Home Reelin’ In the Hits, Part III: A Crackpot Jackpot for Sure!

Reelin’ In the Hits, Part III: A Crackpot Jackpot for Sure!

”Get Larry Foutz in here!” comes your desperate cry.

“What’s up?” he asks, entering your office and sitting down at the table in the corner.

“Trying to figure out this machine,” you reply, walking over to the table and setting down a folder and several papers. “Crackpot Jackpots. We had a complaint from a customer that the machine has been ‘fixed,’ and I’m the lucky person to do a report on it.”

“Interesting,” Larry replies. “What have you done so far?”

“Well, I started with the usual,” you explain. “I checked the flash memory and compared it to our library. The operating system is fine, the game personality is fine, there’s nothing wrong with the programming in the machine. It’s a perfect match to what was put in place two years ago. The hardware is in perfect working order.”

“Good, good. What about faults? Did you check the system for errors?”

“Yes. I went right back to the installation. The only faults we’ve had are ticket jams and the printer running out of paper. I even looked at the MEAL book, and it matches the system.”

“You feel confident that the machine is functioning properly, then,” Larry surmises.

“You bet! I ran a report and looked at the game play statistics. Everything looks good,” you tell him.

“That’s the progressive in the big-bertha case, isn’t it?” Larry asks.

You nod your head in agreement.

“What was the complaint based upon?”

“The player has been watching the progressive jackpot and knows that it has never been paid,” you say. “He has played the game every week for the past two years and thinks that he should have hit the jackpot by now.”

“What does the PAR sheet say and how do the game statistics compare?”

Opening the folder on the desk, you slide the information to Larry. “The cycle is 884,736 games,” you tell him. “There are three reels with 96 stops each. It’s a quarter machine with a base jackpot of $10,000. The increment is 1 percent and currently stands at $8,918.136, giving us a current progressive jackpot of $18,981.13. I did the math, and that is correct. We have 1,238,630 games played so far. The jackpot is increasing properly—it just hasn’t hit.”

“Have there been any hand-pay jackpots?” Larry asks.

“Sure thing. They seem to be relatively proper. I did some math on the PAR sheet. It appears that all of the hand pays are within a few percent of what they should be. For example, two Crackpot Jackpot symbols and a red 7 should have 125 hits so far, and we’ve had 126.”

“That’s good,” Larry nods. “Since we have completed a cycle, then your boss also feels that the progressive jackpot should have hit by now?”

“That’s right. And I have to come up with a report giving actual proof that it’s OK.”

“Good, well, let’s get to work on this. Since the game is random, there is no guarantee that the jackpot will hit within the cycle,” Larry explains. “It could hit in the first week, or it may take months. If we look at the game with 1,238,630 plays and divide by 721 days—that’s the time the game has been in service—then we have a play of approximately 1,718 games played per day, or 71.6 games per hour. That’s just over one game per minute.”

“The report shows heavier play through the day, but not much late at night, so that figure seems correct,” you agree.

“Here is what we need,” Larry says. “We have to show not only that this is normal, but also give a value to quantify this. Let’s also throw in some other numbers. We can see how likely it is to hit during the first month, during the first six months, and during the time it has been in play, and also how likely it is to not hit.”

“That’s exactly what I need to do,” you say. “That’s why I called you. How can I do this? I wrote down some thoughts here.”

You slide another sheet of paper across the table and Larry takes a look. “I see what you have done,” he says. “At 500,000 games, you say that the probability of hitting the jackpot was 56.5 percent. You divided 500,000 games by 884,736 games in the cycle, right?”

“Correct. So, at that point, it’s quite possible that the jackpot hadn’t hit. Almost a 50-50 chance,” you explain.

“Well, that’s not exactly correct,” he says. “You see, you’ve determined that at 500,000 games, you’re 56.5 percent through the cycle—but that doesn’t mean the probability is the same. Remember, the games are random. What you calculated was not probability.”

