Facing the odds

The Gambler’s Fallacy Facing the Odds: The Mathematics of Gambling

Activity 8 The Gambler’s Fallacy

Objectives

• Apply the probability formula

• Compute combined probability

• Define dependent and independent events

• Determine the probability of two or more events

• Define the gambler’s fallacy and relate it to mathematical prob-

ability

Materials paper, pencils, calculators

Time 30 minutes

Math Idea The belief that because a certain event has not occurred many

times in a row that is more likely to occur the next time is known

as the gambler’s fallacy.

Prior Understanding

Students should know how to convert among fractions, decimals,

and percents. They should also know how to find simple theoretical

probabilities using the probability formula.

Introduction: Gambling Connection

You can use or adapt the following scenario as an introduction to

the problem. Then have students do the activity and discuss the solu-

tion.

The Gambler’s Fallacy Facing the Odds: The Mathematics of Gambling

Sasha has three younger brothers, and her mother is about to

give birth again. She is betting her brothers that this time her mother

will have a girl. Assuming the chances of being born a boy and the

chances of being born a girl are the same, how certain is Sasha of

winning her bet?

Discussion

Students should realize that Sasha is a victim of the gambler’s fal-

lacy. She is confusing a single event with a sequence of four events

and is mistaking independent events for dependent events. An

outcome of boys on the three previous births does not remove “boy”

from the pool of possible outcomes for the next birth, nor does it guar-

antee “girl” as the only possibility. No matter what has happened be-

fore, the probability of her mother giving birth to a girl (or a boy) at any

one time remains 50%.

Exercise 1

Have students list the possible outcomes for tossing a coin four

times. Then have them determine the probability of getting four tails in

a row.

Have students list the possible outcomes for tossing a coin the

fourth time after three tails have already come up. Then have them de-

termine the probability that tails will come up again.

Have students compare these two problems, identifying any simi-

larities and differences.

The Gambler’s Fallacy Facing the Odds: The Mathematics of Gambling

Discussion

When determining the probability of two or more events, it is im-

portant to know (a) whether you are determining the probability of a

single event or the probability of a group of events, and (b) whether

the outcomes of the events are independent or dependent.

The tossing of four coins in a row consists of four events with a to-

tal of 16 possible outcomes: HHHH, HHHT, HHTH, HHTT, HTHH,

HTHT, HTTH, HTTT, TTTH, TTTH, TTHT, TTHH, THTT, THTH, THHT,

THHH. Only one of these outcomes shows four tails, so the probabil-

ity of getting four tails is 1/16. The four tosses are independent—the

outcome of the first toss does not affect the outcome of the second,

third, or fourth in any way. On each toss, the probability of getting

tails (T) is 1/2, so the combined probability of getting four tails is 1/2

× 1/2 × 1/2 × 1/2 = 1/16 or 6.25%.

If three tails have come up, the tossing of the next coin is a single

event whose outcome is independent of what has already hap-

pened. The probability of getting tails again is still 1/2 or 50%.

Here it does not matter what the probability of getting four heads

in a row is: at this point, we are only dealing with the next toss. Stu-

dents should note that although determining the probability of an out-

come for a group of coin tosses is different from determining the

probability of an outcome for one coin toss, the individual outcomes

are independent.

The Gambler’s Fallacy Facing the Odds: The Mathematics of Gambling

Exercise 2

Have students determine the probability of drawing two aces in a

row from a well-shuffled, complete deck of cards, if the drawn cards

are replaced and the deck is shuffled again after each draw.

Have students determine the probability of drawing two aces in a

row from a well-shuffled, complete deck of cards, if the drawn cards

are not replaced and the deck is shuffled again after each draw.

Have students compare these 2 problems, identifying any similari-

ties and differences.

Discussion

The drawing of two aces in a row consists of two events. When the

drawn cards are returned to the deck, the two events are independ-

ent events. Each time there are 4 aces out of 52 cards to choose

from, so the combined probability is 4/52 × 4/52 = (1/13)

= 1/169

or about 0.59%. Students should note that it is possible to draw the

same ace twice.

