Basic Gambling Mathematics (Instructor Solution Manual, Solutions) [1 ed.] 9781482208955, 9781482208931, 1482208938

Citation preview

solutions MANUAL FOR BASIC GAMBLING MATHEMATICS The Numbers Behind the Neon by

Mark Bollman

K21613_SM_Cover.indd 1

9/22/14 4:20 PM

K21613_SM_Cover.indd 2

9/22/14 4:20 PM

solutionS MANUAL FOR BASIC GAMBLING MATHEMATICS The Numbers Behind the Neon by

Mark Bollman

Boca Raton London New York

CRC Press is an imprint of the Taylor & Francis Group, an informa business

K21613_SM_Cover.indd 3

9/22/14 4:20 PM

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2014 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20140918 International Standard Book Number-13: 978-1-4822-0895-5 (Ancillary) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

K21613_SM_Cover.indd 4

9/22/14 4:20 PM

Solutions

1

Chapter 1 1.1. This is the set of numbers that are both red and odd: R ∩ O = {1, 3, 5, 7, 9, 19, 21, 23, 25, 27}. 1.2. These numbers are either black or even: B∪E = {2, 4, 6, 8, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20, 22, 24, 26, 28, 29, 30, 31, 32, 33, 34, 35, 36}. 1.3. This set contains all numbers that are red, high, and even—all three criteria must be met. R ∩ H ∩ E = {30, 32, 34, 36}. 1.4. R′ consists of the numbers that are not red—this set includes all of the black and green numbers. (Note very carefully: R′ is not simply equal to B, and similarly for the other “paired” sets: O′ ̸= E, and so on.) It follows that R′ = {0, 00, 2, 4, 6, 8, 10, 11, 13, 15, 17, 20, 22, 24, 26, 28, 29, 31, 33, 35}. 1.5. The question asks for the set of all numbers that are both low and high. Since those sets, by design, are disjoint, it follows that L ∩ H = ∅. 1.6. We can approach this set by constructing L ∪ O ∪ B directly (it has 32 elements) and forming its complement, or by reasoning based on the definition of the sets. To be an element of L ∪ O ∪ B, a number must be low, or odd, or black. The complement of L ∪ O ∪ B includes the numbers that have none of these three characteristics—that is, they are high (or 0 or 00), and even (or 0 or 00), and red or green. Besides the zeroes, the only numbers fitting this description are high, even and red; thus, (L ∪ O ∪ B)′ = H ∩ E ∩ R = {0, 00, 30, 32, 34, 36}. 1.7. As in Exercise 1.6, we consider the set H ∪ R first: this is the set of high or red numbers. The complement of H ∪ R is the set of numbers that are low and black, plus the zeroes, so (H ∪R)′ = {0, 00, 2, 4, 6, 8, 10, 11, 13, 15, 17}. 1.8. The numbers generated are 31248, 114040, 130220, 14268, 96421, and 114059. Scaling them to the range 1-8 gives 1, 1, 5, 5, 6, and 4.

Chapter 2 2.1. A double deck contains eight 7’s, from which you must draw three. There are 104 cards at the start of the double deck, so we have P (3 7′ s) =

8 C3 104 C3

=

56 ≈ 3.075 × 10−4 . 182104

2.2. The first hole card can be any card in the deck; we are interested in the probability that the second card is of the same rank. Since there are three cards remaining of that rank, the probability of pulling one of them is 3/51 = 1/17. 2.3. Every card can be connected both above and below, by a total of eight cards. The identity of the first card does not matter, and so the probability of connected hole cards is simply the probability of drawing one of these eight cards as the second card: 8/51. 2.4a. There are 12 face cards among the 52 cards in a deck. The probability

K21613_SM_Cover.indd 5

9/22/14 4:20 PM

2

Basic Gambling Mathematics: The Numbers Behind The Neon

of receiving five face cards in the deal is therefore 12 C5

p=

52 C5

=

792 ≈ 3.047 × 10−4 . 2598960

2.4b. From the 792 hands identified in part a., the only hands that will be called a blaze consist of two pairs of different face cards and a fifth card of the other rank. There are 3 · 4 C2 · 4 C2 · 4 C1 = 3 · 6 · 6 · 4 = 432 of those, so the probability of a blaze that is called a blaze is 432/2598960 ≈ 1.66 × 10−4 . 2.5. There are two colors from which to choose one. Each color contains 26 cards, from which we select 5 without regard to order. Therefore, P(1 color) =

2·

26 C5

52 C5

=

131560 2 · 65780 = ≈ .0506. 2598960 2598960

2.6. The number of possible full houses is (13 · 4 C3 ) · (12 · 4 C2 ) = 3744, where 13 and 12 count the number of possible ranks for the three-of-a-kind and the pair, respectively. Dividing by 52 C5 gives P (Full house) =

3744 ≈ .0014. 2598960

2.7. The number of possible three-of-a-kinds is 13 · 4 C3 = 52. We must now count the number of ways to draw two further cards from the deck that match neither the three cards selected, nor each other, in rank. This may be done by selecting two ranks from the 12 not yet chosen and choosing one card from each rank; this may be done in 12 C2 · 42 = 66 · 16 = 1056 ways. We then have 78 · 1056 P (3 of a kind) = ≈ .0211. 2598960 2.8a. There remain 13 ranks in this enhanced deck. Once a rank is chosen, we must choose five of the six cards of that rank, which may be done in 6 C5 = 6 ways. It follows that there are 78 ways to draw five of a kind (This is precisely the number of cards in the deck, because every card corresponds to a five of a kind consisting of the other five cards of that rank.) The probability of five of a kind is therefore p=

78 78 ≈ 3.695 × 10−6 . = C 21111090 78 5

There are only six possible royal flushes in this deck, so five of a kind would not outrank a royal flush. 2.8b. We begin by choosing five of the six suits, and then we select one card from each suit. P(5 different suits) =

6 · 135 ≈ .1055. 21111090

2.8c. While the order in which the cards are dealt does not matter, we

K21613_SM_Cover.indd 6

9/22/14 4:20 PM

Solutions

3

organize our thinking by considering the cards in increasing rank order. The lowest card in a rainbow straight may have any rank from ace through 10, and with six suits, there are 60 choices to start the hand. The next rank may be chosen from among the five cards of the next higher rank not matching the suit of the first card. Considering both the need to have cards in sequence and avoid repeating suits, the remaining three cards may be chosen, respectively, in 4, 3, and 2 ways. The probability of a rainbow straight is therefore p=

