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Aug 14, 20

0 3

Why do they do that? Start it at 92% and put a percent meter on it. Cheap bastards. That's what I would do at Bob Bartop's Grill & Bartops.

Thank you for working that out. I'd like to know the exact paytable. Interesting subject. Thanks so much. Seriously.

Answer: WHY THEY DO THAT

There are a few reasons:

1.) The minimum jurisdictional return in Nevada is 75%, so they can. Ironically, in most other gambling states with commercial casinos, that machine would be illegal.

2.) People honestly pay so little attention that it likely gets some action.

3.) If this is one of those games upon which you can bet anywhere from $0.25-$2.00, or $2.50, whatever...the Progressive can theoretically get to a positive point anyway if almost nobody bets enough that they should hit 6/6 with the necessary bet. It would take a really friggin' long time, but it's possible. Anything is possible...but it would be a ton of play hours.

4.) The percentage that feeds the Progressive might even go in if you're playing other numbers of Picks. I haven't seen that, per se, but I've seen machines upon which 7/7, 8/8, 8/9 and 9/10 (or maybe it's 9/9) all share the same Progressive, so it's just whichever you want to play.

5.) Because they are bad people.

BONUS:

92% w/1% Meter

We'll go with this, 2-5-100-1600 = 0.918268440420339

Okay, so we want a 1% meter:

(1-0.918268440420339) = (x * 1/7752.84)

x = 633.651704

2-5-100-2233.66

Okay, so you need it to miss 63366 times.

63366/7752.84 = 8.17326295912 Cycles

(7751.81/7752.84) = (0.999871015)^63366 = 0.00028194721 or 1/0.00028194721 = 1 in 3546.76323983

Honestly, if lower bets didn't contribute, then it would almost never happen. This 1 in 3546.76323983 basically represents the probability, from base, of it going enough plays missing to go positive. The binomial distribution (Poisson) agrees.

But, if it's one upon which as low as $0.25 can be bet and 1% contributes, but a bet of $2.00 is necessary for the Progressive, you probably would see it go positive from time-to-time.

Lesson: There is a really simple lesson here for recreational players: ALWAYS DO THE MATH! If you don’t know how to do the math, then ask me and I will help.

As a player, you could encounter a Keno Progressive that seems like it is ridiculously over the base amount of the game and be inclined to play it blindly. Please don’t do that because it is often a, ‘Trap.’

People will often see a 10-Spot Keno Progressive that starts at $1,000 and has gotten up to $2,200, (just picking numbers, here) and they might say to themselves,That’s more than double! This must be good.”

Of course, that doesn’t take into account the probability of it hitting is 1 in 8,911,711.18 on a given play. In other words, that extra $1,200 adds:

(1200 * 1/8911711) = 0.0001346542768274

Which means 0.01346542768274% to the return of the game.

Since I seriously doubt the game started off with a return of 99.98653457231726%, I equally seriously doubt that the game is at an advantage.

This is also a good lesson for slot players who see a Progressive that seems to be, “Much more than usual,” by itself, that doesn’t always mean much of anything. You have to do an actual analysis of a slot machine to determine whether or not it’s at an advantage.

Aug 14, 20

1 1

Is the 6-spot at least 1600? Fwiw, 1/2% meter ain't half bad as far as progressive 6-spots go. I ain't no expert, but from what I've seen anyway, and I do try to look at keno progressives. But you're right, of course, I can't imagine 76%, that's awful. And the thing is, they don't even need to do that. So greedy. There ought to be a law.

I don't think I've seen anything less than 88-87 or so.

Answer: The first thing is that, other than variance, it really doesn't matter how that 76% is derived. Therefore, let's see what we have to do to get a six-spot to 76% assuming your 1600-FOR-1 pay on 6/6 at base.

How about a nice 2-4-58-1600 paytable for a base return of 0.759713704017501, that should be pretty good.

Okay, so we want to determine the breakeven meter, which is really easy:

(1-0.759713704017501) = (x * 1/7752.84)

x = 1862.901207

Of course, that's expressed in credits, so we have to do 1863 for a total of 2-4-58-3463.

The calculator confirms: 1.000012645582266

Okay, so we know that the meter is one-half of one percent. We are betting one unit, so for each miss, .005 units are being added to the meter. In other words, it takes 200 misses for the meter to move up a full unit, whatever a unit is.

The meter must move up 1863 units in order to be positive, which means that the machine must miss 372600 times from base.

We can tell that is 372600/7752.84 or 48.0598077608 cycles, but just how borderline impossible is that?

For that we can dust off a binomial distribution calculator:

N = 372600 Attempts

K = 0 6/6's

P = 1/7752.84

Well, we get the lowest possible result for the calculator that way: <0.000001

So, let's take the probability of it not happening, 7751.84/7752.84 to the 372,600th power and see what happens:

(7751.84/7752.84) = 0.999871015

(0.999871015)^372600 = 1.3382679e-21 or 0.0000000000000000000013382679 --- 1 in 747,234,540,000,000,000,000

Which is, expressed in words, One in Seven Hundred Forty-Seven Quintillion, Two Hundred Thirty-Four Quadrillion, Five Hundred Forty Trillion.

So, the machine isn't going to go positive very often. (Read: Never) Unless it is one of those ones upon which minimum/smaller bets do not qualify for the progressive but juice the meter...but it's still going to take a lot of those.

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