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Oh, New York, Bring Back Those Big Dippers

1 August 1999

By Donald Catlin

    Every Friday afternoon, a few of my cronies and I head to the local watering hole to down a couple of beers and swap the latest jokes.  In this particular establishment, as in saloons all across the state, the Massachusetts State Lottery runs a Keno booth.  I'm always amazed at how many people spend their money playing this miserable game.  I have tried to dissuade friends from playing Keno by telling them about the house edge on the game, but my words fall on deaf ears. I'm not sure they even know what I'm talking about (or perhaps they don't think I know what I'm talking about).  Anyway, the game sure is popular.  Massachusetts isn't the only state that has discovered this money machine.  In our neighboring state New York, the New York State Lottery has also discovered Keno but they don't call it Keno; they call it Quick Draw.  It's good old Keno though.  I am going to concentrate on the Quick Draw game in this article and you'll see why in a moment.

    During the month of November in 1997, the New York Lottery had a promotion using the Quick Draw game.  The following  is a direct quote taken from a table card advertising the special promotion:

"Win a double dip on Big Dipper Wednesdays.  During our 'Big Dipper Wednesday Special' promotion November 5, 12, 19, and 26, prizes for all winning Quick Draw 4-spot tickets will be doubled!"

Is this a good deal for the player or just a come-on for a sucker bet.  Let's take a look.

    Keno is a game in which the player chooses a fixed set of numbers from among 80 numbers, 1 through 80.  The numbers are displayed on an 8 by 10 matrix and the player marks his or her choices.  How many numbers?  Well, in some casinos you can choose 1 to 20 numbers.  The Massachusetts Lottery restricts it to 1 through 12 and the Quick Draw game allows 1 through 10.  Because of the Big Dipper promotion, we are going to look at the 4 Spot Quick Draw game, that is, the player selects 4 of the 80 numbers.  Once the player chooses the 4 numbers, he or she has the numbers 'certified' by the local lottery agent and waits for the lottery to randomly choose 20 of the 80 numbers.  In some Keno games this is accomplished using 80 Ping Pong balls and a blower machine such as that used in Bingo; in others it is accomplished electronically using a random number generator.  The object, for the player, is to match as many of the 20 'house' numbers with the player's four chosen numbers as possible.  In New York, here is how the payoffs are made:
 

4 Spot Game
 Numbers Matched
Prize per $1 played 
4
$55 
 3
 $5
 2
 $1
Overall Chances of Winning  1: 3.86

The New York State Lottery generously gives the player the chances of winning (as does Massachusetts).  Unfortunately, as pointed out in my column last month, this is a useless piece of information; asking the probability of winning is the wrong question.  What is worse, for the 4 Spot Game, both states give the wrong answer to this (wrong) question.  From last month's article you should know the right question.  Here is the mathematics necessary to answer it.

    How many ways can we put n objects into k arrangements?  (Note that we must assume that k is less than or equal to n.)  Here is a specific example.  If we have 7 paintings and 4 places on a museum wall on which to hang them, then how many different arrangements are there?  Let the boxes below indicate the spaces on the wall.
 

Space 1
Space 2
Space 3
Space 4

There are 7 choices we can make to fill space 1, and  for each of those choices there are 6 ways (paintings left) to fill space 2, so altogether there are 7 x 6 ways to fill the first two spaces.  Now for each of the 7 x 6 ways to fill the first two spaces there are 5 choices left for space 3, so there are 7 x 6 x 5 ways to fill the first three spaces.  Finally,  for each of the 7 x 6 x 6 ways to fill the first three spaces there are 4 ways to fill the fourth, so the answer to our question is 7 x 6 x 5 x 4 or 840 arrangements.  One often uses the symbol P(7, 4) to represent this number; it is read "the number of  permutations of 7 objects taken 4 at a time."  What about the general question of calculating P(n, k)?  I think you can see by analogy with our painting example that the answer is:

P(n, k) = n(n - 1)(n - 2)(n - 3) ... (n - k + 1) (1)

There is a compact way to write expression (1) using something mathematicians call a factorial.  Simply put, the symbol n!, which is read "n factorial", is defined by

n! = n(n - 1)(n - 2)(n - 3) x ...x 2 x 1   for n not zero

0! = 1 (looks strange but, as you'll see, it works)
(2)

