GEK1544 The Mathematics of Games

Tutorial 1 – Suggested Solutions

1. In the movie “ 21 ”, Professor Rosa asked the student Ben Campbell the followingquestion.

“There are three doors, and behind one of them is a car, while behind the other two aregoats. If you choose the door with the car behind it, you win the car. Now, say you chooseDoor 1. The host then opens either Door 2 or Door 3, behind which is a goat. (The hostknows what is behind each door, and never opens the door with the car behind it. ) Thehost now gives you the choice: do you want to stick with Door 1, or switch to the otherdoor. What should you do? Or does it matter ? ”

Without hesitation Ben answered this correctly, which convinced Professor Rosa that Benwould be a good addition to their “card counting team”. Explain why Ben’s answer iscorrect, that is, by switching the choice rather than sticking with the original one, theprobability increases to 2/3 .

Suggested solution. Based on the question, there are three scenarios, each with equalprobability.

(* The car is behind door 1, in this case Ben losses, as he will shift to door 2 or door 3.

(* The car is behind door 2. The host can only open door 3, as door 1 is chosen by Ben.In this case Ben wins as he will shift his choice from door 1 to door 2.

(* The car is behind door 3. Likewise, the host can only open door 2, as door 1 is chosenby Ben. Again in this case Ben wins as he will shift his choice from door 1 to door 3.

Hence Ben’s strategy of shifting his choice gives him 2/3 the chance on winning.

2. Solve Chevalier de Mere’s problem of determining the probability of obtaining oneor more double sixes in 24 rolls of a pair of dice.

Answer. 1−[35

36

]24

≈ 0.4914 <1

2.

3. Consider the basic properties on probability :

P ( not A) = 1− P (A) ,

P (C ∪D) = P (C) + P (D) when C ∩D = ∅ ,

show that P (A ∪ B) = P (A) + P (B)− P (A ∩ B) when A ∩ B 6= ∅ . Here A and Bare independent events.

Suggested solution. Write A ∪B as the union of three disjoint sets:

A ∪B = (A \B) ∪ (A ∩B) ∪ (B \ A) .

As(A \B) ∪ (A ∩B) = A =⇒ A ∪B = A ∪ (B \ A) .

we have

(3.1) P (A ∪B) = P (A ∪ (B \ A)) = P (A) + P (B \ A) (why ?) .

On the other hand,

(3.2) B = (A∩B)∪ (B \A) =⇒ P (B) = P ([A∩B]∪ [B \A]) = P (A∩B)+P (B \A) .

That is,

(3.3) P (B \ A) = P (B)− P (A ∩B) .

(3.3) and (3.1) provide a proof.

4. Compute the probability of getting at least 1 six in 3 rolls of an honest die. Thenexpose the flaw in the following argument, which attempts the same problem in the“positive direction” :

The probability of a six in any one roll is 16.

Since we have 3 rolls, P (at least 1 six ) = 16

+ 16

+ 16

= 12.

Answer. 1−[5

6

]3

≈ 0.4.21 <1

2.

Let

C1 = {obtain a 6 in 1st roll ; doesn’t matter what numbers in the other two rolls } ,

C2 = {obtain a 6 in 2nd roll ; doesn’t matter what numbers in the other two rolls } ,

C3 = {obtain a 6 in 3rd roll ; doesn’t matter what numbers in the other two rolls } .

Then

P (C1) = P (C2) = P (C3) =1

6.

However,

P (C1 or C2 or C3) = P (C1 ∪ C2 ∪ C3) 6=1

6+

1

6+

1

6as

C1 ∩ C2 = {obtain two 6 in 1st & 2nd roll ; doesn’t matter what number in the 3rd roll }6= ∅ .

Likewise, C1 ∩ C3 6= ∅ & C2 ∩ C3 6= ∅ .