L.A. Math: Romance, Crime, and Mathematics in the City of Angels
Page 21
You undoubtedly have at least a passing acquaintance with probability through the weather reports. You’ve heard the TV meteorologist make a prediction such as this: The probability of rain tomorrow is 80%. Intuitively, you know that this means it’s pretty likely that it will rain tomorrow. If you were to analyze this more thoroughly, you would conclude something like this: If we were to keep a running total of what the weather is actually like on all the days when the meteorologist says there is an 80% probability that it will rain, we would expect that it would rain on 80% of those days, but it would not rain on 20% of them.
The meteorologist’s prediction helps you to make decisions. When the probability of rain the next day is 80%, you’ll strongly consider taking an umbrella along with you, and you are unlikely to schedule a picnic. Yes, the meteorologist could be wrong—the day could be nice and sunny, and you would regret looking ridiculous as you carry your umbrella, and you might also regret not having a picnic on such a beautiful day. On the other hand, it would be a lot worse if it rained and you didn’t have an umbrella and got soaked to the skin, and of course everyone knows what a disaster rain is when you have a picnic. You have no guarantee of making the right decision, but that 80% number is pretty convincing, so you take the umbrella and cancel the picnic. Of course, if the meteorologist said that the probability of rain was 100%, you’d have no problem deciding because it is certain to rain. Similarly, if the meteorologist said that the probability of rain was 0%, rain would have no chance of occurring, so the umbrella goes back in the closet and the picnic is on.
Probability, as you see it in predictions of rain, is a number that represents the relative frequency with which rain occurs: from a low of 0%, when rain never occurs, to a high of 100%, when rain definitely occurs. The higher the number, the more likely the rain.
THEORETICAL AND EMPIRICAL PROBABILITIES; SAMPLE SPACES
Summer days in Los Angeles belong to three basic types: sunny, rainy, and cloudy. Assume that these three types of days are defined to be nonoverlapping; a day is sunny, rainy, or cloudy. To make sure that the types of days are nonoverlapping, we might define a day to be rainy if any rain falls, cloudy if no rain falls but a cloud appears in the sky, and sunny otherwise. The weather forecast for the next day might be a 20% probability of rain, a 30% probability of sunshine, and a 50% probability that it will be cloudy.
The mathematical terminology that is used in the above example is that observing the next day’s weather constitutes an experiment. The three possible outcomes to the experiment are sunny, which we shall abbreviate as S, cloudy (C), and rainy (R). The three possible outcomes, which must be nonoverlapping and cover all possibilities, constitute the sample space of the experiment.
Notice that the probabilities assigned to the three outcomes add to 100%: 20% + 30% + 50% = 100%. While it is perfectly possible to describe probability theory using percentages, there are technical reasons, which you’ll see in a later section, that make it preferable to use numbers between 0 and 1, which we can obtain by dividing the percentages by 100. The probability of a rainy day becomes 0.2 after this division; this is usually abbreviated P(R) = 0.2. Similarly, we see that P(S) = 0.3 and P(C) = 0.5, and we observe that P(R) + P(S) + P(C) = 0.2 + 0.3 + 0.5 = 1.
Here’s a summary of this.
Experiments, Sample Spaces, and Probabilities
An experiment is the observation of something that actually happens or could happen. The outcomes of this experiment are the different observations that might be made. The outcomes must be nonoverlapping and must cover all possible observations.
The sample space of the experiment is the set of outcomes. A probability function P is an assignment of numbers between 0 and 1 to the different outcomes in such a way that the sum of the probabilities of all the outcomes is 1.
The probability assigned to an outcome represents the relative likelihood that the outcome will actually occur when the experiment takes place. If the probability of an outcome is 0, the outcome cannot occur, and if the probability of an outcome is 1, then the outcome is certain to occur.
Experiments and sample spaces belong to one of three basic types, which depend on how the probabilities are assigned to the outcomes. There are sample spaces in which the probabilities are assigned on a subjective basis. The probability of acceptance of a marriage proposal is such an example, although generally the person offering such a proposal usually feels that this probability is close to 1. Exceptions, of course, exist—celebrities continually receive such proposals from hopeful, though probably not optimistic, strangers.