60 · 5 · 4 · 3 · 2 7200 ≈ 3.411 × 10−4 . = 21111090 78 C5

2.9. You win 100,000 if any 20 of the 32 edge numbers are chosen. Order does not matter, so P (Hit 20 out of 32) =

32 C20 80 C20

=

225792840 ≈ 6.4 × 10−11 . 3535316142212174320

2.10. The following table gives the number of valid tickets containing from 2-15 spots. Spots 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Combination Count 2 2 C1 = 2 3 4 C1 = 4 2+2 2 C2 = 1 5 1 3+2 4·2=8 3+3 4 C2 = 6 5+2 1·2=2 3+2+2 4·1=4 5+3 1·4=4 3+3+2 6 · 2 = 12 5+2+2 1·1=1 3+3+3 4 C3 = 4 5+3+2 1·4·2 = 8 5+3+3 1·6 = 6 3+3+3+2 4·2=8 5+3+2+2 1·4·1=4 3+3+3+3 4 C4 = 1 5+3+3+2 1 · 6 · 2 = 12 3+3+3+2+2 4·1=4 5+3+3+3 1·4=4 3+3+3+3+2 1·2=2 5+3+3+2+2 1 · 6 · 1 = 6

Summing the last column gives a total of 104 valid tickets. 2.11. There are 20 different blocks of 4 numbers, from which we choose 2

K21613_SM_Cover.indd 7

9/22/14 4:20 PM

4

Basic Gambling Mathematics: The Numbers Behind The Neon

without regard to order. The number of 8-spot tickets is then 20 · 19 = 190. 2 2.12. The chance of your ticket including a line with 8 score draws is equal to the probability that you have 8, 9, or 10 of the 11 score draws represented. This probability is 20 C2

p=

[11 C8 ·

38 C2 ]

+ [11 C9 · 49 C10

=

38 C1 ]

+ [11 C10 ]

=

1342 ≈ 1.437 × 10−5 . 93384347

2.13. The probability that exactlyk numbers come from one half of the ticket is 40 Ck · 40 C20−k . 80 C20 We then add up this quantity from k = 0 to k = 7 to cover the case where the numbers are concentrated on the bottom. Since the problem is symmetric in the top and bottom of the ticket, the probability of a winning ticket is double this sum, or 7 ∑ 40 Ck · 40 C20−k P (Win) = 2 · ≈ .1961. 80 C20 k=0

2.14a. The 15 numbers matched must be drawn from the 20 on the ticket, and the remaining 5 must be drawn from the 60 not selected. Order does not matter, so we have P (Match 15) =

(20 C15 ) · (60 C5 ) 15504 · 5461512 = ≈ 2.395 × 10−8 . C 80 20 80 C20

2.14b. In general, the probability of matching x numbers is given by P (x) =

(20 Cx ) · (60 C20−x ) . 80 C20

We have the following probabilities: x 16 17 18 19 20

P(Match x) 6.683 × 10−10 1.103 × 10−11 9.513 × 10−14 3.394 × 10−16 2.829 × 10−19

2.15. The red and white balls are chosen independently, so the Fundamental Counting Principle may be used to count the number of winning combinations. Let r be the number of matched red numbers and w the number of matched white numbers. We then have 0 ≤ r ≤ 2, 0 ≤ w ≤ 2, and P (r, w) =

K21613_SM_Cover.indd 8

(2 Cr ·

24 C2−r )

· ( 2 Cw ·

(26 C2 )

2

24 C2−w )

.

9/22/14 4:20 PM

Solutions

5

Substituting values for r and w yields the following results: r 2 2 1 2 0 1 1 0

w 2 1 2 0 2 1 0 1

P (r, w) Decimal P (r, w) 1/106525 9.387 × 10−6 48/106525 4.509 × 10−4 48/106525 4.509 × 10−4 276/106525 .0026 276/106525 .0026 2304/106525 .0216 13248/106525 .1244 13248/106525 .1244

Chapter 3 3.1. Successive hands are independent, so the first Multiplication Rule can be used: 40000 P(No royal flushes) = P(Not a royal flush) . The probability of not drawing a royal flush, by the Complement Rule, is 1 − P(Royal flush) = 1 −

1 39999 = . 40000 40000

The desired probability is therefore P(No royal flushes) =

(

39999 40000

)40000

≈ .3679.

3.2. There are 50 cards remaining, and 3 of them are aces. The desired probability is 3 3 C2 p= ≈ .0024. = 1225 50 C2 3.3. The probability of a winner in 65 or fewer draws is 1 - P(No winner). Since the three cards are independent, this becomes 1 − P (A card does not win in 65 draws)3 . Using the Complement Rule gives P(Winner) = 1 − (1 − P(Win in 65 or fewer draws))3 . From Table 3.2, we have P(Win in 65 or fewer draws) = .015415, so P(Winner) = 1 − (1 − .015415))3 = 1 − .9845853 ≈ .0455. 3.4. The numbers on the first card don’t matter; what we are interested in the probability that none of those numbers also appear on the second card. It is not enough simply to compare sets of 24 numbers, as we must take

K21613_SM_Cover.indd 9

9/22/14 4:20 PM

6

Basic Gambling Mathematics: The Numbers Behind The Neon

into account the fact that not every set of 24 numbers from the range 1-75 corresponds to a valid bingo card. For the B, I, G, and O columns, there are 15 C5 = 3003 ways to pick the five numbers. Since we’re just looking for common numbers, the order in which a column is filled does not matter here. The N column may be filled in 4 15 C4 = 1365 ways, which gives 3003 · 1365 = 111007923832370565 different choices for the numbers on a single card. To ensure that your second card has no numbers in common with your first, we simply remove those numbers from the n side of the n Cr formula: For B, I, G, and O, you have 10 numbers left from which to choose 5; for the N column, there are 11 numbers remaining, and you must choose 4 of them. It follows that the number of ways to fill your second card without duplicating your first card is 4

(10 C5 ) · (11 C4 ) = 2524 · 330 = 1330810145280, and so the probability of no matches is p=

1330810145280 1 ≈ 1.1988 × 10−5 ≈ , 111007923832370565 83414

which makes it extremely likely that two cards chosen at random will have one or more numbers in common. 1 3.5. The probability that two cards are disjoint is p ≈ . We find 83414 that P(≥ 2 disjoint cards) = 1 - P(All cards overlap) = 1 − (1 − p)3000 ≈ .0353. 3.6. The probability of exactly k edge numbers occurring among the 20 keno numbers drawn is P (k) =

32 Ck ·48

C20−k

80 C20

.

