Reelin’ In the Hits, Pt. IV: Free Software Day

Sitting at your desk, reading the latest issue of Casino Enterprise Management, you hear a knock at your door. Looking up, you see Larry Foutz, your helper and go-to-guy who saved the day when you needed to explain why the Crackpot Jackpot machine had never hit the progressive jackpot.

“Hey, come on in. Have a seat. I was just about to call you. I had a question about the calculation,” you say.

“Sure thing. And afterward, I have a small present for you.”

“For me? Really?”

“Absolutely. Did you get your report filed? Is management happy?” Larry asks.

“Yes they are. Management’s happy. The gaming commission’s happy. Everyone’s happy. Well, I’m not so sure that the customer believes us, but that’s the way things go. And I couldn’t have done it without you. Thanks again for all of your help.”

“My pleasure.”

“I think I’ll be called upon to do a lot more analysis on the machines from now on. But I am just a little worried.”

“Worried? What’s wrong?”

“I’m not sure the formula works correctly—or that I’m using the formula correctly. I mean, when you went over everything it was great. But I tried a few things out on my own. For the most part everything worked, but there’s something that I just don’t understand.”

“What’s that?” asks Larry.

“Well, one of the first calculations that we did determined the probability of a jackpot not occurring within one cycle. Or, more accurately, the probability of zero jackpots in one cycle. It came out to 36.79 percent. I remember that you wrote down ‘P(0) =’ meaning the probability of 0 jackpots.”

“Correct. And ‘p’ equalled 0.3678797, which, when rounded, gives you 36.79 percent.”

“Yes, so there is just over a one-third chance that there will be no jackpot in a cycle that mathematically states there is one jackpot possibility in that cycle.”

“True,” continues Larry. “‘There is one jackpot outcome in that cycle. However, given the random nature and that the slot machine does not work only within a prescribed cycle, you could have two or three jackpots occurring in that cycle, or none at all.”

“Absolutely. You then showed me how to calculate if there would be a jackpot. Instead of writing ‘P(1) =’ and calculating the answer, you wrote ‘1 – P(0)’ and subtracted the 0.3679 from 1 giving an answer of 0.6321. You said that if there is a 36.79 percent chance of no jackpot, then the chance of having a jackpot would be 63.21 and that both answers add up to 100 percent, meaning that there must either be a jackpot or not be a jackpot.”

“Correct. I’m not sure what your question is, however.”

“Aahh, well, I see how you got the result. And it makes sense, too. But I tried some calculations on my own and I came up with something different. I can’t explain it and it doesn’t seem to be right. I calculated ‘P(1)’ for the probability of having a jackpot and got 0.3679, which is the exact same result of no jackpot. It doesn’t make sense. The probability of zero jackpots and the probability of a jackpot are the same. But if I add them up, I get 0.7358. There’s still 0.2642 missing. I’m confused.”

“No, you’re pretty much right on the mark,” comforts Larry. “What you are doing is using subtle variances in wording. P(0) means?”

“The probability of zero jackpots,” you answer.

“And P(1) means?”

“The probability of a jackpot.”

“Correct. But when you say ‘a jackpot’ you are specifically meaning ‘one’ jackpot.”

“OK, but earlier we calculated the probability of one jackpot to be 0.6321. And now I calculate it to be 0.3679.”

“No, that’s not quite right. You calculated the probability of ‘one’ jackpot to be 0.3679. You calculated the probability of ‘a’ jackpot to be 0.6321. And when you said ‘a’ you meant ‘any’ jackpot, which includes one, two, three or more jackpots.”

“I did?” you answer, looking confused.

“It’s not as difficult as you think. It’s all in the wording. So far, your calculations are correct:

P(0) = 0.3679
P(1) = 0.3679
P(n) = 0.6321

P(n) means ‘any number of.’ We just need to make more calculations and you will see what is happening. Here, let’s work it out on paper.”

Larry makes several calculations, as shown in Figure 1.

“Now, these calculations will explain everything,” Larry says. “See on the left, I have written P(n) and below the numbers from 0 through 8. Beside these numbers are the answer from the Poisson Distribution Calculation. If we calculate P(2) then the answer is 0.1839.”