Probabilities can also be assigned on an empirical basis. If a quarterback has thrown 100 passes, completed 60, had 5 interceptions, and thrown 35 incomplete passes, one can assign probabilities by using the empirically determined relative frequencies: P(complete) = 60/100 = 0.6, P(interception) = 5/100 = 0.05, P(incomplete) = 35/100 = 0.35. In this instance, the probabilities can be interpreted as averages: for each pass thrown, the quarterback has averaged 0.6 of a completion, 0.05 of an interception, and 0.35 of an incompletion.
The third, and most readily analyzable, assignment of probabilities is the theoretical model. An easy example is the flip of a coin, in which the outcomes are heads and tails, denoted H and T, respectively. The symbol P(H) denotes the probability of a head occurring as a result of the flip.
Example 1: The Fair Coin: One of the simplest experiments is the flip of a fair coin. There are only two outcomes: heads (H) and tails (T). To compute P(H) and P(T), the assumption that the coin is fair means that heads and tails are equally probable, so P(H) = P(T).
Since H and T are the only possible outcomes, we must have P(H) + P(T) = 1. Since P(H) = P(T), we obtain P(H) = P(T) = 1/2. This probably (appropriate word here!) comes as no surprise. ■
Expressed in terms of percentages, the probability of heads is 50%, and the probability of tails is also 50%. This idea has worked its way into the language: equal chances are sometimes called fifty–fifty chances.
The fair coin is a specific example of a uniform probability space. In a uniform probability space, all the outcomes are assumed to be equally likely (one of the meanings of “uniform” is “the same”). There are many interesting sample spaces in which the outcomes are all equally likely, such as the roll of a fair die or the drawing of a single card from a deck of 52 cards. It’s easy to see, similar to the fair coin, that in a uniform probability space with n outcomes, the probability of each outcome is 1/n.
When a fair die, which has six sides, is rolled, the probability of rolling a 4 is 1/6 (as is the probability of rolling a 1, or a 2, etc.). When a card is drawn from a deck of fifty-two cards, the probability of drawing the ace of spades is 1/52, which is the same as the probability of drawing the king of spades or the deuce of clubs.
(Probabilities for two-child family continued from p. 83)
Even Pete can make a mistake! Once March sees a boy, since he has no way of knowing whether he is seeing the younger or older child, it is twice as likely that he is seeing a family with two boys as either the BG or the GB family. There are four equally likely families: BB (he sees the younger boy), BB (he sees the older boy), BG, and GB—and in half those families the child March does not see is a girl. So DiStefano was offering odds of 11 to 10 on a 50–50 bet—but got lucky!
In the story, Pete was interested in the probability that the man March asked had a sister. This is not just a single outcome, such as BG, because we do not know whether we are asking the firstborn child. What Pete was interested in was whether the actual outcome belonged to a predetermined set of outcomes, which in this instance consisted of both BG and GB.
An event in a sample space is a set of outcomes; in other words, it is a subset of the sample space. In order to compute the probability that the actual outcome of an experiment belongs to a particular event, one adds the probabilities of all the outcomes in the event, just as Pete added P(BG) + P(GB) = 1/3 + 1/3 = 2/3 to (erroneously) get a probability of 2/3 that the man’s
sibling was a girl. It is customary to use capital letters to denote events. If we use the event from the story we have been discussing (the sibling of the man March asked was a girl), we might let A = {BG, GB}. So Pete computed that P(A) = P(BG) + P(GB) = 2/3. Generalizing this example, the probability of an event E, written P(E), is the sum of the probabilities of all the outcomes in the event E.
The null set, ∅, also represents a collection of outcomes: no outcomes, to be precise. When we perform an experiment, it is understood that something will be observed, so the probability that no outcome will be observed is 0. The null set is sometimes called the impossible event, or the null event, and P(∅) = 0.
If S is the entire sample space, the sum of the probabilities of all the outcomes in S is equal to 1, and so P(S) = 1. The entire sample space is sometimes called the certain event because it is certain that, when an experiment is performed, one of the outcomes will be observed.