The probability of a losing Outside Eagle’s Edge bet is then P (6) + P (7) + P (8) + P (9) =

215025122532807 ≈ .6939. 309897978805415

3.7. There are five ways to roll an 8, four ways to roll a 9, three ways to roll a 10, two ways to roll an 11, and one way to roll a 12, so by the First Addition Rule, the probability of winning an Over 7 bet is p=

5+4+3+2+1 15 = ≈ .4167. 36 36

By symmetry, this is also the probability of winning the Under 7 bet. 3.8. Adding the six probabilities from Exercise 2.12b gives P (Win 100,000) = 2.463 × 10−8

K21613_SM_Cover.indd 10

9/22/14 4:20 PM

Solutions

7

—roughly one chance in 40.6 million, and not all that different from the probability of matching 15, which accounts for roughly 97.2% of this winning probability. 3.9a. A complete bet on 23 covers all 9 numbers from 19-27 with varying numbers of bets: Number 19 20 21 22 23 24 25 26 27

Straight Split

Street

1

1

1 4 1 1

1 1 1

Corner 1 2 1 2 4 2 1 2 1

Dbl. St. 1 1 1 2 2 2 1 1 1

Total 2 4 2 6 12 6 2 4 2

9 . It should be noted that the 38 minimum payoff if one of these nine numbers hits is 13, so you are guaranteed a profit if one of your numbers is spun. 3.9b. The complete bet on 33 is somewhat smaller than the complete bet on 23, because fewer split and corner bets are required. A total of nine bets, covering all numbers from 28-36, must be placed: one straight bet on 33; three split bets also involving 29, 32, and 36; one street bet on 31, 32, and 33; two corner bets, and two double street bets. The bet still covers nine numbers, so 9 the probability of winning this bet remains . 38 3.10. The first number can be any of the 38, so the chance of getting a “new” number on the first spin is 38 38 . The second number can be any of the 37 that were not spun first, so P (No match on second spin) = 37 38 . The conditional probability of then getting another new number on the third spin is 36 38 , and so on. Multiplying gives The probability of winning this bet is thus

p=

1 38! 38 37 36 · · ··· = 38 ≈ 4.861 × 10−16 . 38 38 38 38 38

3.11. We have the following probabilities for a single roll of 2d8: 1 = P (16) 64 2 = P (15) P (3) = 64 8 . P (9) = 64 P (2) =

On the come-out roll, then, the Addition Rule gives the following probabilities for the Pass line:

K21613_SM_Cover.indd 11

9/22/14 4:20 PM

8

Basic Gambling Mathematics: The Numbers Behind The Neon

P (Win) = P (9) + P (15) =

10 = .15625. 64

4 = .0625. 64 3.12. Simple counting yields the following table: P (Lose) = P (2) + P (3) + P (16) =

Point P(Point rolled) 4 or 14 3/64 5 or 13 4/64 6 or 12 5/64 7 or 11 6/64 8 or 10 7/64

P(Point wins) 3/11 4/12 5/13 6/14 7/15

P(Win on this point) 9/704 16/768 25/832 36/896 49/960

The probability of winning a pass line bet at spider craps is then .15625, the probability of an immediate win (Exercise 3.10), plus double the sum of the last column in this table, to account for the fact that each line of the table corresponds to two possible points. We have ( ) 10 9 16 25 36 49 37319 P (Win) = +2· + + + + = ≈ .4660. 64 704 768 832 896 960 80080 3.13. Using exercise 3.11, we see that the probability of winning a bet on 37319 ≈ .5340 if no numbers were barred on the Don’t Pass line would be 1 − 80080 1 the come-out roll. Barring only the 16 means that we must subtract 64 from this value, which gives P (Win) ≈ .5184 > .5. 1 If the 2 is also barred, we subtract a further 64 from this value, giving P (Win) ≈ .5027. 3.14. Barring both the 2 and 3, but not the 16, from winning a Don’t Pass bet on the come-out roll gives ( ) 37319 1 2 42761 3 156029 P (Win) = 1 − − − = − = ≈ .4871 < .5, 80080 64 64 80080 64 320320 as desired.

Chapter 4 4.1a. Denote the faces of the first Sicherman die as 1, 2a, 2b, 3a, 3b, and 4. The following combinations lead to a sum of 7: 1-6, 2a-5, 2b-5, 3a-4, 3b-4, and 4-3; it follows that P (7) = 16 , the same as for two standard dice. 4.1b. The only ways to roll doubles on Sicherman dice are 1-1, 3a-2, 3b-3, and 4-4, making the probability of doubles 4/36 = 1/9. 4.1c. Careful counting of the possibilities will reveal that the probability distribution of the sum of a pair of Sicherman dice is the same as that for a pair of standard d6’s.

K21613_SM_Cover.indd 12

9/22/14 4:20 PM

Solutions

9

Sicherman dice are the only nonstandard d6’s which give the standard distribution of sums from 2 through 12. 4.2. The expectation of this game is E=

1+2+3+4+5+6 = $3.50, 6

so charging 3.50 to play would make it a fair game. 4.3. There are two ways to roll an 11, so we have ( ) ( ) 34 4 2 + (−1) · =− ≈ −.1111. E = (15) · 36 36 36 The HA is 11.11%. 4.4a. The Big Six bet is resolved by one of eleven rolls: the six ways to roll a 7 and the five ways to roll a 6. The bet wins if a 6 is rolled before a 7; 5 this event has probability 11 . The expectation is ( ) ( ) 6 1 5 + (−1) · =− ≈ .0909. E = (1) · 11 11 11 The house advantage, therefore, is 9.09%. The same advantage applies to the Big 8 bet. 4.4b. Let A be the new payoff. We need ( ) ( ) 5 6 5A − 6 E = (A) · + (−1) · = = 0. 11 11 11 Solving for A gives A = $ 65 = $1.20, so paying the bet at 6-5 odds would yield a fair game. 4.5. Like the Big Six, a place bet on 8 is resolved by one of eleven rolls: the six ways to roll a 7 and the five ways to roll an 8. The bet wins if an 8 is 5 rolled before a 7; this event has probability 11 . The expectation is ( ) ( ) 6 1 5 + (−6) · =− . E = (7) · 11 11 11 Dividing by the 6 bet gives a HA of 1.515%, considerably less than the 9.09% HA on the Big 8. 4.6. The possible outcomes of this wager are as follows: The bet is won on the first spin. The payoff is 1 and the probability is 18/37. The bet is lost on the first spin. The payoff is - 1 and the probability is 18/37. The bet is imprisoned and then wins on the second spin. The payoff is 0 and the probability is 18/1369.

K21613_SM_Cover.indd 13

9/22/14 4:20 PM

10

Basic Gambling Mathematics: The Numbers Behind The Neon The bet is imprisoned and loses on the second spin. The payoff is - 1 and the probability is 19/1369.