“What about the other column. You have written ‘Difference from 0.6321’”

“We calculated the probability of having a jackpot as 0.6321, meaning ‘any’ jackpot. What I’m doing is subtracting the answer from P(1) from 0.6321 to get 0.2642, which is what you said was still missing. Next I calculate the probability of two jackpots in the cycle, P(2), which is 0.1839. Subtracting that answer, the missing amount is reduced to 0.0803. Subtracting the probability of three jackpots reduces this amount to 0.0190. By the time we’ve calculated seven jackpots, the probability is so low that it becomes insignificant to us, and our missing amount is 0. This is what has happened. We have a 36.79 percent chance of having one jackpot in the cycle. The chance of two jackpots is, understandably, less; it’s only 18.39 percent. The chance of three jackpots is even less—only 6.13 percent. We continue and when we get to eight jackpots, it’s only 0.001 percent. In a long-way around, that’s your answer. The chance of ‘any’ jackpot is the cumulative sum of one jackpot, plus two jackpots, plus three jackpots, etc. Instead of calculating P(1) + P(2) + P(3) … we just subtract P(0) from 1 giving us the probability of ‘any’ jackpot occurring. Remember that even though the probability of 10 jackpots occurring in one cycle is infinitely tiny, it still could happen.”

“When I calculated P(1), then,” you answer, “I hadn’t taken into account two jackpots. But when I calculated 1-P(0) I was taking into account the probability of any number of jackpots occurring.”

“That’s right, and that’s where your discrepancy arose. It really was all a matter of using the correct words to describe what you were calculating. Your calculations were correct.”
“Now I get it. Thanks, it’s clear now. Hey, you said something about a surprise? I like surprises!”

“Well, you’ll like this one, then. I wrote a small program for you. It’s an executable that runs under Windows®. There are no dlls, libraries or anything required. It’s called POISSON.EXE and you can download it and put it in a folder or on your desktop. You don’t even need to install it. Just run the program directly. If you want to remove it at any point, and for the life of me I can’t understand why you would,” jokes Larry, “just delete the file and it’s gone. It doesn’t save data to your hard drive or write to the registry. It’s actually quite simple.”

“Great. What’s it do?”

“It calculates the Poisson distribution for you. You simply provide up to four pieces of information and hit the Calculate button. It shows you the formula on-screen, as well as the intermediate calculation and the final answer. Let me show you.”

This download is available from Save the file to the desktop or to a folder of your choice. To run the program, simply double click on the program and it loads. The screen looks like that shown in Figure 2.

As a special bonus, I have created a simple four-function calculator with a percentage calculation. It works just like a normal calculator. If you click on the calculator itself, you can use your keyboard. The equals (=) key is accessed by pressing the Enter key on your keyboard.

The basic Poisson calculator is quite simple. The formula is shown near the bottom of the screen and when it first loads, it will show the equation. After you press the Calculate button, the formula will be displayed with the appropriate numbers.

There are two values that you must provide for the formula. You need to provide “μ” and “x”. The top three boxes are used for “μ” and the bottom box for “x”.

We have to indicate how many units are going to be used in the calculation. If we know this value, such as one cycle, we can put that in the box to the right of the “Units” label. Otherwise, we can have the program calculate this value for us. In the example we looked at previously, when calculating Crackpot Jackpots, we had 1,238,630 games played and the game cycle was 884,736. Enter “1238630″ in the box beside the “Number of games in target” label. Beside the “Number of games in cycle” box, enter “884736″. The program will automatically calculate the “Units” value, changing it as we type. If we enter these values, the units will be 1.4, representing 1.4 cycles.

Below this we would enter the value of “x,” which means we are looking for the probability of the event occurring exactly “x” times. In our previous example we were looking at the probability of no jackpots, so we enter “0” in this box.

To the left is a button marked Calculate. Press this and the answers are displayed. Figure 3 shows the results of the calculation.

The formula changes to reflect the values which are being used in the calculation. The intermediary calculations are shown as well. The result of “2.71878 raised to -1.4″ is 0.246597. Below this the final answer is shown. The calculations are also displayed in the small box to the left, directly under the Calculate button.

“Have fun with the utility and see how it works for you,” Larry says.

“Looks good. I’ll try it out,” you say.

I could add additional features, such as the calculation of multiple probability (e.g., probability of four or fewer jackpots), as well as the ability to print out the calculations. If there is interest in this, I will expand the functionality of the utility. Please report any bugs or send your comments to

Leave a Comment