Expectation
Now that we know how to compute probabilities, how do we use them to make plans? Many situations arise in which payoffs are associated with the occurrence of an outcome or an event. Sometimes these payoffs are measured in terms of money, as in the bet between March and DiStefano. Sometimes they are measured in other units. One can, for instance, compute the probability of a worker being sick and measure the payoffs in terms of hours worked, or the probability of a shipment of items containing defective items and measure the payoffs in terms of defective items.
Let’s start with a simple example. Suppose that you draw a card from a deck of 52, and a benevolent individual offers to pay you $5 for each time you select a face card, provided that you pay her or him $1 every time you don’t. The game goes as follows: you draw a card, pay or get paid, put the card back in the deck, shuffle the deck, and draw again.
If you were to play this game 52 times and happened to draw each card in the deck once, your balance sheet would show a gain of $5 for each of the 12 face cards, and a loss of $1 for each of the 40 other cards. Your net would be (12 × $5) − (40 × $1) = $20. Since you played the game 52 times, your average win per play would be $20/52, which is approximately $0.38. We say that your expected gain per play is $0.38.
Let’s look at a different way of computing the same number, $0.38. If we let F denote the event “draw a face card” and N denote the event “draw a nonface card,” we see that our expected gain per play, which we know is [(12 × $5) − (40 × $1)]/52, can also be written
(12 × $5 − 40 × $1)/52 = 12/52 × $5 + 40/52 × (−$1)
= P(F) × V(F) + P(N) × V(N)
where V(F) denotes the value of getting a face card ($5), and V(N) denotes the value of getting a nonface card (-$1).
This setup gives us the “recipe” for computing the average gain per play associated with any game. Break the game up into events such that each outcome in a selected event has the same payoff. In the above game, the two events (F and N) were determined by the fact that each outcome in F had a payoff of $5, and each outcome in N had a payoff of −$1. Multiply the probability of each of the events by its associated payoff and add the result. This average gain per play is called the expectation of the game and is usually denoted by E. A game that has an expected value of 0 is called a fair game.
Definition of Expectation
Suppose that an experiment has outcomes x1, …, xN.
If P(x) denotes the probability associated with outcome x, and V(x) denotes the payoff (value) associated with outcome x, then the expectation E is defined by
E = P(x1) V(x1) + … + P(xN) V(xN)
Payoffs are positive from the point of view of the person or persons for whom the expectation is being computed. If several different outcomes have the same payoff, it is natural to group these together as an event. The above formula would then apply, with x1 denoting the first event, and so on.
Once again, this is what Pete did in the story when he explained March’s expectation by assuming that March bet $10 three separate times, winning once and losing twice. Denote the two events as S (the sex of the sibling is the same as the person being asked), and O (the sex of the sibling is opposite to the person being asked). According to the rules of the game, if March bet $10, V(S) = $11 and V(O) = −$10. You saw earlier that P(S) = 1/3 and P(O) = 2/3. Therefore, the expectation is
E = P(S) × V(S) + P(O) × V(O) = 1/3 × $11 + 2/3 × (−$10) = −$3
You may recall that Pete described the expectation as 30% (from DiStefano’s point of view). It wasn’t clear whether each bet was $10, $100, or $1,000, so one might just as well assume that each bet was 10 units. Then the expected loss per bet would be 3 units. Considering that 10 units was the amount of the bet, the percentage expected loss per unit bet would be 30%. Percentages therefore give a convenient way to describe the expectation of a game.
Odds
When a game has only two outcomes, sometimes the payoffs are quoted in terms of how much the player will win as compared with how much the player will lose. Odds of 3 to 1 means that the player stands to win three units if he or she wins, but will lose one unit if he or she loses. Odds of 3 to 1 describes those situations in which the player stands to win three times as much as he or she stands to lose.
Odds of 2 to 5 means that the player stands to win two units but risks losing five units. It is customary to quote odds in whole numbers at the simplest possible ratio. Odds of 1 to 1 are sometimes called even money.