The expectation is ( ) ( ) ( ) 19 19 18 18 18 E = (1) · + (0) · + (−1) · + =− ≈ −.0139. 37 1369 37 1369 1369 The bet has a HA of approximately 1.39%. 4.7a. For the straight bet, the expectation is ( ) ( ) 5000 1 9999 E = (4999) · + (−1) · =− = −$.50, 10000 10000 10000 and the HA is therefore 50%. For a wheel bet, we have ( ) ( ) 24 9976 12000 E = (4976) · + (−24) · =− = −$12, 10000 10000 10000 or half the amount wagered. Once again, and as was the case in the Daily 3 straight bet, the HA is 50%. 4.7b. If you make an n-way boxed bet, there are n numbers that win and 10000 − n that lose. Let x be the net amount you win if a boxed bet hits; your expectation for any of these bets is then ( ) ( n ) 10000 − n (x + 1)n − 10000 E = (x) · + (−1) · = 10000 10000 10000 Using the values for x given in the problem yields the following: n x E HA 24 207 - .5008 50.08% 12 415 - .5008 50.08% 6 832 - .5002 50.02% 4 1249 - .5000 50.00% All four bets have a house edge close to 50%, as we expect from a lottery. There’s a slight increase in the expectation as you bet fewer combinations, but it’s scarcely enough to recommend one bet over another. 4.8. We have the following table of prizes and probabilities: k 3 4 5 6

K21613_SM_Cover.indd 14

P(Match k ) .0199 .0011 2.291 × 10−5 9.313 × 10−8

Prize 5 100 2500 1,000,000

9/22/14 4:20 PM

Solutions

11

Multiplying the probabilities by the payoffs gives .3499, from which we must subtract the 1 cost of the ticket. The expectation is then - .6501, or about -65 . 4.9. The only way that switching the order of these two lines matters is if a player is holding a hand that contains both combinations. Since no five-card poker hand can contain both a full house (2 different ranks) and a 4-card royal flush (4 or 5 different ranks), these lines can be interchanged without changing the play of the game for a player using the table correctly. 4.10a. Let p be your probability of picking a game correctly. Then if the wager is fair, E = (5) · p3 + (−1) · (1 − p3 ) = 6p3 − 1 = 0, which is true if

√

1 ≈ .5503. 6 4.10b. To make the 9 for 9 bet fair, we require that p=

3

E = (249) · p9 + (−1) · (1 − p9 ) = 250p9 − 1 = 0, which is true if

√

1 ≈ .5415. 250 Given this probability of success, the expectation in the 10 for 10 wager is p=

9

E = (699) · p10 + (−1) · (1 − p10 ) = 700p10 − 1 ≈ .5161, resulting in a 51.61% player advantage. 4.11. For a bet on the 2 spot: ( ) ( ) 39 9 1 15 + (−1) · =− = − ≈ −$.1667. E = (2) · 54 54 54 6 This is the best of the remaining bets. For a bet on the 5 spot: ( ) ( ) 47 12 2 7 + (−1) · =− = − ≈ −$.2222. E = (5) · 54 54 54 9 For a bet on 10: ( ) ( ) 10 5 50 4 + (−1) · =− =− ≈ −$.1852. E = (10) · 54 54 54 27 For a bet on 20: ( ) ( ) 52 12 2 2 + (−1) · =− = − ≈ −$.2222. E = (20) · 54 54 54 9

K21613_SM_Cover.indd 15

9/22/14 4:20 PM

12

Basic Gambling Mathematics: The Numbers Behind The Neon For a bet on either one of the two logos: ( ) ( ) 53 13 1 1 + (−1) · =− = − ≈ −$.2407. E = (40) · 54 54 54 9

This is the worst wager on a wheel full of bad bets. 4.12. If the 90% payback figure is accurate, there are 67 spots which hold a number corresponding to a 1 prize. (The 5 payoff for punching out the last spot is not an option for the first player.) Since the 10 paid to punch is not refunded with a win, the two possible outcomes are winning a net total of .90 or losing .10. The expectation is therefore ) ( ) ( 733 130 67 + (−.10) · = −$ = −$.01625. E = (.90) · 800 800 8000 Dividing by the 10 of the original bet gives a house advantage of 16.25%. 4.13. Two cases must be considered: 67 1. The first punch wins, an event with probability . There are 66 win800 ning spots remaining. The expected return in this case is ) ( )] [ ( 733 66 E1 = (.90) · + (−.10) · . 799 799 733 . All 67 winning 2. The first punch loses, an event with probability 800 spots remain in play. Here, the expectation is ) ( )] [ ( 732 67 E2 = (.90) · + (−.10) · . 799 799 The total expectation is ) ) ( ( 67 733 · E1 + · E2 = −$.01625. E= 800 800 The house advantage is the same 16.25% that the first player faced. 4.14. The number of wins is a binomial random variable X with parame1 39999 ters n = 40000 and p = . With q = 1 − p = , we have 40000 40000 P (X = 2) = (40000 C2 ) · p2 · q 39998 ≈ .1839. 4.15. Let X be the number of wins. X is binomial, with parameters n = 10 and p = .5. The probability of picking 5 of the 10 games correctly is P (X = 5) =

K21613_SM_Cover.indd 16

10 C5 210

=

252 ≈ .2461. 1024

9/22/14 4:20 PM

Solutions

13

Though this is the most likely outcome, its probability is still only about 25%. 4.16. The number of 7’s is a binomial random variable X with parameters n = 400 and p = 16 . For a result to be statistically significant, it would need to be at least 2 standard deviations above the mean. The mean of X is µ=n=

400 ≈ 66.6667 6

and the standard deviation is √ 1 5 √ √ σ = npq = 400 · · ≈ 55.5556 ≈ 7.4536. 6 6 The required number of 7’s is then µ + 2σ ≈ 81.5738, so if the shooter rolls 82 or more 7’s in the 400 throws, he or she has attained a statistically significant result. 4.17. The expectation is ( ) ( ) 1 63 8 E = (55) · + (−1) · =− = −$.125, 64 64 64 and so the HA is 12.5%, comparable to but slightly less than the value for standard craps. 4.18. The 10 winning rolls cover 30 different combinations of the 2d8. One of these, the 16, pays off at 2-1 and the others pay off at 1-1. We have the following probability distribution for the winnings X: x P (X = x)

-1

1

2

34 64

29 64

1 64

The expectation is ( ) ( ) ( ) 29 34 3 1 + (1) · + (−1) · =− ≈ −$.0469, E = (2) · 64 64 64 64 and the house edge is thus 4.69%. 4.19. There are two bets, 2 and 16, for which the net payoff is 55-3 = 52 and two, 3 and 15, for which the net payoff is 25. The corresponding probability distribution is x P (X = x)

-4

52

25

58 64

2 64

4 64

The expectation is ( ) ( ) ( ) 4 58 28 2 + (25) · + (−4) · =− ≈ −$.4375. E = (52) · 64 64 64 64 Dividing by the 4 wagered gives a house edge of 10.94%.

