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Best of Donald Catlin  # Right Question - Right Answer (Hopefully)

7 July 1999

Last month, I looked at a letter sent in to the popular Ask Marilyn column of Parade Magazine.  This letter was about the old game of Chuck-a-Luck.  The writer, who I called Chuck, thought that the chances of winning the game were 50% or better.  Chuck's argument contained an age-old error which we'll review in a moment.  I solved the question of calculating the winning probability and near the end of the article pointed out that it was the wrong question.  In this article we are going to see why that is so and figure out what the right question is.  First a bit of background.

Last month I rather loosely defined the notion of a probability density on a sample space S (the set of outcomes).  This month I want to be more specific.  Given a sample space S = {x1 , x2, x3, ... , xn}, a probability density function (pdf) p on S, written as p (xi), such that the following two conditions hold:

 (i)   For each i, p(xi) is a number between 0 and 1.   (ii)    p(x1) + p(x 2) + p(x3) + ... + p(xn ) = 1 (1)(2)
p(xi) is read as "the probability of x i ."  For finite sample spaces like ours, 0 probability represents impossibility and probability 1 represents certainty.  In view of our discussion last month, condition (1) certainly makes sense.  What about (2), what is that for?  The answer lies with some other ideas that we introduced last month, namely, an event and the probability measure of an event.  An event is just a subset of S.  Suppose that the event E = {y1, y 2, y3, ... , ym}, that is, y 1, y2, y3, ... , ym are all of the outcomes that comprise E.  Then we define the probability measure of E, written P(E), by the expression
 P (E) = p(y1) + p(y2) + p(y3) + ... + p(ym) (3)
Put simply, (3) says that P(E) is just the sum of the probabilities of all of the outcomes in E.  This said, let me raise the question, what is P(S)?  As I said in my column in The New Chance and Circumstance while addressing this question (Spring 1999 issue), there is an old joke about a guy who comes home and finds a man standing in his bedroom closet.  He says, "What the hell are you doing in my closet?"  The man answers, "Everyone's gotta be somewhere."  The point is that S contains every possible outcome, so whatever outcome occurs, the event S has occurred with certainty.  Thus we would want
 P (S) = 1 (4)
Referring to (3), condition (2) assures us that (4) holds; that is exactly why condition (2) is stipulated and it is the only reason why.

Returning to the Chuck-a-Luck game, I'll let you refer to last month's article for the rules and payoffs; you can also find this information in the Summer 1999 issue of NC&C.  Last month I defined three events E1, E2, and E3 as the events that, respectively, exactly 1, exactly 2, or exactly 3 fives occur on the three dice.  I also showed that P(E1 ) = 75/216, P(E2) = 15/216,  and P(E 3) = 1/216.  In addition, I made the general observation that if any collection of events E1, E2, ... , Em is pairwise disjoint, that is no two sets have any outcomes in common, then the event E1or E 2 or ... or Em (the set of outcomes that are in at least one, maybe more, of the sets in the collection) can be calculated by:

 P (E1or E2 or ... or Em ) = P(E1) + P(E2) + ... + P(Em) (5)
The hypothesis of pairwise disjointness is essential; if even one pair of sets have an outcome in common, the right side of (5) will be larger than the left.  Chuck, though not explicitly, did implicitly define three events D1, D2, and D3 as being the occurrence of a 5 on, respectively, die 1, die 2, or die 3.  He then correctly observed that P(Di) = 1/6 for each i.  Chuck then tried to calculate the probability of a win, namely P(D1 or D2 orD 3), by adding the probabilities of the individual events and in doing so obtained 1/2.  Unfortunately, these events are not pairwise disjoint, so (5) doesn't hold.  This was Chuck's classic error.  On the other hand, since the three sets I defined above, E1, E2 , and E3, do include all of the winning outcomes for a bet on five in the Chuck-a-Luck game and are pairwise disjoint as well, we can calculate the probability of a win at Chuck-a-Luck by calculating P( E1 orE2 orE3) using (5) and obtain
 P (E1 or E2 orE3) = 75/216 + 15/216 + 1/216 = 91/216 (6)
Note that this result is considerably less than Chuck's 50%.   This is essentially where last month's column ended.