You frown, realizing that this is going to be more difficult than you thought. “I’m really lost, then. I don’t know how to figure this out. I know there should be a way to do it, but I have no idea what it is.”

“It’s not as difficult as it seems,” Larry reassures you. “Your graph shows a rough idea of what you’re looking for. The bell curve indicates that more jackpots will occur around the cycle mark, fewer when the game is first installed, and varying levels before and after the cycle mark.”

“I guess that I could try to read a value off the graph, but that isn’t very scientific …”

“Or accurate,” Larry says. “You’re on the right track, though. What you need is called a ‘Poisson Distribution Analysis.’ That will give you the numbers that you’re seeking.”

“Really?” you ask, smiling for the first time.

“Yes. Siméon Poisson was a mathematician who lived in the late 1700s and the 1800s. Basically, his formula studies events that take place, and it does so over a specific period. It could be a period of time, such as a day, or a period of units, such as boxes of something produced. Or it might be events, such as operations performed by a machine.”

“Or something like games played on a slot machine?” you ask.

“Exactly,” Larry says. “We have an event that is to occur and a period of time that it occurs within. In our case, we’re looking at the event of the jackpot hitting, and the time is our cycle. We know the probability of it occurring during a cycle from the PAR sheet. In this case, we have one jackpot in 884,736 games. Dividing the numbers, we get 1/884,736 = 0.0000011312217. The probability is therefore .00000113, or 0.000113 percent. It’s a rare event. This game won’t hit the jackpot frequently. It’s just over 1 in three-quarters of a million games.”

“Yes, that’s correct,” you say. “But—”


“I know that already,” you say. “And it doesn’t really mean that much in reality. It’s fine on paper, but on the floor it doesn’t work out.”

“Correct,” Larry says. “That is the probability, but on the floor we won’t see this as an absolute. What the Poisson formula will tell us is the probability of the jackpot hitting in one cycle. On the PAR sheet we will have a jackpot during a cycle. In reality, we may have two. Or none. This gives us more information so that we can see how often we will have no jackpots in a cycle. It gives us mathematical information that is a bit more meaningful to us.”


“Alright, let me continue. So we need an event and a unit—a jackpot in a cycle, in our case. But there are four considerations we need to understand.”

You pick up a pencil and begin to make notes. “Great! Continue.”

“The first thing we need to consider is that the experiment has to result in outcomes that we can look at logically,” Larry explains. “By that I mean that it’s either a success or a failure. It happens or it doesn’t. We don’t award a partial jackpot. The player either wins the progressive amount or they don’t. And we don’t need to consider any other event. They may win a smaller jackpot from having a red 7 and a couple of two-times multipliers, but we’re only considering the one event. The rest is irrelevant to our studies. The game could have only one winning combination—the progressive jackpot—and it wouldn’t make any difference.”

“Alright, I get that,” you say. “And that’s exactly what we’re trying to determine—why the jackpot has not hit.”

“Yes. Now the second consideration is that we must know the average number of successes in the unit region. We know that there is one jackpot, and only one jackpot, in a cycle. There is only one jackpot stop on each reel. We will call this value ‘mu,’ and it’s a lower-case u with a lower descender at the beginning. Here, let me write it for you. It looks like this: µ.”

“Alright. I have that,” you nod. “The first variable we need is called mu, it’s a funky-looking u, and it is how many times the event will happen in a cycle.”

“Right,” Larry says. “Now, the third consideration is this: The probability that a success will occur is proportional to the size of the region.”

“Say what?”

“Well, if we have two cycles, then we have two jackpots. It works in proportion,” he explains. “If we have two cycles, we don’t have four jackpots, for example. If we have one-half of a cycle, then the probability of the jackpot is one-half. That’s all that it means. In other words, the probability won’t change as the region changes. Consider this: You enter a draw for a widescreen TV. If 10 people enter the draw, you have a 10 percent chance of winning it. If two people enter the draw, you have a 50 percent chance. If 1 million people enter—well, you get the idea. With the slot machine, the probability does not change. For every game played, the probability of you winning the jackpot is 1-in-884,736. That doesn’t change.”

“Right! It’s constant,” you say.

“Yes. And proportional to the size of the region,” Larry nods. “Finally, the fourth consideration is that the probability of the event occurring in a very small region is virtually zero. It must be a rare event. That makes it suitable for a jackpot or an infrequently occurring event. It won’t work for something like mixed bar symbols. We need it to be an event that is rare. If you sit down at the machine and play one game, the probability of you hitting the jackpot is virtually zero. You may hit it, but it’s exceptionally rare.”

“Perfect. We match each of the four items, then,” you say.

“There are a few other pieces of information that we need first,” Larry explains. “First of all, remember the number 2.71828.”

“That sounds familiar. That’s called g or something like that, right?” you ask.

“Actually, it’s e. It is the base of the natural logarithm system. All that we need to remember is that it’s a constant. Every time we do this calculation, no matter what other numbers we use, we always have to plug 2.71828 into the equation.”

“Got it. 2.71828. OK.”

“We also have to know x,” Larry explains. “That is the number of actual successes in a specific region. µ is the number we are supposed to have, in theory, and x is how many we actually will have. So, in the base of this machine, µ is 1 since we have one jackpot per cycle statistically—that’s right from the PAR sheet. x is 0 because, in reality, we haven’t hit that jackpot yet.”

“Alright,” you say. “What else?”

“That’s it. It’s a simple equation. It looks like this.” Larry draws the equation on a piece of paper (see Figure 2).

“It looks complicated,” he continues, “and if you saw it in a textbook, you might just flip the page. But it’s simple to use. Let’s take a look at our game. We are trying to determine the probability that we won’t have a jackpot in a cycle. Remember that e is 2.71828. We raise it to the exponent of -1. You’ll need a scientific calculator for that. Then multiply it by 1 raised to the exponent 0.”

“If I remember correctly, any number raised to the exponent 0 is 1,” you say.

“Yes. Then divide by a factorial.”

“Say what?”

“A factorial,” he repeats. It has an exclamation mark after the number. It’s quite simple, too. It means that you take the number, multiply it by the number one less, then multiply that value by the number two less, and continue until you multiply by 1. For example, 2! means that you multiply 2 x 1. Five factorial means you multiply 5 x 4 x 3 x 2 x 1. So, 50! would be … ?”

“50 x 49 x 48 x 47 and continue until you get to 1,” you say. “So, 5! equals 120?”

“You got it! See, it’s not hard,” Larry says. “So, in the example I showed you, the answer is 0.24659.” (See Figure 3).

“What’s that mean?” you ask.

“Well, that’s the probability of no jackpot occurring so far. We have had 1,238,630 games played, and dividing that by the cycle size of 884,736 gives us 1.4 cycles played. Since µ us the amount of cycles, we use the value 1.4 for µ, and x is 0 because we want the probability of no jackpot, which is what has actually happened on the floor. This means that if we take this game and put it on our floor, we can expect that in roughly one out of four cases, there won’t be a jackpot yet, even though on paper there will be one each cycle.”

“And that takes into account the randomness,” Larry continues. “Even though we have completed more than one cycle, there’s still a 25 percent chance that we won’t see a jackpot. Remember the bell curve that you drew? We’re on the far side of it, and there’s obviously still a chance for the jackpot to occur in this region.”

“Wow! That’s an actual number I can give to management,” you exclaim. “So it’s not wrong that we didn’t hit the jackpot in just over one cycle, then?”

“No. As long as you know that the game is functioning properly, mathematically you can show that it won’t necessarily have hit a jackpot up yet,” he explains. “If you had three more identical games, and each one had a cycle’s worth of games played, statistically three of them would have hit a jackpot and one wouldn’t have.”