K21613_SM_Cover.indd 17

9/22/14 4:20 PM

14

Basic Gambling Mathematics: The Numbers Behind The Neon 4.20. Let the desired payoff be x to 1. We seek to solve the equation (x) · (.0955) + (−1) · (.9045) = −.04.

Isolating the variable gives x=

.8645 ≈ 9.05. .0955

Setting the payoff at 9-1 would produce a wager with a HA of 4.5%, which is the closest to 4% that is achievable without using fractions. 4.21. Assuming no commission on a Banker bet, the probability distribution for either bet will then be x P (X = x)

-1 .5539

1 .4461

(see Example 4.2.21) and the corresponding expectation is E(X) = (−1)(.5539) + (1)(.4461) = −$.1078, so the house advantage is 10.78%. A good game for the gambler has been reduced to one with a carnival-game level of house edge.

Chapter 5 5.1. There are seven rolls that will resolve the All Day 2 bet: any 7 or a 1-1. The bet wins if 1-1 is rolled before any of the 7’s. ( ) ( ) 6 1 1 + (−1) · = − ≈ −.1429, E = (5) · 7 7 7 so the HA is 14.3%. The same percentage applies to All Day 12. 5.2. Having completed the calculations for chuck-a-luck makes this expected value easy to compute. Suppose that we bet 1 on 5. The probability 1 of rolling two 5’s is 36 , and the probability of rolling exactly one 5 is 10 36 . A 1 bet on 5, therefore, has an expected return of ( ) ( ) ( ) 10 25 1 1 + (2) + (−1) =− ≈ −$.028, E = (4) 36 36 36 36 for a 2.8% house advantage—considerably less than the 7.9% HA for standard chuck-a-luck, and in line with many standard craps bets. 5.3. For any number chosen, one die contributes a factor of 1 to any probability, since the number cannot appear on it. We have the following probability distribution for X, the number of times that number appears: x P (X = x)

K21613_SM_Cover.indd 18

5 6

·

5 6

0 =

1 25 36

2·

1 6

·

5 6

=

10 36

1 6

·

1 6

2 =

1 36

9/22/14 4:20 PM

Solutions

15

The resulting expectation is ( ) ( ) ( ) 10 1 13 25 + (1) · +2· =− ≈ .3611. E = (−1) · 36 36 36 36 The HA is 36.1%. 5.4a. The probability of rolling a 2 (or a 12) before a 7 is 1/7, so the odds against this event are 6-1. An odds bet should pay 6-1 when the point is 2 or 12. 5.4b. The probability of rolling a 3 (or an 11) before a 7 is 2/8 or 1/4, so the odds against this event are 3-1. An odds bet should pay 3-1 when the point is 3 or 11. 5.4c. Using the results from parts a. and b. gives E = 5·

6 2 2 2 4 50 7303 +305· +155· +105· +80· +65· −55· ≈ −.2691. 36 252 72 36 45 396 13860

Dividing by the 55 wager yields a HA of .489%. 5.5. If p is the probability of a given number being rolled before a 7 is rolled, then independence tells us that the probability of that number being rolled twice before a 7 is rolled once is p2 . If the return on a 1 bet is $X, we have the following expectation: E = (X) · p2 + (−1) · (1 − p2 ) = p2 · (X + 1) − 1. The following expectations apply to the different rolls: Rolls 2, 12 3, 11 4, 10 5, 9 6, 8

p 1/7 2/8 3/9 4/10 5/11

X 45 14 7 5 3.5

E −4/49 ≈ −.0816 −1/8 = −.1250 −1/9 ≈ −.1111 −1/25 = −.0400 −17/242 ≈ −.0702

The best (least negative) expectation arises when betting on the 5 or 9. 5.6. If X is the number of 7’s appearing in four tosses, then X is a binomial random variable with parameters n = 4 and p = 16 . We have ( )4 5 625 = P (X = 0) = 4 C0 · 6 1296 and thus E = (1) ·

(

625 1296

)

+ (−1) ·

(

671 1296

)

=−

46 ≈ −$.0355. 1296

5.7. The expected value of a 1 bet is ( ) ( ) 35 1 1 + (−1) · =− =≈ −$.0278, E = (34) · 36 36 36

K21613_SM_Cover.indd 19

9/22/14 4:20 PM

16

Basic Gambling Mathematics: The Numbers Behind The Neon

which is considerably better than the 1/19 or - .0526 expectation on an American wheel and about the same as the 1/37 or - .0270 expectation on a European wheel. 5.8. ( ) ( ) 1 168 1 E = (1200) · ≈ .1227. + (−1) · 1 − 2 = − 2 37 37 1369 The HA is therefore approximately 12.3%. 1 5.9a. P(Blue) = 25 , and so ( ) ( ) 24 1 + (−1) · = 0. E = (24) · 25 25 (Yes, this bet is fair!) 5.9b. For any other color, let X be the return on a 1 bet. We have the following probability distribution for X: x P (X = x)

-1

2

3

4

19 25

2 25

3 25

1 25

and the corresponding expectation is ( ) ( ) ( ) ( ) 2 3 1 2 19 + (2) · + (3) · + (4) · =− = −$.08. E = (−1) · 25 25 25 25 25 5.9c. In light of the results in parts a. and b., the best strategy is to bet only on Blue. The expected return is 0, which is better than the return for any other color. 5.10. For the 1-1 bets, the common expectation is ( ) ( ) 16 20 4 1 E = (1) · + (−1) · =− = − = −0.1111. 36 36 36 9 For a single-number bet, the expectation is ( ) ( ) 32 4 1 4 + (−1) · =− = − = −0.1111. E = (7) · 36 36 36 9 In either case, the house advantage is 11.1%. 5.11. For a single-zero wheel, ( ) ( ) 2 7 30 E = (4) · + (−1) · =− ≈ −.0540. 37 37 37 This is double the HA of the standard single-zero roulette bets. For a double-zero wheel, replacing the 37’s with 38’s and the 30 by 31 in 3 this formula gives E = − 38 ≈ −.0789. This is the same as the HA for the basket bet.