Example 2: A British bookmaker (bookmaking is not only legal in England but respectable) has had punters (British for bettors) bet 300 pounds on Oxford and 200 pounds on Cambridge on the upcoming boat race. If the bookmaker plans to take 50 pounds for his or her commission, what odds should he or she offer on each school?
Solution: A total of 500 pounds has been bet, and after he or she takes his or her commission, 450 pounds are left. Those who have bet on Cambridge stand to lose 300 pounds, but they could win 450 pounds. So the odds the bookmaker should offer for a bet on Cambridge are 450 to 300, or 3 to 2. Similarly, since 200 pounds were bet on Oxford, the odds on Oxford are 450 to 200, or 9 to 4. ■
A reasonable assumption in example 2 is that because 0.6 of the total money wagered was bet on Oxford, the probability that Oxford will win the race is 0.6. In this case, the expectation of the Oxford bettors is
E = (0.6 × 150) + (0.4 × −300) = −30
In this case, though, the percentage expectation of the Oxford bettors is therefore
100 × −30/300 = −10%
But it is not just games that have expectations associated with them. When an insurance company sells a policy or a company brings out a new product, they compute expectations. To decide whether an investment is a good one, they must calculate the expectation involved. Sometimes the probabilities or payoffs must be estimated based on lack of complete knowledge. A company usually brings out a new product only if it can project a positive expectation.
APPENDIX 10
CONDITIONAL PROBABILITY IN “ONE LONG SEASON”
(Conditional probability continued from p. 93)
Let’s analyze Julie’s dilemma (to stand pat or to switch her choice) from the story at the start of chapter 10. Recall that Julie has initially selected Wyatt as the man she thinks Debbie will marry. The scriptwriters have already decided whom Debbie will marry before Julie even got involved. There would be no drama, from the standpoint of Silktex shampoo, if Wyatt was in a coma up in Monterey—nobody would watch to see Julie’s choice because Wyatt was now clearly out of the picture. So the scriptwriters were told to involve either Ellison or Lowell, neither of whom was Julie’s choice, in the car crash.
Since we have no advance information about which one Debbie is likely to marry, we can assume that she is equally likely to marry Ellison, Wyatt, or Lowell.
Case 1: Debbie marries Wyatt. Either Ellison or Lowell would be in the car crash. It is wrong for Julie to switch (she would lose the $5,000 cost to switch, as well as the grand prize of $100,000).
Case 2: Debbie marries Lo
well. The scriptwriters were obviously told to involve Ellison in the car crash. Now Julie wins the $100,000 by switching.
Case 3: Debbie marries Ellison. In this case, the scriptwriters would have been told to make Lowell the victim of the car crash. Again, it is right for Julie to switch.
The key point here is that the scriptwriters are guaranteed to put either Ellison or Lowell in the car crash once Julie selects Wyatt as Debbie’s intended. We can therefore see that, in two out of these three cases, it is right for Julie to pay the $5,000 and switch choices.
The scriptwriter was given the following condition: Involve Ellison or Lowell in a car crash. In so doing, the sample space of the experiment has been modified from the original sample space (Debbie might marry any of the three suitors) to a subset of the original sample space (Debbie will only marry either Wyatt or Lowell). Because Ellison was in a car crash, the original sample space {Wyatt, Ellison, Lowell} has been changed; it’s now {Wyatt, Lowell}. The study of sample spaces that have been modified as a result of additional information forms the subject of conditional probability.
The argument that Julie should switch is far from obvious and indeed goes against the grain for most people. When a variation of this problem (known by mathematicians as the Monty Hall Problem) was posed in a nationally syndicated magazine column, the author of the column mentioned that it generated incredible amounts of write-in commentary, and some very highly educated people came to the wrong conclusion!
CONDITIONAL PROBABILITY
The poker game of five-card stud was the de facto money poker game for almost a century until televised big-stakes poker made Texas Hold ’Em the game of choice.
In Texas Hold ’Em, each player is initially dealt two cards face down so that only the player to whom those cards are dealt can see them. After a round of betting, three cards are simultaneously turned face up in the center of the table (this is called “the flop”), and each player regards those three cards as augmenting the two they were originally dealt.