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Probability in action

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Probability in action

  1. 1. PROBABILITY IN ACTION : CAR INSURANCE , PROMOTION,GAMBLING & THE US PRESIDENCY A1. Introduction.Probability theory is very important in the study of many areas of life. Itsimportance to the betting industry is perhaps most explicit. Upwards of $15billion is staked every year in the India with bookmakers and on person-to-person betting exchanges, where those wishing to bet, offer and take oddsfrom each other directly.2008 was a very interesting and unusual year in this industry. In the worldof football we saw a series of unexpected results in the FA Cup. It was thefirst time in 21 years that there was a semi final which did not includeArsenal, Chelsea, Liverpool or Manchester United.2008 was also the year of a very hard fought nomination for the democraticpresidential candidate in the U.S.A. While most of the money bet in the UK ison horse racing, dog racing and football, there is also a rapidly growinginterest in betting on politics. The biggest political betting market is theelection to choose the President of the United States. In this case study weuse this race to illustrate how we can infer probabilities from the odds thatbookmakers offer and to deepen our understanding of probability theory.Probability theory is important in other areas of economics, such as in theinsurance and financial markets. Probability theory is highly relevant inmany aspects of life. We consider briefly the example of car insurance andso demonstrate the significance of the concept of conditional probability2. Car Insurance and conditional probabilityThe Association of British Insurers reports that in 2006 the net writtenpremiums of car insurance amounted to close to £10.3 billion. An importantmarket segment is that of young drivers. But, many young people pass theirdriving test prior to going to university and consider the possibility ofrunning their own car only to be put off by the very high insurancepremiums they will be required to pay. An 18 year old that has just passedtheir driving test can pay several times as much as their parents forapparently identical insurance.Is this discrimination against the young? No, it is merely a consequence ofconditional probability theory.Now let us return to the premiums of young drivers. The main reason aninsurance company charges, for example, 5 times the premium for a newyoung driver is that they believe that the new young driver is about 5 timesas likely to make a claim. 1
  2. 2. There are 2 reasons for this. Firstly, the conditional probability of having anaccident if you are 18 years old is significantly higher than the conditionalprobability of having an accident if you are over 40 years old.Secondly a person over 40 may have a significant “no claims discount”whereas an 18 year old will not have had the chance to build up a “no claimsdiscount”. The logic here is that probability of having an accident given youhave not claimed for 5 years is significantly less than the probability ofhaving an accident if you are opening a completely new policy.Differences in the prices of car insurance occur mainly as a consequence ofthe above 2 types of conditional probability.3. Union and Intersection rules of Independent events andPromotions:Great Air Service is a small regional airline.It provides commuter service betweenRaipur and Ranchi – twenty seater planes commuting three round trips daily (total ofsix flights per day). Promotional contest awarding a large prize to be run one day permonth on each flight.The marketing manager of the airlines, Mr Das was looking forward tointroduce a new promotion, which will be innovative yet not too hard onbudget of airlines. He knew that there is little chance to find out thelikelihood of any person sharing his birthday with his friends in a group of 20persons.Probability of birthday of any person in a year= P(A)= 1/365The complement of event A :P(A’)=1- 1/365= 364/365Independent events are events in which the occurrence of the events will not affect theprobability of the occurrence of any of the other events. Example: The date of birth oneperson has no bearing on other’s birth date.With Intersection rules of Independent events of probability, the chance thatany one person in this room sharing his birthday with other person in thegroup is 1 – {(364/365) 20 times} = 5.3%.With this backdrop, the manager run a promotional contest . The day eachmonth for contest to be run will be selected randomly on the first day of eachmonth. On each flight that day, all passengers will write down their birthday(month and day).If any two people on the plane have the same birthday, they willplace their names in a hat and one name will be selected to receive the grandprize. 2
  3. 3. Capacity of each flight is a maximum of 20 passengers (plus crew).The MarketingManager believes there is a very low chance of a birthday match, so only a smallchance of giving away the large prize.The owner wants to know: What is the probability of one or more birthday matches onflights of 20 passengers ?,Looked at another way, what is the probability of 20 passengers not sharing a birthday?20 independent events with each event as a corresponding person not sharing theirbirthday with any of the previously analyzed people.P(A) = the probability of at least two passengers having the same birthdayP(A’) = the probability of there not being any two passengers with the same birthday.P(1) = for one person, there are 365 distinct birthdays = 365/ 365P(2) = for two people, there are 364 different ways that the second could have abirthday without matching the first: = 364/ 365 .P(3) = if person 3 is born on any of the 363 days of the year other than the birthdays ofpeople 1 and 2, person 3 will not share their birthday.= 363/ 365For 20 passengers:P(A’) = 365/365 * 364/365 * 363/365 * 362/365 * …*346/365 = .588528Therefore, the probability of at least two passengers having the same birthday is:P(A) =1 - .588528 or 41.15%.Consider the number of flights on the one day per month the contest is run (3 roundtrips, 6 flights total)Multiply by the probability of a birthday match assuming group size (20 passengerpossibilities)=6 x .4115 = 2.47 prizes/month for 20 passenger flights .The owner immediately directs to withdraw the promotion 4.Probability and betting oddsThere is no example where the importance of probability is more explicitthan that of the betting industry. But, before considering the odds beingoffered during the US Presidential race in 2008 will must first consider threealternative ways of expressing probability: the standard approach, theBetfair approach and the bookmakers’ odds approach. 3