K21613_SM_Cover.indd 20

9/22/14 4:20 PM

Solutions

17

5.12.The expectation on a 2 white bet is ( ) ( ) 30 4 8 + (−2) · =− , E = (7) · 38 38 38 and dividing by 2 gives the familiar roulette house advantage of 5.26%. 5.13. The three dice may land in 216 ways. The following chart shows the number of ways to roll each number from 11 to 17, without rolling the excluded triples of 4-4-4 and 5-5-5: Roll Ways

11 27

12 24

13 21

14 15

15 9

16 6

17 3

There are 105 winning combinations. Since the bet pays off at even money, the expectation is ( ) ( ) 105 111 6 1 E = (1) · + (−1) · =− =− ≈ −.0278, 216 216 216 36 and so the HA is about 2.78%. 5.14. There are six triples. The expectation is ( ) ( ) 6 186 6 1 E = (30) · + (−1) · =− =− ≈ −.0278, 216 216 216 36 for a HA of about 2.78%. 5.15. For the first game, we have the following probability distribution for your net winnings: If zero or one of your three numbers are drawn, you lose your 1 bet. This event has probability P (−1) =

77 C20

+

3 C1

80 C20

·

77 C19

=

3481 ≈ .8474. 4108

If exactly two of your numbers are drawn, you break even—note that the probabilities stated are all “x for 1”. This probability is P (0) =

3 C2

·

77 C19

80 C20

=

285 ≈ .1388. 2054

Finally, if all three of your numbers are drawn, your net profit is 40. P (40) =

77 C17 80 C20

=

57 ≈ .0139. 4108

Your expectation on a 1 wager is therefore E = (0) · P (0) + (40) · P (40) + (−1) · P (−1) = −

K21613_SM_Cover.indd 21

1201 ≈ −.2924, 4108

9/22/14 4:20 PM

18

Basic Gambling Mathematics: The Numbers Behind The Neon

and so the HA is 29.24%. For the second game, your only winning option is when you match all three numbers. Your net win is 125.50; we see from above that this event has probability 57 77 C17 P (125.50) = ≈ .0139. = 4108 80 C20 The expected value of this second wager is E = (125.50) · P (125.50) + (−2.50) · (1 − P (125.50) ≈ −.7240. We must divide this by the amount of the bet to get the HA, and so it follows that .7240 HA = ≈ 28.96%. 2.50 The second bet has a lower house advantage—but not by much. 5.16. The probability distribution for the field bet with a 100 nonnegotiable chip is x P (X = x)

0

100

200

20 36

14 36

2 36

.

The expectation is E = (200) ·

(

2 36

)

+ (100) ·

(

14 36

)

= $50,

exactly half the face value of the chip.

Chapter 6 6.1. The count after the first hand is complete is 0—there were four low cards, two neutral cards, and four high cards dealt. At the start of the second hand, you have seen three high cards, and so the running count is -3. With 13 cards dealt, 3/4 of the deck remains, and so the true count is -4. The count alone suggests that insurance is not a wise bet—see Table 6.8. The remaining 39 cards include 11 ten-count cards that will complete a dealer natural. If you do not make the insurance bet, your expectation is ( ) ( ) 11 28 E = (0) · + (15) · = $10.77. 39 39 If you make the 5 insurance bet, you win 10 whether the dealer has a natural or not: 10 from the insurance bet combined with a push on your initial bet if the dealer turns over a ten, and 15 from your own natural minus the 5 insurance bet if not. Either way, your expectation is less than the value above. You should not make the insurance bet. 6.2a. In all three cases, there are 101 cards left unseen. If n denotes the

K21613_SM_Cover.indd 22

9/22/14 4:20 PM

Solutions

19

number of aces that you hold, it follows that the probability that the dealer has a natural is 8−n p= , 101 and the expectation of this bet is then ) ( ) ( 13 + 9n 93 − n 8−n + (−1) · =− . E = (10) · 101 101 101 If n = 0, then the expectation is - .1287. If n = 1, the expectation is - .2178, and if n = 2, the expectation is - .3069. 6.2b. If you have a natural, then n = 1 above. Assume that your initial bet is for $x and that you also make the insurance bet for $.5x. There are two possible outcomes: The dealer does not have a natural, in which case you lose your insurance bet and win at 3-2 on your initial bet, for a total return of $x. The dealer does have a natural. Here, your initial bet pushes and your insurance bet pays off a total of $5x. Making the bet gives you an expected value of ( ) ( ) 7 94 129x E = (5x) · + (x) · = ≈ 1.277x. 101 101 101 If you do not make the insurance bet, then you either win 0 or 1.5x, according as the dealer has or does not have a natural. The corresponding expectation in this case is ( ) ( ) 94 7 141x E = (1.5x) · + (0) · = ≈ 1.396x. 101 101 101 As with the standard insurance bet, your expectation is higher if you do not make this insurance bet. 6.3. We seek the probability of drawing three identical 7’s. In the infinite deck approximation, this probability is ( ) ( ) ( ) 4 1 1 1 p= · · = ≈ 2.845 × 10−5 , 52 52 52 35152 where the first factor is the probability of drawing any 7 and the next two factors are the probability that the other two match the first card drawn. 6.4. The probability distribution for X, the random variable measuring your net winnings, is x P (X = x)

K21613_SM_Cover.indd 23

-1 0 1/2 1/26

1 12/26

9/22/14 4:20 PM

20

Basic Gambling Mathematics: The Numbers Behind The Neon

and the corresponding expectation is ( ) ( ) ( ) 1 12 1 1 + (0) · + (1) · =− ≈ −$.0385. E = (−1) · 2 26 26 26 The house advantage on this bet is therefore 3.85%. It is worth noting that this side bet is susceptible to a modified form of card counting that tracks the colors of dealt cards (see Blackjack Attack for details). A system that counts red cards as +1 and black cards as -1 indicates an advantage to the player making this bet if the true count is greater than 2 (bet red) or less than -2 (bet black); keeping a secondary count for the number of deuces already played can further enhance this edge. 6.5. The correct strategy, as outlined by Thorp in Beat The Dealer, is to bet as much as you can on the next hand. Borrow money if you have to, for you are a guaranteed winner. You and the dealer will each be dealt a total of 14, 15, or 16, with one card remaining to be dealt. You stand, and the dealer must draw. If the dealer has 14, the remaining card must be an 8, and the dealer busts. If the dealer’s total is 15 or 16, the remaining card—whether 7 or 8—is a certain bust card. Enjoy your win. 6.6. The appeal of depth charging is that by the time you get to your last hand—the one with the largest wager—you have seen all of the cards for the previous six hands, plus the dealer’s upcard. An average blackjack hand uses 2.7 cards, so approximately 19 cards, counting the two that comprise the seventh hand, have been revealed. If any earlier hand has split a pair, this number goes up. With this knowledge of over a third of the deck, you are in a good position to adjust your play based on your information. Smaller advantages, of course, apply to hands 4, 5, and 6. 6.7. In a fresh six-deck shoe, there are 312 cards, and so 312 C4 = 387278970 ways to draw four cards. We shall denote this number by N . The distribution of the four cards between the two player hands is immaterial. The probabilities for the four payoff hands are as follows: 136401408 13 · (24 C2 ) · (12 C2 ) · (24 C1 )2 = . N N 5941728 (13 C2 ) · (24 C2 )2 = . p2 = P(2 pairs) = N N 7577856 13 · (24 C3 ) · 12 · (24 C1 ) p3 = P(3 of a kind) = = . N N 138138 13 · (24 C4 ) p4 = P(4 of a kind) = = . N N p1 = P(1 pair) =

