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Bingo pinball

Mr. Phil Bogema, on how to properly play the Bingos

One day not to long ago Phil writes me on how surprised he is to have stumbled across my Bingo pages and see some entries he made years ago on the Usernet site rec.games.pinball

Although true, the surprise was actually mine.

In one of his emails he mentioned that he has never really seen any reference on how to play properly these machines.

Dah! So I immediately took him up on his kind offer to explain the mechanics.

Please remember folks these are skill games. Yes! Skill games - even though many people would argue this fact since the Bingos never had flippers.

As you historians and trivia buffs know: The advent of the flipper was really all about crushing the myth and life of the illicit gambling machines.

Don Hooker (The father of Bingos) was known to talk about how part of his vision was to entice the player into playing the Bingos by bringing them into the game.

By involving them. By have odds and features that they had to choose and decide upon to increase the odds and make the needed combinations to win.

He felt that this made someone proud when they won, that they felt it wasn't pure chance but their decisions that influenced the game.

I some what embellish and paraphrase, but I think that was the point of his message.

When I read Phil's text below - to me it simply inforces this:


Hi Danny,

I have been thinking about the "playing the bingos" problem. I just used to do it, so it is difficult to put into writing. Here is a short dissertation I wrote about the base probabilities of winning on a bingo. Please let me know if you need any more info to make this clearer - as I said, I am no writer.
Here goes:

The calculation of the basic probability of winning on a bingo pinball machine is an interesting (and moderately complex) problem. Most players never bothered; I'll bet most never even thought about it. I have always approached games of chance with an abnormal curiosity about the math involved, mainly because I don't enjoy losing.

Since I played the 25 hole machines, that is what will be tackled first. Then we'll look at the 20 hole machines.

First, I have to lay the foundation. For purposes of this discussion, I am assuming every number has an equal probability of being "hit". I know, I know. That's patently absurd, and anyone who ever played knows it, too. The thing is, no two machines are alike, and this is the most straightforward and understandable explanation I can make. (In fact, the math involved when the chances of the various numbers differ becomes outrageously complex. Good luck if you decide to undertake it.)

Okay, that said, the way to start is to figure how many ways the balls can be arranged. In other words, how many ways can you take 25 things 5 at a time? The answer is 25! / (20!*5!) = 53,130. If you don't understand the notations, don't worry about it. This isn't a math lesson. Just keep in mind that there are 53,130 ways you can arrange 5 bingo balls in 25 holes. We don't care what order they drop in the holes, just what the configuration is when all is said and done.
That was easy. Now we need to see how many winners there are out there.

Starting with the five-baggers, we can count on two hands the vertical and horizontal winners - 5 up and 5 across. Oh, there are those two diagonal lines, too, so that makes 12 ways in all. Now you know why you almost never made a natural five-in-line. The odds against it are 4,426.5 to 1.
Now the fun starts. If you approach the problem of the four-baggers in somewhat the same manner, you will count 10 up, 10 across, and 4 on the diagonals, or 24 ways. But what about that fifth ball? It has to land somewhere. Even though there are 21 holes left to hit, there are only 20 holes it can land in and keep the result a four-in-line - the other one makes it a fiver and we already counted it. So, now we multiply the 24 ways times the 20 useless holes and get 480 combinations. Odds ? Try 109.6875 to 1.

Even that wasn't too bad, but now things get really hairy. The three-in -lines present a real challenge. Counting how many winning combinations of three balls there are, we note that each line has three potential winners - the first, second and third positions; the second, third and fourth positions; and the third, fourth, and fifth positions. Since there are 12 lines (again, 5 horizontal, 5 vertical, and 2 diagonal), we take 12 times 3 and come up with 36 ways to win.
Of the 36 ways, 12 sit in the middle of the line (i.e., the second, third and fourth positions), and 24 start or end with a number on end of the line.

Of the set of 24, there is one number that can't factor into the final equation, and that is the one that would make a four-in-line. Remember, we have already counted that. So, there are 21 numbers left to fill and 2 balls to do it with. More notations : 21! / (19! * 2!) = 210. There are 210 ways to drop the other 2 balls without improving the three bagger, and 24 times 210 equals 5,040 ways to hit the outside three-in-line.

The 12 winning combinations that sit in the middle have only 20 numbers that don't improve the situation, because the two numbers on the end of the line would make a four-in-line. Using the same logic, 20! / (18! * 2!) = 190; 190 times 12 = 2,280 ways to hit the inside three-in-line.
So, adding the 5,040 and the 2,280 means there are 7,320 ways to hit a three-in-line. Right? No, now things get even more confusing. See, of the 7,320 ways, there are a number of double hits - 2 three-in-line hits in the same game. How many? At this point, my head about exploded since I am not a mathematician. I contemplated the analytical solution for some time before I gave up and counted. It turns out there are 166 ways this can happen, so the 7,320 has to be reduced by 166 to arrive at the final answer.

There are 7,154 different combinations resulting in at least one three-in-line hit. That means in the long-run, you'll hit a 3 bagger 13.4651% of the time.