  4. 4. In the standard approach a probability of winning of 0.4 (alternatively 2 or 540%) implies an individual is expected to win 40% of the time and not towin 60% of the time. The sum of probabilities should add up to 1(100%). Inthe bookmaker’s odds approach this will be expressed as odds of 3 to 2. Thisimplies you win Rs 3 for a Rs 2 stake. This implicitly assumes the ratio oflosing to winning is 3:2 implying you win 2 times out of 5.Betfair is the world’s biggest betting exchange. In the Betfair approach thesame probability would be expressed as 2.5. If you make a Rs 1 bet youreceive Rs 2.50. As the Rs 2.50 includes the original stake you win Rs 1.50on a Rs 1 bet which is equivalent to winning Rs 3 on a Rs 2 stake.One interesting thing to note is that while actual probabilities add up to 1,the sum of the probabilities offered by betting companies may add up to atotal which is different to 1. The probabilities they offer are not totallyobjective but depend a little on market situations.5. US party nominations in 2008By examining betting odds for the 2008 US presidential race we illustrate theimportance of probability theory to bookmakers. The US election iscontested by at least two candidates, the candidate nominated by theDemocratic Party and the candidate nominated by the Republican Party. Itwas the media’s attention on the race for the Democratic nomination thatdominated the early coverage of the 2008 US Presidential race. This saw ahard fought contest between Senators Hillary Clinton of New York andBarack Obama of Illinois.We begin by considering what could have been inferred from the odds takenon 8th February, 2008 for the probability of Clinton or of Obama beingnominated for the Democrats. Further, we consider two bookmaker’s oddsand compare the implied probabilities.Betfair was offering odds for Obama of 1.58 and for Clinton of 2.72. A Rs 1bet on Obama would pay Rs 1.58 if Obama was the Democratic nomination.Similarly, a Rs 1 bet on Clinton would pay Rs 2.72 if nominated. Each pay-out includes the original Rs 1 stake so Obama winning would see theindividual profit by Rs 0.58 on a Rs 1 bet, while Clinton winning would seethe individual win Rs 1.72.Obama’s odds imply that the ratio of losing to winning is 0.58:1. Hence, theimplied probability in these odds of Obama winning the nomination was1/(1+0.58) = 1/1.58. This is equivalent to a 63.29% probability of winningthe nomination. Clinton’s odds imply that the ratio of losing to winning is 4
  5. 5. 1.72:1. The implied probability in these odds of Clinton winning thenomination was 1/(1+1.72) = 1/2.72. This is equivalent to a 36.76%probability of winning the nomination.We noted earlier that the probabilities offered by bookmakers may not addup to 1 or 100%. In this case, the percentages sum to 100.05%. If wedeflate the raw probabilities by 1/1.0005 (=100/100.05) they will then sumto 100%. The adjusted implied probability of Obama winning was 63.26%(=63.29/1.0005) while that for Clinton winning was 36.74%(=36.76/1.0005).6. Probability of being PresidentAt the same time as betting on a Democratic or Republican candidate gettingtheir party’s nomination an individual could simply bet on a candidatewinning the Presidential race. Hence, the individual would be betting on twoevents occurring: being the party’s nominee and the President. The oddsoffered would be reflecting a joint probability resulting from an intersectionof events.Betfair was offering odds of 2.42 for Obama and 4.2 for Clinton beingelected President. We can use these odds to calculate the impliedprobabilities that each would win the Presidency. Betfair’s odds include theoriginal stake of Rs 1. So Obama’s implied ratio of losing to winning was1.42:1 while Clinton’s was 3.2:1. Based on these odds, Obama’s probabilityof winning the Presidency was (1/(1+1.42) = 1/2.42. This is equivalent to a41.32% chance of winning the Presidency. The odds for Clinton suggestedshe only had a 1/4.2 (=1/(1+3.2)) or 23.81% chance.The 2 Republican candidates running for the Presidency at the time wereSenator John McCain of Arizona and Governor he odds are expressed as a(losing) to b (winning) the implied probability in Mike Huckabee of Arkansas.Their respective Betfair odds of 3.05 and 70 implied that McCain has a32.79% chance of winning the Presidency and Huckabee a 1.43% chance.The percentages across the 4 candidate sum to 99.35%. We can inflate theraw probabilities to sum to 100 by applying an adjustment factor 1/0.9935(100/99.35). Our focus is on Obama and Clinton. The adjusted impliedprobability of Obama becoming the Presidency is 41.59% (41.32/0.9935)while that for Clinton is 23.97% (23.81/0.9935).For each candidate to run as their party’s nominee they needed to first benominated by their party. We can use the implied joint probabilities forObama and Clinton of being elected as President and of being the 5
  6. 6. Democratic nomination to calculate the conditional probability of each beingelected President IF nominated by their party.We will do this using the Betfair odds. Betfair was offering the adjustedimplied probability of 41.59% of Obama being elected President. At thesame time they were offering the adjusted implied probability of Obamabeing the Democratic nominee of 63.26%. From these two figures we cancalculate that the conditional probability of Obama being elected PresidentIF nominated by the Democrats was P( A B)= P( A / B) P( B)=0.4159/0.6326= 65.74%Where P(A/B) = Probability of Obama getting elected as president IFnominated by the DemocratsP(A B) = Joint Probability of Obama nominated by the Democrats andgetting elected as presidentP(B) = Probability of Obama nominated by the DemocratsSimilarly, we now calculate the conditional probability of Clinton beingelected President IF nominated by the Democrats. To do we use theadjusted joint probability of 0.2397 of her being a Democratic president andthe adjusted probability of 0.3674 of her winning her party’s nomination.Therefore, the conditional probability is 65.24% (0.2397/0.3674).Obama was clearly the favourite to get his party’s nomination. But,bookmakers’ odds infer that the likelihood of Obama or Clinton becomingPresident, if nominated by the Democrats, was practically the same. ------------ 6

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