“Suppose we have two cycles completed,” he continues. (See Figure 4.) “Once you start to use the formula, it’s actually fairly simple. With two cycles completed, the probability of having no jackpot drops to 13.5 percent. After four cycles, it’s almost never going to happen, at only 1.83 percent. That means that if we have 10 rows of five machines each of the same game, after four cycles there will only be one machine without a jackpot having hit. It’s still possible, but not very probable.”

“Now, we can also look at this a little differently,” Larry continues on. “Suppose the opposite had occurred, where there have been three jackpots hit in 1.4 cycles and management is complaining that the game is not playing correctly. We can look at that in our formula as well. In this case, µ would be 1.4, x would be 3, and our answer would be 0.112777. (See Figure 5.) There is just over a 10 percent chance that the game could have hit three times. It’s twice as likely to have hit 0 times. But all of these possibilities could occur. Some are more likely, and the formula lets us determine how likely each scenario is.”

“This formula could also be used for other calculations,” he noted. “It could be used for staffing requirements, for example. If we need one staff member for 100 machine faults in a day, and we know that machine faults occur once every so-many games, then we can determine the probability of having, say, 200 machine faults on a busy day.”

“We can also determine a cumulative probability,” he says. “Suppose this same Crackpot Jackpot game is used to look at different questions. For example, what is the probability that this jackpot will occur less than three times after three cycles? We need to determine the probability of it occurring no times, one time and two times. Our formula could be written like this: P(0,1,2) = P(0) + P(1) + P(2).” It’s actually a pretty straightforward calculation. Let me draw it for you.” (See Figure 6.)

“We can determine that the probability of having less than three primary jackpots is 42.32 percent,” he explains. “If we use the same formula, we determine that having five or fewer jackpots will occur almost 92 percent of the time. Therefore, having six or more jackpots would occur only 8 percent of the time. We start to get a good ‘feel’ for how things should fall into place, and our game certainly is working within statistical norms.”

“Just remember that we are studying one particular game,” Larry says, summing up his explanation. “When we have more games involved, such as the example of having 50 games I mentioned earlier, then we will see a bunch of them playing right where they should, and some playing on both sides of the standard. In other words, most games will hit the proper number of jackpots, some will hit more and some will hit fewer. Remember that at four cycles, the probability of no jackpot was 0.0183, or 1.8 percent. That means that the probability of one (or more) jackpot is 98.2 percent, calculated by subtracting 1.8 from 100. The probability of 100 of these games all hitting a jackpot would be 98.2^100, or 0.1626. You can see that there is only a 16 percent chance that all 100 games would hit the jackpot. That’s pretty low, and the reason is that when we have a large population of games, it is expected that at least one of them is going to be playing at the outside of the statistical norm. In this case, we are over 83 percent (100-16.26 = 83.74) sure that at least one game will not have hit the jackpot after three cycles. This is vital to remember because it looks bad. If you walk through the casino floor and look at the progressive jackpot meters of these games, then one will stand out. Where most of them are closer to the $10,000 starting mark, one will stand out at $16,000. At a casual glance it will appear that something is wrong with this machine. And the fact that this meter is so high could compound the problem. When people see that it has risen ‘too high,’ they may feel that there is a greater probability of it hitting, increasing the play level, and increasing the meter amount even quicker. Of course, just because the jackpot hasn’t hit, doesn’t change the probability of it hitting next game. It’s a constant. But we can discuss this later. For now, you have a way to give some accurate, quantifiable numbers to management to show that the game is working fine. Remember the law of probabilities: They will work themselves out if left alone long enough.”

Author’s Note: Using this formula lets us determine if a game is playing within a statistical norm and also to assess our risk. We can look at a very large, very infrequent jackpot and determine the risk of it occurring within the first month, for example. We can also take a situation where one meter sticks out and looks incorrect, and mathematically show that it is acceptable. By providing an actual numeric value, it gives credibility to our report and doesn’t rely on feelings or guesses.

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