The answer's right in front of us now. All we need to do is multiply the winning combinations by the number of games and add them. Okay, 12 times 75 games for the five-in-lines equals 900; 480 times 16 for the four-in-lines is 7,680; and 7,154 times 4 for the three-in-lines comes to 28,616. The total expectation is 900 + 7,680 + 28,616 = 37,196. Divide that by 53,130 and we find the payout percentage to be 70.0094%. That is for the older games without triple deck scoring. The newer games cause even more headaches.

The calculation we just did ignores what happens when there are two three-in-line hits in two different colors. Starting with Big Show in 1956, all of the games had the triple decks. Therefore, it would be nothing short of negligence to not perform this calculation as well. Again, I know of no way to do this easily with mathematical formulas. I didn't feel like racking my brain through all of the counting again, and I was worried I may have made an error, anyway. To do it this time, I decided to cheat and write a computer program to do the work for me. It works like sort of a digital search routine, and goes through every single one of the 53,130 combinations, recording the score in a database.

Here are the results for three-in-line hits:

Red line, single: 2,375
Yellow line, single: 2,375
Green line, single: 2,310
Red and Yellow: 18
Red and Green: 38
Yellow and Green: 38

The total adds to 7,154, which is the total number of hits we calculated above (Good Thing!). However, now we have 7,060 hits paying 4 games and 94 paying 8 (i.e., 2 three-in-line scores). Multiplying again, the number is 7,060 times 4 plus 94 times 8, or 28,992.

With that number, it is an easy proposition to figure the payout percentage on a triple deck game, and it turns out to be 70.7171%, an improvement of .7077%.

What about the 20 hole machines? After all of the problems I had calculating the 25 hole results, this was nothing more than a little relaxation.

There are 15,504 ways to drop 5 balls into 20 holes. That's 20! / (15! * 5!), if you want to revisit the formula.

This time, there are only 4 ways the balls can be combined for a five-bagger. (That's in the color sections - red, green, blue, and yellow).

To achieve a the four-bagger (I still want to call it a four-in-line, but the 20 hole machines are just sections), 4 balls must be in one color and the remaining ball has to land in a useless hole, meaning anything but the associated star. That means there are 15 holes to keep it from being a five-bagger (which we already counted). Multiplying 1 way to arrange the 4 balls in the color, times 4 color sections, times 15 losing numbers on the fifth ball, and the answer is 60 ways.
The three-bagger follows the same logic. There are 4 ways to arrange 3 balls in four holes (the color sections).

There are 120 ways to arrange the other 2 balls in the 16 holes that don't improve the winner. Remember, that star number is no good unless all four of the other numbers are plugged. Doing the multiplication again, 4 ways to arrange the 3 balls in the color, times 4 color sections, times 120 ways to arrange the losers equals 1,920 three-bagger winners.

Now, doing the same thing we did on the 25 hole machines, 4 times 75 games for the fiver is 300; 60 times 16 games is 960 for the four-baggers; and 4 time 1,920 is 7,680 for the three-baggers. Adding, we see the normal expectation after 15,504 games is a return of 300 + 960 + 7,680, or 8,940 games. Divide 8,940 by 15,504, and the answer is 57.6625%. A lousy 57.6625%! This is a full 13.0546% less than the triple deck 25 hole games. If you ever played the 20 hole machines and wondered why you were having so much trouble winning anything, now you know.

With those odds, why bother with the 20 hole machines at all? I never really liked the concept behind them, anyway, so?back to the 25 hole machines again. The next trick is to see what happens to those odds when more coins are deposited. I have to add the further assumption now that no features will come on when these extra coins are played.

With 2 coins in, and odds at 6/20/75, the payout percentage on one-color games is 50.2767%; on triple-deck games it is 50.8075%. That's starting to look pretty atrocious! What happens when another coin is dropped in? The odds go to 8/24/96, and your return dwindles to 43.8572% for older games, and 44.3290% for the newer ones. This is so bad, it's almost offensive. Just for laughs, I wanted to see what the pay-back period was for the operators at this level. The Bally price list for Laguna Beach, dated June 1, 1960, demands $1,075 to buy the machine. With three coins in and that 44.3290% tugging away, the player can expect to lose 8.3507 cents per game (on a nickel machine). Divide the $1,075 by .083507, and the answer is about 12,873 games. I would guess that's about 430 hours of play, so if the operator could keep the game going for 4 hours a day, he could recoup his investment in about 3 1/2 months if he didn't have to split the take with the establishment owner.

Then it's all gravy! Did you ever notice how a lot of the bingo operators drove around in nice cars and wore big rings? No mystery, eh?

If you use the blue pick-a-play button on your second coin to jump the odds to 8/24/96, your expected return "only" dips to 66.4935%. But you only have to put in two coins! Now you can play 32,083 games before the operator breaks even, and that will take about 8 1/2 months. (Laguna didn't have pick-a-play?.let's pretend this was a Circus Queen and the price was the same.) At any rate, you now have an operator starting to get agitated with you, because even though you are losing, you aren't losing as quickly.

Thanks for your interest,
Phil Bogema
Denver, CO


Last Updated 4/2/2002


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