The probability of losing this wager is therefore p5 =

K21613_SM_Cover.indd 24

237219840 ≈ .6125, N

9/22/14 4:20 PM

Solutions

21

and the expectation is E = (1) · p1 + (8) · p2 + (5) · p3 + (40) · p4 + (−1) · p5 = −

9869808 ≈ −.0255, 387278970

giving a HA of about 2.55%. 6.8. For the Perfect Pairs bet, we are looking for the probability that the second card matches the rank of the first, and optionally that the color or suit are also matched. The first card does not matter; all we must do is count the appropriately matching cards from the remaining 259. The Mixed Pair pays off on the 10 cards of the same rank but opposite color. For the Colored Pair, there are 5 cards of the same color but the other suit. Finally, 4 identical cards remain for the Perfect Pair. 240 cards of different ranks, for which the bet loses, remain in the shoe. The expectation is ( ) ( ) ( ) ( ) 4 5 10 240 20 E = (30) · + (10) · + (5) · + (−1) · =− , 259 259 259 259 259 and the HA is 7.72%. 6.9. A six-deck shoe contains 12 red 5’s. The probability of drawing four of them is p=

12 C4 312 C4

=

495 1 ≈ 1.278 × 10−6 ≈ . 387278970 782382

For n > 2 decks, this probability becomes p=

2n C4 52n C4

=

(2n)! (2n−4)!·4! (52n)! (52n−4)!·4!

=

(2n)! · (52n − 4)! (2n − 4)! · (52n)!

(2n)(2n − 1)(2n − 2)(2n − 3) (52n)(52n − 1)(52n − 2)(52n − 3) 16n4 − 48n3 + 44x2 − 12n . = 7311616n4 − 843648n3 + 29744n2 − 312n

=

In the infinite deck approximation, the probability of drawing a red 5 is 1 fixed at 26 , regardless of how many cards have been dealt, so the probability of drawing four red 5’s is ( )4 1 1 p= ≈ 2.188 × 10−6 . = 26 456976 This can also be found by taking the limit of the general expression for n > 2 decks as n → ∞. Either way, it’s not very likely. It is worth noting that a player pursuing this bonus would be opting not to double down on a dealt total of 10, and so would be passing up a favorable play for a longshot. That forgone opportunity makes this side bet even less attractive for the player—and possibly more attractive for the casino, although this wager is not currently offered anywhere.

K21613_SM_Cover.indd 25

9/22/14 4:20 PM

22

Basic Gambling Mathematics: The Numbers Behind The Neon

6.10a. The high-low running count before the hand is dealt is 45-38 = +7. After the deal, the running count is +6. Dividing by the two decks remaining gives a true count of +3, indicating a surplus of high cards. Table 6.8 indicates that you should stand on 16 against a 10 if the true count is ≥ 0, so the correct play is to stand on your 16. 6.10b. The K-O count for a 4-deck game starts at -12. The count at this point is −12 + 45 + 4 − 38 = −1, counting the 7’s as low cards. Once the cards have been dealt, the count remains -1, as the 7 and 10 cancel each other out. Referring to Table 6.10, we see that the index number for 16 vs. 10 in a four-deck game is -1, and since this is the current count, you should deviate from basic strategy and stand on 16. 6.11a. For a six-deck game, the first card may be any of twelve: six A♢’s and six J♢’s. The second card must be one of the six cards of the rank not dealt first, so we have P (Pinkjack) =

12 6 72 · = ≈ 7.42 × 10−4 . 312 311 97032

6.11b. From the top of a six-card shoe, the probability that the dealer has a pinkjack, given that the player has a pinkjack, is p=

10 5 50 · = ≈ 5.22 × 10−4 . 310 309 95790

72 . By the We know that the probability that the player has a pinkjack is 97032 General Multiplication Rule, we have P (Two pinkjacks) =

72 50 3600 · = ≈ 3.87 × 10−7 . 97032 95790 9294695280

6.11c. In the infinite deck approximation, the probability of a single pinkjack is 2 2 1 · = . p= 52 52 2704 The probability of two pinkjacks is simply )2 ( 4 2 2 ≈ 5.47 × 10−7 . = p = 2704 7311616 6.12. The Instant 18 wager wins if the dealer’s total is 17 or if the dealer busts, which has probability .1458 + .2836 = .4294. The wager is a push if the dealer has an 18 (p = .1381) and loses otherwise (p = .4325). The expectation is E = (1) · (.4294) + (−1) · (.4325) = −$.0031, and the house advantage is .31%. The HA will increase with the number of decks in play and will rise if the dealer hits soft 17. 6.13. With 8 hands in play and an average of 2.7 cards per hand, there

K21613_SM_Cover.indd 26

9/22/14 4:20 PM

Solutions

23

will be approximately 22 cards dealt out on the first round. There is one way to choose the four aces and 48 C18 ways to choose the other 18 cards. The probability that all four aces are dealt is thus p=

48 C18 52 C22

≈ .0270,

a small, but not insignificant probability. 6.14. The only way to beat a dealer 20 is with a 21, which requires drawing a 7 to your total of 14. Since two of the four sevens in the deck are accounted 2 for in your hand, your chance of drawing a third seven is only 49 with no other cards known. Drawing two smaller cards that add up to 7 is also an unfavorable proposition. Standing on 14 vs. 10 is a defensive move designed to lose less in the long run; you may be unlikely to win, but you have more of a chance if you stand and take your chances that the dealer does not have 20 and will bust.

