Presentation is loading. Please wait.

Presentation is loading. Please wait.

- How to Be A Winner - The Maths of Race Fixing and Money Laundering John D Barrow.

Similar presentations


Presentation on theme: "- How to Be A Winner - The Maths of Race Fixing and Money Laundering John D Barrow."— Presentation transcript:

1 - How to Be A Winner - The Maths of Race Fixing and Money Laundering John D Barrow

2 Why is Probability Theory Not Ancient? Religious beliefs Or No concept of equally likely outcomes ? ? ?

3 “And they said every one to his fellow, Come, and let us cast lots, that we may know for whose cause this evil is upon us. So they cast lots and the lot fell upon Jonah.” Book of Jonah 1 v 7 St. Augustine: “We say that those causes that are said to be by chance are not nonexistent but are hidden, and we attribute them to the will of the true God”

4 Sheep's’ ankle bones, 6-sided, numbered, asymmetrical Divination with sets of 5 in Asia Minor from 3600 BC Eventually replaced by dice Astragali

5 Ancient Dice The most popular dice game of the Middle Ages: was called “hazard” Middle Ages: was called “hazard” Arabic “al zhar” means “a die.” Roman icosahedral die 20 faces

6 Western dice are right-handed: if the 1-spot is face up and the 2-spot is turned to face the left then the 3-spot is to the right of it. Chinese dice are left-handed: they will have the faces the opposite way round. Right and Left-handed Dice

7 The Problem of the Points Chevalier de Méré and Blaise Pascal and Pierre de Fermat 1654 Two people play a fair game The first to win six points takes all the money. How should the stakes be divided if the game is interrupted when one has 5 points and the other 3? HHH, HHT, HTH, TTT, THT, TTH, THH. HTT Player with 3 points has to win all the next 3 games. He has 1/8 chance of doing that. His opponent has a 7/8 chance of winning 1 more game. Give 7/8 of prize money to the one with 5 and 1/8 to the other

8 More Chevalier de Méré He won lots of money betting on at least 1 six in 4 rolls of a die based purely on experience Probability of no 6 is 5/6 Probability of no 6 in four throws is 5/6  5/6  5/6  5/6 = (5/6) 4 = 625/1296 Probability of one 6 is 1 – 625/1296 = 671/1296 = > 1/2 So he thought that he should bet on one or more double 6’s occurring in 24 rolls of 2 dice Probability of no double sixes in 24 throws is (35/36) 24 = Probability of one double six is 1 - (35/36) 24 = < 1/2 After a while he stopped doing this !

9 Winning The Toss Australian Open January 2008

10 Playing Fair With a Biased Coin Unequal probability of H and T: p  ½ Probability of H is p Probability of T is 1-p Toss twice and ignore pairs HH and TT Probability of HT is p(1-p) Probability of TH is (1-p)p Call combination HT ‘Newheads’ Call combination TH ‘Newtails’ Newheads and Newtails are equally likely Efficiency is poor (50%) – discard the HH and TT s

11 Faking Random Sequences 1. THHTHTHTHTHTHTHTHTTTHTHTHTHTHTHH 2. THHTHTHTHHTHTHHHTTHHTHTTHHHTHTTT 3. HTHHTHTTTHTHTHTHHTHTTTHHTHTHTHTT Do these look like real random sequences ?

12 4. THHHHTTTTHTTHHHHTTHTHHTTHTTHTHHH 5. HTTTTHHHTHTTHHHHTTTHTTTTHHTTTTTH 6. TTHTTHHTHTTTTTHTTHHTTHTTTTTTTTHH Some More Candidates With 32 tosses Are they random?

13 4. THHHHTTTTHTTHHHHTTHTHHTTHTTHTHHH 5. HTTTTHHHTHTTHHHHTTTHTTTTHHTTTTTH 6. TTHTTHHTHTTTTTHTTHHTTHHHHHHTTTTH Some More Candidates With 32 tosses The chance of a run of r heads or r tails coming up is just ½  ½  ½  ½  ….  ½, r times. This is 1/2 r If we toss our coin N > r times there are N different possible starting points for a run of heads or tails Our chance of a run of length r is increased to about N  1/2 r A run of length r is going to become likely when N  1/2 r is roughly equal to 1, that is when N = 2 r. Note that 32 = 2 5

14 Winning (and Losing) Streaks The Nasser Hussain Effect England cricket captain During Atherton took over for one game after he had lost 7 and won the toss Normal service was then resumed There is a 1 in 2 14 = chance of losing all 14 tosses But he captained England 101 times and there is a chance of about 1 in 180 of a losing streak of 14 “Flipping useless, Nasser!” BBC