Let me convince you that Chuck-a-Luck is not a very good game for the player.  Rather than a single bet on the five, let's put a dollar bet on every one of the numbers 1 through 6 and see what happens.  Suppose that all three dice turn up different from one another.  Then the house pays out \$3 on three of the numbers and collects \$3 on the other three numbers for a push.  On the other hand, if two of the dice are the same and one is different from the other two, the house pays \$2 on one number, \$1 on the other number, and collects \$4 on the remaining numbers, a \$1 profit.  If all three dice are the same, then the house pays out \$3 on one number and collects \$5 on the remaining numbers, a \$2 profit.  Not very good for the player, is it?  The argument I just gave is the one Ms. vos Savant used to convince Chuck that Chuck-a-Luck favors the casino.  Suppose, however, that some casino decided to pay \$5 for doubles and \$10 for triples.  Would you play this game?  I would hock my house, my car, and my dog to play this game.  Note, though, that the probability of winning this second game is still 91/216, the same as the first.  This should make it clear to you that asking for the probability of winning is to ask the wrong question.  What is the right question?

I should like to propose what I believe is a reasonable empirical procedure to evaluate a gambling game having a fixed unit bet (we'll address games having a multilevel betting structure later - they are a bit tricky).  We play the game over and over again a large number of times and keep track of how many units we lose how many units we win.  We then compute the statistic

 Average Profit per Bet = (units won - units lost) / number of bets (7)
Make sense?  Now, if you and I each carried out such an experiment, it is unlikely that we would come up with the same number.  But for a large number of trials our numbers would probably be close.  It would seem, therefore, that each of us would be making an experimental approximation to some number that represents the theoretical average profit per bet.  Let's see if we can figure out what this theoretical number is.

In order to address the issue of the theoretical average profit per bet, it is convenient to define an additional event in the Chuck-a-Luck game.  Let E0 represent the event of losing the game, that is, not rolling any fives.  Since there are 5 ways to roll a non-five on a die, there are 5 x 5 x 5 = 125 ways to roll a loser, so P(E0) = 125/216.  Note that the collection of four events E0, E1, E2, E3  satisfy two conditions:

 (i )   E0, E1, E2 , E3 is a pairwise disjoint collection.    ( ii)  E0 or E1 or E2 or E3 = S (8)
Any collection of events that satisfies both (i) and (ii) of (8) is called a partition of S.  This is very descriptive terminology.  When one partitions something, he or she breaks the thing up into distinct separate pieces.  Well, every point in S is in one of the four sets E0, E1, E 2, E3 (condition (ii)) but no point is in more than one (condition (i)); they form four distinctly separate pieces of S.  Why do we care?  The answer, given below, is a consequence of the following calculation.  I'll show you the calculation for the four sets above and let you generalize it to an arbitrary partition.  Simply put:

 P (E0) + P(E1) + P(E 2) + P(E3) = P(E0 orE1 or E2 or E3) by (i) and (5) (9) = P(S) by (ii) = 1 by condition (4)

The argument in (9) works for any partition of a sample space S .  This allows us to make a very subtle but very useful shift in perception.  Given a partition E1, E2, ... , E n of S, we mentally construct a new sample space by thinking of the events E1, E2, ... , E n as outcomes and the numbers P(E1 ), P(E2), ... , P(En) as forming a probability density on this set of  'outcomes'.  Wait a minute, we can't just declare these numbers a probability density by decree; both (1) and (2) have to hold.  Well, (1) obviously holds, but what about (2)?  Aha!  That is exactly what we just demonstrated in (9).  So, as long as we start with a partition, this shift in perception always works.  Thinking of the elements of a partition as outcomes and their corresponding probabilities as a pdf will be referred to as a reduced sample space.  Let us use the Chuck-a-Luck game to see why this terminology is reasonable.  Often it is convenient to represent a reduced sample space (or any sample space for that matter) in tabular form and that is what I'm going to do here.  Recalling that

 P (E0) = 125/216    P(E1 ) = 75/216    P(E2) = 15/216     P(E3) =1/216 (10)
here it is:
 (11)
 Outcome Probability E 0 125/216 E 1 75/216 E 2 15/216 E 3 1/216

Here we have reduced the sample space for Chuck-a-Luck from a space with 216 outcomes to one with just 4 outcomes and it contains all of the information that we need.  After all, we don't care whether we roll a (5, 4, 2) or a (3, 1, 5); all we care about is that event (now outcome) E1 occurred.  By the way, notice that as predicted in (9), the probability column in table (11) adds up to 1.

