long as it isn't a winning combination as I understand it. Example: games which don't multiply top symbol in bonus such as Cleopatra II.



Yeah, this. But the more I think about it...Nevada may be 100M to 1, but I am failing to find the right legal blurb. I know Missouri is 50M to 1. And even jackpots rarer than 100M to 1 are legal in NV IF the probability is displayed on the payglass. However, that rule makes them practically illegal since slot machine designers are not idiots.
jwblue
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April 11th, 2013 at 11:26:13 PM permalink
Quote: tringlomane

Yeah, this. But the more I think about it...Nevada may be 100M to 1, but I am failing to find the right legal blurb. I know Missouri is 50M to 1. And even jackpots rarer than 100M to 1 are legal in NV IF the probability is displayed on the payglass. However, that rule makes them practically illegal since slot machine designers are not idiots.



Here is the answer straight from the NGC.

Quote:

2.070 Jackpot Odds. If the odds of hitting any advertised jackpot that is offered by a gaming
device exceeds 100 million to one, the odds of the advertised jackpot must be prominently
displayed on the award glass or video display.
(Adopted: 12/04. Effective 1/1/05.)



http://gaming.nv.gov/modules/showdocument.aspx?documentid=3309

Some one did an analysis of the odds that didn't make much sense to me. Does this sound right?

I am not sure why the number of reel positions has any relevance. What is the PAR sheet he is referring to?

Quote:

I recently read an article that quoted Frank Scoblete that the odds of hitting megabucks is 49,836,032 to 1. I was puzzled at how he arrived at this since if each reel has 512 positions, 512 * 512 *512 would be 134,217,728 to 1. If one reel has two megabuck symbols and the other two have one each, the odds would be 67,108,864. The closest I come is 44,564,480 for a 1 to 1 to 3 weighting.

Thanks,
Jim

Dear Jim,

You said, "...if each reel has 512 positions...." What if each reel does not have 512 stops?

What if each reel has 368 stops and there is one Megabucks symbol on each reel? You would have one chance in 49,836,032 to hit the Megabucks, and these are the number on the Megabucks par sheet I have.

Best of luck in and out of the casinos,
John



http://robison.casinocitytimes.com/article/megabucks-odds-13763

Someone on this forum said the following. This I find impossible to believe.

Quote:

To add another quirk to all you math guru's heads, I read how the megabucks and most other wide area progressives inculding the former jumbo jackpot (stations) work. An amount is randomly selected for the jackpot to hit at and as soon as that amount is achieved the first slot machine signal received will display the jackpot. In other words the jackpot will never come up randomly as you guys are suggesting. It is predetermined and whoever is there at the right time will hit it.

tringlomane
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April 11th, 2013 at 11:41:41 PM permalink
Quote: jwblue

Here is the answer straight from the NGC.



http://gaming.nv.gov/modules/showdocument.aspx?documentid=3309

Some one did an analysis of the odds that didn't make much sense to me. Does this sound right?

I am not sure why the number of reel positions has any relevance.



Thanks for finding that legal bit...my search skills suck apparently.

The reels are very important. how many stops the jackpot symbol has (1) versus the total number of stops (368) helps determine the probability. If John Grochowski, has a PAR sheet for one of them, then it's likely to be the correct odds even today. And (1/368)*(1/368)*(1/368) = 1 in 49,836,032 IGT could change them to any other probability greater than 1 in 100M in theory, but unless that is easily done remotely, it will not happen since all the Megabucks machines in the state need the same jackpot probability. The jackpot probability is very likely still 1 in 49,836,032.

The PAR sheet is the "Percentages and Returns" sheet given to gaming personnel from the slot manufacturer. It tells in detail the expected probability of each outcome, its corresponding payout, and its contribution to the slot machine's expected return.
onenickelmiracle
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April 11th, 2013 at 11:46:02 PM permalink
It's just the way it is. Reminds me of this, https://www.youtube.com/watch?v=iUZvO36zm6Y .
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jwblue
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April 11th, 2013 at 11:46:10 PM permalink
Quote: tringlomane

Thanks for finding that legal bit...my search skills suck apparently.

The reels are very important. how many stops the jackpot symbol has (1) versus the total number of stops (368) helps determine the probability. If John Grochowski, has a PAR sheet for one of them, then it's likely to be the correct odds even today. And (1/368)*(1/368)*(1/368) = 1 in 49,836,032 IGT could change them to any other probability greater than 1 in 100M in theory, but unless that is easily done remotely, it will not happen since all the Megabucks machines in the state need the same jackpot probability. The jackpot probability is very likely still 1 in 49,836,032.

