GEK1544 The Mathematics of Games

Suggested Solutions to Tutorial 3

1. Consider a Las Vegas roulette wheel with a bet of $5 on black (payoff = 1 : 1) anda bet of $2 on the specific group of 4 (e.g. 13, 14, 16, 17 ; payoff = 8 : 1). What is thebettor’s expectation on this combined bet?

Suggested solution. Let

A = {outcome is black } and B = {outcome is 13, 14, 16 or 17} .

Then

P1 := P (A and B) = P ({outcome is 13 or 17 }) =2

38,

P2 := P (A and [not B]) = P ({outcome is black but not 13 and 17 }) =18− 2

38,

P3 := P ([not A] and B) = P ({outcome is 14 or 16 }) =2

38,

P4 := P ([not A] and [not B])

= P ({ outcome is white, 0 , or 00, and not 14 or 16 } )

=20− 2

38=

18

38.

These result

X = P1 · (5 + 2 ∗ 8) + P2 · (5− 2) + P3 · (2 ∗ 8− 5) + P4 · (−5− 2)

=2

38· (5 + 2 ∗ 8) +

16

38· (5− 2) +

2

38· (2 ∗ 8− 5) +

18

38· (−5− 2)

=(

18

38· 5− 20

38· 5)

+(

4

38· 2 ∗ 8− 34

38· 2)

= −14

38.

That is, the bettor’s expected loss is14

38· 1

5 + 2≈ 5.263 cents for every dollar bet.

2. A mythical slot machine has three wheels, each containing ten symbols. On eachwheel there is 1 JACKPOT symbol and 9 other non-paying symbols. You put in 1 silverdollar ($1) in the slot and the payoffs are as follows :

3 JACKPOT symbols – $487 in silver is returned (including your $1).

2 JACKPOT symbols – $10 in silver is returned (including your $1).

1 JACKPOT symbols – $1 in silver is returned (you get your wager back!).

Define what it would mean to say that this slot machine is fair and then show that it isindeed a fair “one-armed bandit” !

Suggested solution. Assuming that the slot machine is random.

P1 = P (1 Jackpot symbol exactly) =1

10· 9

10· 9

10+

9

10· 1

10· 9

10+

9

10· 9

10· 1

10=

243

1000;

P2 = P (2 Jackpot symbols exactly) =1

10· 1

10· 9

10+

1

10· 9

10· 1

10+

9

10· 1

10· 1

10=

27

1000;

P3 = P (3 Jackpot symbols) =1

10· 1

10· 1

10=

1

1000;

pN = P (No Jackpot symbol at all) =9

10· 9

10· 9

10=

729

1000= 1− (P1 + P2 + P3) [please verify] .

The profits are 0, 9 and 486, respectively. We infer that

X = P1 ∗ 0 + P2 ∗ 9 + P3 ∗ 486 + pN ∗ (−1) =27 ∗ 9 + 486− 729

1000= 0 .

3. The martingale strategy had the gambler double his bet after every loss, so thatthe first win would recover all previous losses plus win a profit equal to the original stake.Since a gambler with infinite wealth will with probability 1 eventually flip heads, theMartingale betting strategy was seen as a sure thing by those who practised it.

Of course, none of these practitioners in fact possessed infinite wealth, and theexponential growth of the bets would eventually bankrupt those who choose to use theMartingale. Moreover, it has become impossible to implement in modern casinos, due tothe betting limit at the tables. Because the betting limits reduce the casino’s short termvariance, the Martingale system itself does not pose a threat to the casino, and many willencourage its use, knowing that they have the house advantage no matter when or howmuch is wagered.

Let one ‘round’ be defined as a sequence of consecutive losses followed by a win,or consecutive losses resulting in bankruptcy of the gambler.

After a win, the gambler “resets” and is considered to have started a new round.A continuous sequence of martingale bets can thus be partitioned into a sequence ofindependent rounds. We will analyze the expected value of one round.

Let ` be the probability of losing (e.g. betting for “even” for roulette has 20/38chance of losing). Let y be the amount of the commencing bet, and n the finite numberof bets you can afford to lose. Show that the expected profit (a loss if the number isnegative) per round is given by

y · [1− (2`)n] .

Suggested solution. In the sequence of numbers

20y = 1 ∗ y, 2y, 22y, · · ·, 2n−1 y, 2ny,

the chance that the better loses the first n games is `n. The loss is

1 ∗ y + 2y + 22y + · · ·+ 2n−1y = y[1 + 21 + · · ·+ 2n−1

]= y · 2n − 1

2− 1= y (2n − 1) .

The chance that the better does not lose all n games is (1 − `n). In this case, disregardwhere is the win in the ‘round’, the profit is y. Hence

X = `n [−(y (2n − 1))] + (1− `n) y = y · [1− (2`)n] .

As a side note, X < 0 once the probability of losing ` > 12.