A random variable on a sample space S is just a real valued function f on S.  This means that to each outcome x in S the function f associates with x a unique real number that is written as f(x ).  The symbol f(x) is read "f of x" or sometimes "the value of f at x."  A classic example of a random variable is the payoff function for a gambling game.  For example, the payoff function for our five bet in the Chuck-a-Luck game is

 f (E0) = -1    f(E1) = 1    f(E2) = 2      f(E3) = 3 (12)
Here the function f associates to each outcome in our reduced sample space the payoff (or loss) to the player if that outcome occurs.  This should seem quite natural to you.  We'll see lots of payoff functions as well as other random variables in future articles.  Now a FANFARE!  Ta Da!  Here is the most important mathematical definition there is in the world of gambling; you'll see why shortly.  Given a particular sample space S = {x1, x2, x 3, ... , xn} equipped with a pdf p and a random variable f, we define the expected value of f, written exp(f), by the formula
 exp (f) = p(x1)f(x1) + p(x2)f(x2)  + p(x3)f(x3)  + ... + p(xn)f(xn) (13)
So, the prescription is easy enough; just take the probability of each outcome, multiply it by the value of f at that outcome, and add them all up.  Okay, but why is this so important to gamblers?

Recall the experimental average profit per bet we introduced in (7).  There we agreed that this is an estimate of some number representing the theoretical average profit per bet.  Let's see if we can derive such a number using the Chuck-a-Luck game.  Suppose we play the Chuck-a-Luck game 216 times, betting on five, and that everything happens exactly the way probability theory predicts.  Is this likely?  No, but if we played 216,000,000 times we would be close to this situation.  I'll stick with 216 and my idealistic fantasy.  In such a scenario, the outcome E 0 would occur 125 times and we would lose \$1 each time, E1 would occur 75 times and we would win \$1 each time, E2 would occur 15 times and we would win \$2 each time, and E3 would occur once with a win of \$3.  In other words, using the idea in (7), we would have

 Theoretical Profit per \$ = (125 x (-1) + 75 x 1 + 15 x 2 + 1 x 3)/216 (14)
Now I'll calculate (14) in just a moment, but first I want to rearrange the right hand side by dividing the 216 into each term as follows:
 Theoretical Profit per \$ = (125/216)(-1) + (75/216)(1) + (15/216)(2) + (1/216)(3) (15)
Recalling (10) and (12), we see that (15) can be rewritten as
 Theoretical Profit per \$ = P (E0)f(E0) + P(E1 )f(E1)  + P(E2)f( E2) + P(E3)f(E 3) (16)
But comparing (16) to (13) we see that the right hand side of (16) is just exp(f ).  This exact argument would work for any gambling game with a finite sample space.  In other words, we have shown
 Theoretical Profit per \$ = exp(f ) (17)

I said I would calculate the above number in (14) and (15) and I will.  Here is a tabular way to calculate expected values; I'll use (15)

 (18)
 Outcome Probability Function's Value Product No Five 125/216 -1 -125/216 One Five 75/216 1 75/216 Two Fives 15/216 2 30/216 Three Fives 1/216 3 3/216 Total -17/216

So, the theoretical profit per dollar risked is negative, meaning that the game (not surprisingly) favors the house.  It is customary to throw away the minus sign and report this figure as a percentage; this is called the house edge.  In the Chuck-a-Luck game this is approximately 7.87%  This means that, in the long run, the house will collect approximately \$7.87 for each \$100 wagered.  Not too good, Chuck.

What is the house edge for the field bet in Craps?  Easy!  Recall that the Field pays even money on 3, 4, 9, 10, 11 and 2 to 1 on the 2 and 12.  Here is the table:

 (19)
 Outcome Probability Payoff Product 2 or 12 2/36 2 4/36 3,4,8,10,11 14/36 1 14/36 losers 20/36 -1 -20/36 Total -2/36

So, the house edge is approximately 5.56%.  Better than Chuck-a-Luck, but not great.

We now see that the right question when evaluating a gambling game is not to ask the chance of winning, but rather, "What is the house edge?"  In the gaming world the probability of winning is called the hit frequency and is an important number when judging the psychological appeal of a game to the player.  The house edge, on the other hand, is an important number when judging the effect of the game on one's wallet.

As these articles proceed, we will be constructing lots of tables like those in (18) and (19), so make sure you understand why these tables are constructed the way they are.  There is a lot of interesting gambling material ahead using these ideas.  Next month we'll look at a technique for for counting the number of outcomes for events such as those in Keno and we'll look at the game of Keno.  Then on to Video Poker, Let it Ride, Craps, Slots and other games and gambling curiosities.  Oh, by the way, can you now calculate the house edge for the 5 number bet in Roulette at a payoff of 6 to 1?

See you next month.

Recent Articles
Best of Donald Catlin
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
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