When the drawn cards are not returned to the deck, the two events

are dependent events— the result of the first draw affects the poten-

tial outcome of the second draw. If the first draw is an ace, there are 3

aces left out of 51 cards in the deck. The probability of drawing an-

other ace is now 3/51 and the combined probability is 1/13 × 3/51 =

1/221 or about 0.45%.

The Gambler’s Fallacy Facing the Odds: The Mathematics of Gambling

Activity 8 The Gambler’s Fallacy Teacher Support

Vocabulary

probability a number from 0 to 1 that expresses the likelihood that

a given event (or set of outcomes) will occur

combined probability the probability that two or more events will

all occur

independent events events in which the outcome of the first

event does not affect the outcome of subsequent events

dependent events events in which the outcome of the first event

does affect the outcome of subsequent events

Ongoing Assessment

What is the probability of drawing four clubs in a row from a well-

shuffled, complete deck of cards if the drawn cards are not replaced?

(13/52

12/51

11/50

10/49 = 0.00264)

The Gambler’s Fallacy Facing the Odds: The Mathematics of Gambling

Added Practice 8 The Gambler’s Fallacy

Name Date

1. Suppose you have already drawn two cards from a well-shuffled, complete

deck and neither one is a five. What is the probability that you will draw a five

the next time if the first two cards are not replaced?

2. Suppose you have already drawn a four and a five from a well-shuffled, com-

plete deck. What are the chances of drawing a five the next time if the first two

cards are not replaced?

3. A friend of yours has a cousin who was in a minor plane crash and survived.

Whenever your friend has to fly in an airplane, she insists on being accompa-

nied by her cousin, figuring that the probability of her cousin being in two plane

crashes is very small. Is your friend's reasoning correct? Why or why not?

4. An acquaintance of yours is concerned about a recent terrorist bombing of an

airline flight. He decides that on all future flights he takes, he will bring a bomb

in his suitcase. He reasons that the probability of two bombs being on a plane

is very, very low. Will his plan decrease his chances of being killed by a terror-

ist bombing while flying? Why or why not?

The Gambler’s Fallacy Facing the Odds: The Mathematics of Gambling

Answer Key Added Practice 8 The Gambler’s Fallacy

1. On the next draw there are 50 cards left in the deck, 4 of which are fives. The

probability is 4/50 = 8%.

2. On the next draw there are 50 cards left in the deck, only 3 of which are fives.

The probability is 3/50 = 6%.

3. Your friend's reasoning is not correct. Plane flights are not dependent events—

the outcome of one flight does not have an effect on the outcome of subsequent

flights. Assuming that the probability of a commercial flight crashing is one in a

million, a person will be subjected to that one-in-a-million risk each time he or

she takes a commercial flight, whether or not he or she has been in a previous

crash. In reality, there are specific causes of plane crashes: faulty or damaged

airplane parts, engine failures, weather conditions, pilot incompetence, etc.

However, because you cannot evaluate all of the variables each time you take a

flight, you have to assume that plane crashes are more or less random occur-

rences. That is, because the factors leading to a crash are unknown and unpre-

dictable before the crash, you have to approach crashes as rare chance out-

comes. Clearly, the presence or absence of someone who has been in a previ-

ous crash does not influence the outcome of the flight in any way.

4. Your friend's plan will not change his chances of being blown up by a terrorist. A

terrorist choosing to blow up the plane that your friend is on is a random occur-

rence. Let's say, hypothetically, that the chance of this happening on any par-

ticular plane is 1/5,000,000. If your friend knows that he is carrying a bomb on a

particular flight, then he can be 100% confident that there is at least one bomb

on his plane. The combined probability of his bomb and a terrorist’s being on

his plane is the product of the individual probabilities, that is 1 x 1/5,000,000 =

1/5,000,000. Although it is true that the probability of two people putting a bomb

on the same plane independently is very, very small, what your friend is carrying

in his suitcase has no effect on the situation.