Monday, December 31, 2007

An Introduction to the Mathematics of Bingo - Part I: An Investigation for Grades 7-12

While you're celebrating New Year's Eve (meaning you're probably not reading this blog!), or thinking about your favorite math teacher (and probably keeping it to yourself), or considering clicking on the new subscriber chiclets in the sidebar, I thought I would kick off 2008 with something different.

As we were playing family Bingo a few days ago with about three dozen families (my wife was reluctant to go and, of course, she won the first game!), I began thinking about the underlying mathematics of the game and its many variations. I know what my wife would be thinking: "Dave, Why can't you just enjoy the game without analyzing it!"


Here were some thoughts running around in my head, when I should have been concentrating on the two boards I was playing (I won nothing BTW):
(1) Historically: What are the origins of this game? Designed by some brilliant mathematician or was it just some game of chance that evolved?
(2) The number of possible boards must be astronomical. How many different boards are supplied by the companies that manufacture this product?
(3) If I buy a bingo game that comes with, say, 36 boards, are these same 36 boards in every box? If I buy a set from a different manufacturer will the boards overlap or be entirely different?
(4) Are all boards randomly generated by software these days? If a bingo game is to be played in a large hall, with hundreds or even thousands of players, how likely is it that there will be multiple winners in a single game? Do the boards all have different winning lines or are there lots of overlap among boards?

Some probability thoughts:

(5) What is the probability of a winner after the minimum number of balls drawn from the bingo cage, namely four numbers (don't forget the free space!). Since I've never seen this happen, I'm assuming the chances are virtually zero!
(6) More realistically, for about two dozen players, each playing a single board (or one player using 24 boards!), what the expected number of balls drawn before a winner occurs? I was conjecturing less than half of the seventy-five numbers, maybe low thirties.
(7) How did the probabilities change as one increases players and boards? I assumed many mathematicians had already solved all of the intricacies regarding the probability of a winner after 10 numbers, 20 numbers, 30 numbers, etc. Instinctively, I felt that this was a very sophisticated problem, probably beyond my comprehension, but I wanted to know more.

Of course, when we returned home, I did some online research of the game -- fascinating stuff: Origins in Italy, Lotto, Beano, Bingo, Mr. Lowe (the toy manufacturer), Professor Leffler from Columbia University, the fund-raising aspect that started in a church in Wilkes-Barre, PA, and so on...
You can easily find these same sources so I'll leave that for our readers. However, there was a dearth of serious mathematical analysis of the probabilities and the combinatorial aspects. I only found a couple of these and neither went into much explanation of the underlying theory, other than to suggest it is complicated, oh, and data tables generated by some software. Of course, I'm sure I missed some wonderful references that my readers will find.

So I decided to do what I usually do when facing a complicated task (a la Polya): Reduce it to a much simpler problem! Not only to understand it better for myself, but, in the back of my mind, I was thinking of how a middle schooler could begin to understand the complexities of all this.

What could be easier than a 2x2 board - just 4 little numbers on a card and to really oversimplify it, only four numbers will be available: 1,2; 3,4. The semicolon separates the possible values for the first column on the card from the 2nd column. I will use this notation from now on. So here's the first elementary question for the reader and for the student:

STUDENT/READER QUESTION #1:
Assume there is one player with one card using the numbers above. Explain why the probability of winning this simple 2x2 version after two numbers are called is 1, that is, 100%.
You're thinking: Way too obvious a place to start, right! Too boring for the student...

STUDENT/READER QUESTION #2:
Ok, let's dial it up a tad. We'll still keep it a 2x2 card, but, this time, there are three numbers available in each column: 1,2,3; 4,5,6. Remember this notation means that 1,2,3 are the possibilities for the first column and so on.
Again, one player with one card: What is the probability of a win after two numbers are called?
Comment: There are many methods here from listing all of the possibilities to permutations and combinations to multiplication of probabilities (one number at a time without replacement), etc. I believe, pedagogically, it is important for the student to see more than one way!

I could stay with the 2x2 game and add more numbers but the student and our readers are an impatient lot and want to move on to something more interesting, right? So let's move on to a 3x3 board which is much more like the 5x5 board in that it has a free square in the middle. But we have to start slowly here - trust me!

STUDENT/READER QUESTION #3:
Now we have a 3x3 board. The available numbers will be simply 1,2,3; 4,5,6; 7,8,9. Again, one player, one card. Couldn't be easier, right?
(a) What is the probability of a win after TWO numbers are called? That's the minimum number with the free space covered.
(b) A little harder now: What is the probability of a win after THREE numbers are called?

Ok, we'll ask the same two questions with more numbers available:
Suppose the possible numbers are: 1-6; 7-12; 13-18


(c)
Now, what is the probability of a win after TWO numbers are called?
(d)
What is the probability of a win after THREE numbers are called?

I better stop here! This is enough for Part I. As usual, any results I've stated need to be verified by my readers and don't forget to give proper attribution if using any of this in a classroom setting.

HAPPY 2008!

11 comments:

mathmom said...

(2) The number of possible boards must be astronomical.

Ok, that part is easy at least. Each column has 15 numbers that my be used in it. Four of the columns have five numbers, and one has four (with the free space). So the number of possible bingo cards is (15P5)^4 * (15P4) which is approximately 5.52 * 10^26 (which agrees with what wikipedia says).

I'm off for now, but I'll look back at the rest of it later.

Dave Marain said...

Happy'08, Mathmom!
I'd be surprised if anyone else gets to look at this today. Your analysis of the number of possible boards was dead-on. I chose not to make that one of the questions because it is searchable, however, I may add it to the questions for 2x2 and 3x3 boards or should I say matrices. I figured those questions are more standard and I was looking to delve into the tougher probability issues.

Also, it is important for students to recognize the significance of order in counting the number of possible boards since the columns are labeled. Of course one might argue that a B-column containing 1-2-3 is equivalent to 2-1-3 if these 3 numbers are called but changing the order affects other rows and diagonals. This gets very complicated, very quickly.

jd2718 said...

Dave, all of a sudden your posts reappeared in my reader. A New Year's gift?

The probability that there is a winner after 4 balls.

Method: we are going to add the probability that this happens with a vertical stripe (easier to calculate) with the probability that this happens with a horizontal or diagonal stripe (harder to calculate).

Vertical stripe (all N's)
We are looking for the probability of drawing 4 N's, times the probability of having the right set of N's.
P(all N's) = (15*14*13*12)/(65*64*63*62)
P(right group of N's) = 1/(C(15,4))

I will (foolishly?) simplify a bit:
1/[(65*64*63*62)(4*3*2*1)]

For the horizontal and diagonal cases, I will take the probability that we get BIGO and multiply by the probability that the B is an eligible spot on the board, times the probability the I is in the (only one now) eligible spot on the board...

P(one of each letter except N) = (15^4)/(65*64*63*62)

P(B is in an eligible spot) = 3/15
P(I,G,O are in the right places) = (1/15)^3

Multiplying and simplifying: 3/(65*64*63*62)

Well the sum of the cases is 4/[P(65,4)], suggesting I missed something obvious!

I kind of like when it ends clean like that, even if there were a ton of effort along the way.

Happy New Year!

Jonathan

Dave Marain said...

Happy New Year to you too, Jonathan!
Ok, here goes (this is one heck of a long response!):

1. I had redirected my post feed through Feedburner several weeks ago. Unfortunately, I burned more
than one feed as I was experimenting with the process and didn't update Blogger properly with the latest feed address. Those using Google Reader (including myself!) stopped receiving the feed. I probably lost a lot of readers that way. I corrected the problem yesterday and all should be ok - I hope!

2. I really appreciate your detailed explanation of the probability for winning after the minimum number of calls (4) in the 5x5 Bingo game. The final result suggests a one-step approach but, in solving probability problems, we often don't see the most efficient method the first time around. Pedagogically, students benefit much more from playing around with several approaches. In fact, even if I think I've solved a probability or combinatorial problem, I generally try to verify my results via another approach!

Now for the excruciating details:
First, I think you intended to use '75' wherever a '65' appeared in your calculation, but that's a minor point here.

Your explanation for P(Vertical) was essentially:
P(Vertical) = P(4 N's called in succession)⋅ P(matching the 4 N's on the card) =
[C(15,4)/C(75,4)] ⋅ [1/C(15,4)]
= 1/C(75,4).
This simplified form makes sense since you only have ONE chance that your N column will match the 4 numbers called (I'll explain below how we could also use permutations here rather than combinations). Thus, we really don't need to first compute the probability that 4 N's were called -- you simply have to determine the probability that your column of 4 numbers matches the 4 that are called in any order!

This logic applies equally well to the horizontal or the two diagonals. Thus, we could have simply argued that a single card has exactly FOUR chances of winning in 4 calls! This is then divided by C(75,4) to determine the probability. This result differs from 4/P(75,4) as you obtained. I believe if we use 4 in the numerator, we must use combinations in the denominator as I will now explain:

Let's apply the approach to the simplest case of a 2x2 card with exactly 4 numbers available. I think you would agree that any card in this game has a 100% chance of winning after 2 numbers are called! This is critical when we analyze whether to use permutations or combinations.

One can argue that a single card in this simplified 2x2 game has exactly 6 chances of winning after 2 numbers are called (2 columns, 2 rows, 2 diagonals) or 12 ways to win if order counts. The key is what we divide by:

Via permutations: 12/P(4,2) = 1
Via combinations: 6/C(4,2) = 1.
I know that many students (not to mention myself) often overlook this subtle point about whether one should use P or C when computing probabilities. Here's what I've generally told my students: If the numerator uses P (i.e., takes order into account), then your denominator better do the same! Similarly, for combinations.

For the 5x5 Bingo game, I chose to use combinations since it seemed more natural to me to cover the four squares in any order. However, one would obtain the same answer using permutations:
For a vertical win, you would have P(4,4) ways to win out of P(75,4) possible calls. This ratio reduces to 4!/[(75!)/(71!)] = 1/C(75,4).

Sorry for the long-winded explanation, Jonathan, but I believe these are very subtle ideas and that's why I reduced the problem to the simplest case which I could verify!

jd2718 said...

ok, in the New Year 5*15 = 75

That'll be a good start.

Thanks for untangling my strands. Once we got to 4 * easy number, I knew that there was untangling to do, but didn't have the concentration left to do it.

I should write about the teacher who taught me to do this stuff. P's, C's, !'s all over the place, just grab what's convenient. Or I'll do a full post...

I am much more careful when I teach. There are subtle differences between the quotient of the products and the product of the quotients (see how I miswrote the probability of getting 4 N's? It was easier to type, but that is not my actual calculation)

Jonathan

Dave Marain said...

Jonathan--
I'll forgive your one mental error if you forgive my hundreds!

If you want to acknowledge your teacher, pls do it in the comments section of my post from the other day -- I'd really like this post to grow over time as an ever-expanding tribute to our math teachers. We live at a time when there's an award for almost everybody, but Teachers of the Year, Presidential Award Winners for Math and Science, etc., only recognize a miniscule number of the wonderful dedicated teachers out there. Some of the most truly deserving educators are often overlooked. I'll probably copy this paragraph into my original post!

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