15 Can You Always Win? Or avoid ever losing ?

16 The Win-Win Scenario The odds for the runners are a 1 to 1, a 2 to 1, a 3 to 1, and so on, for any number of runners in the race. If the odds are 5 to 4 then we express that as an a i of 5/4 to 1 Bet a fraction 1/(a i +1) of the total stake money on the runner with odds of a i to 1 If there are N runners, we will always make a profit if Q = 1/(a 1 +1) + 1/(a 2 +1) + 1/(a 3 +1) +….+ 1/(a N +1) < 1 Winnings = (1/Q – 1)  our total stake Example: Four runners and the odds for each are 6 to 1, 7 to 2, 2 to 1, and 8 to 1 and. Then we have a 1 = 6, a 2 = 7/2, a 3 = 2 and a 4 = 8 and Q = 1/7 + 2/9 + 1/3 + 1/9 = 51/63 < 1 Allocate our stake money with 1/7 on runner 1, 2/9 on runner 2, 1/3 on runner 3, and 1/9 on runner 4 We will win at least 12/51 of the money we staked (and of course we get our stake money back as well).

17 Race Fixing ‘101’ The favourite is always the largest contributor to Q because a 1 is the smallest of the a i s We could have Q > 1 with all runners included Q = 1/(a 1 +1) + 1/(a 2 +1) +….. > 1 But if you know the favourite has been hobbled then you calculate Q excluding a 1 which can result in Q fix = 1/(a 2 +1) + 1/(a 3 +1) + …. < 1

18 If there are 4 runners with odds 3 to 1, 7 to 1, 3 to 2, and 1 to 1 Q = 1/4 + 1/8 + 2/5 + 1/2 = 51/40 > 1 So we can’t guarantee a winning return Dope the favourite and place you money on the other three runners only, betting 1/4 of our stake money on runner 1, 1/8 on runner 2, and 2/5 on runner 3 You are really betting on a 3-horse race with Q fix = 1/4 + 1/8 + 2/5 = 31/40 < 1 Whatever the outcome you will never do worse than winning your stake money plus {(40/31) -1}  Stake money = 9/31  Stake money

19 OutcomeBookmaker 1’s odds Bookmaker 2’s odds Oxford win Cambridge win Q of Bookie >1He gains 5.6% Q of Bookie 2He gains 5.1%1.051 > 1 A Mixed Strategy Back Oxford with Bk 2 and Cambridge with Bk 1 Q = Q = < 1 You can earn 4.6% Bet 100 on Oxford with bookie 2 and 100 x 1.43 / 3.9 = on Cambridge at bookie 1. If Oxford win, you collect 100 x 1.43 = 143 from bookie 2. If Cambridge win, you could collect x 3.9 = 143 from bookie 1. You invested and collect 143, a profit of 6.33 (4.6%) no matter what the outcome. When Bookies Disagree

20 What About the Q > 1 Situations This is the money-laundering case You are guaranteed a loss of (1 - 1/Q) of your stake money That is the cost of the laundering and carries no risk of greater loss

21 Weird Judging Means Ice Skating Ladies Figure Skating Salt Lake City Olympics

22 SkaterShortLongTotal Kwan Hughes Cohen Slutskaya1.0?? Before the last competitor skates… Lowest scores lead

23 SkaterShortLongTotal Hughes Slutskaya Kwan Cohen And after Slutskaya skates… Hughes wins by tie-break! Slutskaya has changed the order of Hughes and Kwan

24

25 This is scored as a draw Even though Lewis has won on rounds

26 The Moral Don’t add preferences or ranks If A best B and B beat C It doesn’t mean A beats C Preference votes ABC, BCA, CAB imply A bts B 2-1 and B bts C 2-1 But C bts A 2-1

27 The Three-Box Trick Monty Hall – 2 goats and 1 car

28 1 2 3 Prob = 1/3 Prob = 1/3 Prob = 1/3  Prob = 2/3  1 2 now open 3 Prob = 1/3 Prob = 2/3 Prob = 0 You choose Box 1: he opens Box 3 So you should switch from Box 1 to Box 2

29 You are Twice As Likely to Win if You Switch than if You Don’t !

30


Download ppt "- How to Be A Winner - The Maths of Race Fixing and Money Laundering John D Barrow."

Similar presentations


Ads by Google