For example, 4! would be 4 x 3 x 2 x 1 or 24.  Using this notation, P(7, 4) can easily be written as follows:

P(7, 4) = 7!/3! = (7 x 6 x 5 x 4 x 3 x 2 x 1)/(3 x 2 x 1) = 7 x 6 x 5 x 4 (3)

where the 3 x 2 x 1 on the top and bottom of the fraction cancel leaving us with the correct number.  By analogy with (3), P(n, k) can be written as

P(n, k) = n!/(n - k)! (4)

Now from (1) we see that the number of orderings of n objects in n spaces, P(n, n), is just n!  Notice that (4) gives us this same result as well:

P(n, n) = n!/(n - n)! = n!/0! = n!/1 = n! (5)

Now you see why we adopted the convention that 0! = 1 in (2).

    Very well, we have solved the problem of how many ways we can arrange n objects in k positions.  Suppose we now ask a different question: How many ways can we choose k objects from n objects?  In this case we don't care about the ordering, only which objects are picked.  We will go at this question in a somewhat indirect fashion; I'll use the 'paintings on the museum wall' to illustrate the idea. 

    Imagine that we are still interested in counting the number of ways to arrange the 7 paintings on 4 wall spaces, but the paintings are stored in a warehouse that is some distance across town from the museum wall on which they are to be hung.  In this case each arrangement is achieved by selecting 4 paintings from the 7, loading them in my van, hauling them across town, and then arranging them on the wall once I arrive. 

    There is a principle in mathematics that says if you count the number of elements in a set two different ways, then you damn well better get the same number either way.  What if I count P(7, 4) as follows?  First I'll count how many ways I can select 4 paintings from the 7 in the warehouse to load in my van.  Since I don't know this number let's just use the symbol C(7, 4) for it.  Now, after I arrive at the museum I have 4 paintings to arrange in 4 spaces on the wall.  We already know that there are 4! ways to do this.  So, for each of the C(7, 4) possible choices to load in my van there are 4! museum arrangements that I can make with this particular choice of pictures. Altogether there are C(7, 4) x 4! arrangements I can make with 4 paintings chosen from the 7.  Wait a minute, that is just the number P(7, 4) that we calculated earlier.  In other words, we have just argued that

C(7, 4) x 4! = P(7, 4) (6)

or by (4)

C(7, 4) x 4! = 7!/(7 - 4)! (7)

How about that?  All we have to do is divide both sides of (7) by 4! and we'll have a formula for C(7, 4).  The result is:

C(7, 4) = 7!/[4!(7 - 4)!] = 7!/[4!3!] = 35 (8)

I think you can see that if there were n paintings in the warehouse and k spaces on the museum wall, the argument just given would produce the formula

C(n, k) = n! / [k!(n - k)!] (9)

Now we can count the number of  (unordered) sets of size k that can be selected from a set of size n.  For example, how many different 5 card Poker hands can one select from a 52 card deck?  Easy, its just C(52, 5).  The number is calculated as follows:

C(52, 5) = 52!/(5!47!) = (52 x 51 x 50 x 49 x 48)/(5 x 4 x 3 x 2 x 1) = 2,598,960 (10)

The number C(n, k) is referred to as the number of combinations of n objects taken k at a time.  The formula (9) will be used over and over in future articles, so you might want to tuck it away somewhere.

    Consider a committee comprised of five men and four women.  If we randomly pick a subcommittee of 5 people, what is the probability that the committee consists of exactly three men and two women? 

    There are nine people on the committee so, according to our discussion above, there are C(9,5) different 5 person subcommittees we could form.  According to (9), this number is 126.  There are C(5, 3) ways of choosing three men and for each of these choices there are C(4, 2) ways of choosing two women, so there  are C(5, 3) x C(4, 2) ways of  choosing a subcommittee with exactly three men and two women.  According to (9), this number is 10 x 6 or 60, so the probability of choosing exactly three men and two women is 60/126 or approximately 0.4762.  In general, if we partition a set S into two pieces R and T  having r and t elements, respectively (note that necessarily the set S is of size r + t), then the probability p of randomly choosing a subset having u elements from R and v elements from T is given by

p = [C(r, u) x C(t, v)] / C(r + t, u + v) (11)

This particular type of probability density has a dollar-and-a-half name: the hypergeometric distribution.  Forget that, we'll just refer to it as the Keno Calculation.  Here's why.