Chapter 7 7.1. The expected return is ( ) ( ) ( ) 22 12 5 4 + (1) · + (−5) · =− . E = (7) · 38 38 38 19 Dividing by the 5 wager gives a HA of 5.26%. 7.2. While it is true that red and black occur equally often, they do not occur 50% of the time each. The 0 and 00 on the wheel mean that the probability of losing a bet on Red is greater than the probability of winning; in practice, this means that you’ll be increasing your bets under the d’Alembert system after a loss slightly more often, in the long run, than you’ll be decreasing them after a win. More time spent making large bets with a negative expectation translates into a diminishing bankroll. This system, like many others, may lead to small wins in the short term, but when the losing spins come around, as they inevitably will, the losses mount rapidly. 7.3a. Let X be the number of times your number hits in 50 spins. X is binomial with parameters n = 50 and p = 1/38. The probability of exactly 1 hit is ( )1 ( )49 37 1 P (X = 1) = (50 C1 ) · · ≈ .3562. 38 38 The probability that you will profit from this system is the probability of at least one hit in 50 spins, or P (X ≥ 1) = 1 − P (X = 0) = 1 −

(

37 38

)50

≈ 1 − .2636 = .7374.

7.3b. If the first hit is on spin 20, you lose 95 before winning 175, for

K21613_SM_Cover.indd 27

9/22/14 4:20 PM

24

Basic Gambling Mathematics: The Numbers Behind The Neon

a total profit of 80. If the first hit is on the 40th spin, then you lose 290 before winning 350, and your net profit is 60. Finally, if it takes until spin 50 before your number comes up for the first time, you have lost 435 before winning 525, and your profit is 90. If your number comes up on any other spin, you will lose less before winning either 175, 350, or 525, and so will still make a profit. 7.3c. You lose if your number does not come up once in 50 spins. The probability of this event is ( )50 37 P (X = 0) = ≈ .2636. 38 If you follow this system, your total loss will be 450, and this will happen roughly one time in four. 7.4. If you put all 500 on a single Field bet, you have a 1 in 18 chance of tripling your money to 1500, 14 chances in 36 of doubling your money, and 20 chances in 36 of losing everything on the first roll. The probability of having at least 1000 after the next roll of the dice is 16 ≈ .4444. 36 This is greater than the probability of .4396 derived in Example 7.6.1. It follows that, though the difference in probabilities is small, the increased payoff on a 2 or 12 is not large enough to justify risking your money across multiple rolls. 7.5. If you place a 1 wager on each of the 12 numbers, then your possible outcomes are a net win of 24 (35-11), with probability 12 38 , and a loss of 12, with probability 26 38 . The expectation of the Red Snake is p=

E = (24) ·

26 24 12 + (−12) · =− . 38 38 38

Dividing by the 12 wagered gives a HA of 5.26%. 7.6. The probability that a designated number will come up at least once in 19 spins is 1 minus the probability that the number will not be spun at all. We have ( )19 37 p=1− ≈ .3975, 38

or 39.75%, rather less than the 50% claimed by the purveyor of this system. 1 7.7. The binomial random variable X has parameters n = 19 and p = 38 . If 00 comes up on k spins, where 0 ≤ k ≤ 19, the total amount won on a 1 wager is 35 for each win, minus 1 for each loss, or 35k − (19 − k) = 36k − 19. The expected value of 19 consecutive bets on 00 is [ ( )k ( )19−k ] 19 ∑ 37 1 E= · = −1. (36k − 19) · (19 Ck ) · 38 38 k=0 �� � � Binomial P (X=k)

K21613_SM_Cover.indd 28

9/22/14 4:20 PM

Solutions

25

Dividing by the 19 wagered gives a house advantage of 5.26%, which should come as no surprise. 7.8.There are three possible outcomes to a Shotwell bet: One of your four single-number bets hits. Your net profit is 35 - 4, or 31. One of the numbers covered by the six-number bet wins. Your net win is 5 - 4 = 1. One of the 28 uncovered numbers is spun, and you lose 5. The expected value of this bet is then ( ) ( ) ( ) 4 6 28 10 E = (31) · + (1) · + (−5) · =− . 38 38 38 38 Since 5 was originally wagered, the house edge is found by dividing E by 1 5, which yields − 19 , or an advantage of 5.26%. 7.9. The only difference between this bet and the previous bet is that a 2 wager on the double street carries a 10 payoff if it hits. With 6 at risk rather than 5, the expectation is ( ) ( ) ( ) 6 28 12 4 + (10 − 4) · + (−6) · =− . E = (35 − 5) · 38 38 38 38 On division by the 6 wager, our friend 5.26% returns. 7.10. Recall that the following four criteria define a “good” combination for a 6/47 lottery: The sum of the numbers is at least 166, with at least five numbers greater than 31. The cluster number is between 2 and 5. The edge number is at least 4. The edge numbers on this slip are 1 through 6, any number ending in a 0,1,5, or 6, and 42 through 47. The AC is at least 6. We begin with an eye toward generating a large sum. The combination {11, 33, 38, 41, 45, 47} has a sum of 215, easily exceeding the minimum. To achieve an arithmetic complexity of at least 6, we need at least 11 positive differences. The positive differences of this set are 2, 3, 4, 5, 6, 7, 8, 9, 12, 14, 22, 26, 30, 34, and 36, and so the AC is 15-5 = 10—the maximum possible. Note that choosing one very low number (11) increases the number of differences—if 11 is replaced by 31 to drive up the sum, the AC of the new combination drops to 7. Plotting these numbers on the bet slip gives the following pattern:

K21613_SM_Cover.indd 29

9/22/14 4:20 PM

26

Basic Gambling Mathematics: The Numbers Behind The Neon 1 6 11 16 21 26 31 36 41 46

2 7 12 17 22 27 32 37 42 47

3 8 13 18 23 28 33 38 43

4 9 14 19 24 29 34 39 44

5 10 15 20 25 30 35 40 45

The bet slip shows a cluster number of 4 and an edge number of 4, so the final two criteria are met. (Other solutions are of course, possible.) 7.11a. From page 55, the probability of losing a single Don’t Pass bet is p = .492, which is just the probability of winning a Pass bet. The probability of losing 4 straight Don’t Pass bets is then p4 ≈ .0586. 7.11b. With p defined in part a., the chance of losing 16 straight Don’t Pass bets is then 1 p16 ≈ 1.17 × 10−5 ≈ . 84832 7.11c. The question asks for the probability of winning at least 1 of 4 Don’t Pass bets, 15 times. The probability of winning at least 1 Don’t Pass bet in 4 tries is (1 - the result from part a.), or .9414. Raising this to the 15th power gives P (Win back all losses) = .941415 ≈ .4042. Your chance of battling back to a break-even level is actually not bad.

K21613_SM_Cover.indd 30

9/22/14 4:20 PM

K21613

w w w. c r c p r e s s . c o m

K21613_SM_Cover.indd 31

9/22/14 4:20 PM