The PAR sheet is the "Percentages and Returns" sheet given to gaming personnel from the slot manufacturer. It tells in detail the expected probability of each outcome, it's corresponding payout, and it's contribution to the slot machine's expected return.



Why does the odds of each reel hitting a stop have to be equal (i.e 1/368)?

Why couldn't IGT remotely change the odds of each machine? Don't they already have a download connection to each machine?

Here is a possible answer to my own question.

Is it possible to remotely change the number of virtual stops?


Quote:

The reels in a 3 reel machine have 22 symbols each. But internally, there are 368 'stops' on a virtual reel. Each of those virtual stops has an equal chance. However, the virtual stops are not evenly distributed among the 22 symbols.

Therefore, the one symbol for the megabucks prize may be programmed to only one of the 368 virtual stops, making the 1 in 368 ^ 3 calculation correct. [q\]

tringlomane
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April 12th, 2013 at 12:19:13 AM permalink
Quote: jwblue

Why does the odds of each reel hitting a stop have to be equal (i.e 1/368)?

Why couldn't IGT remotely change the odds of each machine? Don't they already have a download connection to each machine?

Here is a possible answer to my own question.

Is it possible to remotely change the number of virtual stops?



It doesn't. They could do something like (3/512)*(2/512)*(1/512) = 1 in 22,369,621 instead.

They possibly could change odds/virtual stops if the machines are modern enough. Nevada state law requires machines to be idle for 4 minutes before allowing to change payouts though, so they would need to shut them all down simultaneously.

If they do bother to do this, however, I can almost guarantee it's a rare occurrence. Casinos rarely have to worry about machines that pay < 90% if many of their players play $3/spin on them.
onenickelmiracle
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April 12th, 2013 at 12:48:04 AM permalink
Quote: tringlomane

Quote: jwblue

Why does the odds of each reel hitting a stop have to be equal (i.e 1/368)?

Why couldn't IGT remotely change the odds of each machine? Don't they already have a download connection to each machine?

They possibly could change odds/virtual stops if the machines are modern enough. Nevada state law requires machines to be idle for 4 minutes before allowing to change payouts though, so they would need to shut them all down simultaneously.

My problem is it doesn't mean they can't interrupt play and disable it for four minutes while the customer sees an error message or some counterintuitive terminology which precedes this. This is more to do with SBG than Megabucks perhaps, but still uneasy for me.
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CrystalMath
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April 12th, 2013 at 6:51:12 AM permalink
1: all linked games must have identical odds of hitting the top award. If using SBG, they would need to ensure that all games are updated at the same time.
2: to make a change on a WAP game via SBG, it would involve downloading new software, rather than a paytable change, because WAP games are made with only 1 paytable to ensure the correct progressive odds.
3: because of this, the WAP infrastructure cannot support this; instead, downloads can only be handled by servers at the individual casinos.

I would not expect this to ever happen.
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DRich
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April 12th, 2013 at 10:13:39 AM permalink
Quote: CrystalMath

1: all linked games must have identical odds of hitting the top award.



I don't believe this is true anymore because games with different denominations may share the same jackpot. I believe the relative bet to odds still must be the same.

For example:

Two games are linked to the same jackpot. A dollar game with a $3 bet is a million to one to win the jackpot. I believe you can have a 75 cent quarter game have the jackpot odds at four million to one to share the same jackpot.
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CrystalMath
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April 12th, 2013 at 11:53:16 AM permalink
Quote: DRich

I don't believe this is true anymore because games with different denominations may share the same jackpot. I believe the relative bet to odds still must be the same.

For example:

Two games are linked to the same jackpot. A dollar game with a $3 bet is a million to one to win the jackpot. I believe you can have a 75 cent quarter game have the jackpot odds at four million to one to share the same jackpot.



Yes. I agree, and it has been this way for a while. But, the software is for a fixed denomination/odds, so that any machine linked will comply.
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ThatDonGuy
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April 12th, 2013 at 9:20:37 PM permalink
Quote: jwblue

Why does the odds of each reel hitting a stop have to be equal (i.e 1/368)?


Because the term "stop" is misleading. It does not refer to actual ("physical") locations on each reel, but "virtual" stops.

The reels on a Megabucks machine do not each have 368 symbols on them, but "internally" they do. Each of the 368 "internal" stops points to a particular "real" symbol; needless to say, more point to a particular blank symbol than the Jackpot symbol.

"Can't the internal stops have different probabilities?" Yes, they can, but unless you want to make the machine really complicated, each probability will be a rational number (the ratio of two integers), so by multiplying by all of the denominators, you once again have an integer number of internal stops, each of which is equally likely. (In other words, if virtual stops 1-367 were equally likely but 368 was 1 1/2 times as likely as the others, then you could translate this into 1-367 each having 2 stops and 368 having 3.)
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