    For purposes of calculation it will be convenient to think of the Keno game as being played in a slightly different manner than it really is.  Specifically, let's imagine that the lottery picks its 20 numbers prior to the player picking his or her spots but doesn't announce the 20 numbers until all player cards have been certified.  From the player's point of view, this game is equivalent to the actual Keno game.  Okay, we're all set to Keno Calculate.  What is the probability in the 4 Spot Game of getting 3 matches?  Well, the 20 house numbers partition the 80 Keno numbers into two sets, 20 winning matches and 60 losing matches.  Using (11), the probability p3 of getting 3 of the 20 and 1 of the 60 is just

p3 = [C(20, 3) x C(60, 1)] / C(80, 4) (12)

Let me carry out the calculation for you.  There are some very large and very small numbers possible in calculations arising from the general formula in (11) and, for purposes of accuracy, it helps to carry things out in the following manner.  First, writing out the three terms in (12) we have

p3 = 20 x 19 x 18
3 x 2 x 1
 X  60
1
 X  4  x  3  x  2  x  1
80 x 79 x 78 x 77
(13)

Notice that 20/80 puts a 4 on the bottom, so we can cancel the resulting 4 x 3 x 2 x 1 terms on the top and bottom of the fraction and be left with

p3 = (19 x 18 x 60) /(79 x 78 x77) ~ 0.04325 (14)

Rather than multiplying out the top and the bottom and then dividing, it is safer to divide the 19 by the 79, multiply the result by 18, divide this by 78, multiply by 60, and finally divide by 77.  Calculating in this manner keeps things in ranges where the calculator truncation errors will be minimized.  In (14) this wouldn't really be a concern, but for some of these Keno calculations it can be.  Better safe than sorry.

    The reduced sample space for the Quick Draw 4 Spot Game has five outcomes, 0 matches through 4 matches.  The probabilities for each of the outcomes are calculated as I have just shown you.  The payoff is one less than the prize paid, that is, though you are paid $55 for 4 matches, it costs you a dollar to play so your actual profit is just $54.  Here is a table showing the probabilities, the payoff function, and the house edge calculation as we discussed last month:
 

Outcome
Probability
Payoff
Product
  4 Matches     0.00306          54     0.16524
  3 Matches     0.04325           4     0.17300
  2 Matches     0.21264           0     0.00000
  1 Match     0.43273          -1    -0.43273
  0 Matches     0.30832          -1    -0.30832
    1.00000      Total    -0.40281

Well, there it is; the house edge on the 4 Spot Quick Draw Game is 40.281%.  Not too good, is it?  Let me show you something else.  If we add up the probabilities for 2, 3, and 4 matches the figure is 0.255895.  This represents a 1 in 3.86 chance of getting one of these outcomes.  The New York State Lottery advertises this as the chance of winning (Massachusetts pulls the same stunt).  Wait a minute!  I don't see any win when 2 matches occur; that is a push.  The actual chance of winning something in this game is 1 in 21.59.  Of course, as I pointed out last month, this figure is really irrelevant when it comes to evaluating the game.  The house edge is the relevant figure and we have calculated it.

    Now, what about this Big Dipper Wednesday?  As you can see from the following table, doubling the payoffs on the 4 Spot Game turns two matches into a winner (making the 1 in 3.86 figure correct, by the way).  This has a very dramatic effect on the house edge.
 

Outcome
Probability
Payoff
Product
  4 Matches     0.00306         109     0.33354
  3 Matches     0.04325           9     0.38925
  2 Matches     0.21264           1     0.21264
  1 Match     0.43273          -1    -0.43273
  0 Matches     0.30832          -1    -0.30832
    1.00000      Total     0.19438

The house edge is now - 19.438%, that is, the player enjoys a 19.438% edge over the house in this game.  This game ran every Wednesday in November 1997.  Incredible!  How did I fare in this game?  I learned of this promotion in January of 1998, so my profit was one big goose egg!  The people who tipped me off, however, did just great.  See you next month.

Donald Catlin
Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers