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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:
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:
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
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:
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
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
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
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)
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:
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. This article is provided by the Frank Scoblete Network. Melissa A. Kaplan is the network's managing editor. If you would like to use this article on your website, please contact Casino City Press, the exclusive web syndication outlet for the Frank Scoblete Network. To contact Frank, please e-mail him at fscobe@optonline.net. Recent Articles
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