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Brian Prior - Probability and gambling

Brian Prior presentation to the Isle of Wight Maths Society about - Probability and Gambling

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Brian Prior - Probability and gambling

  1. 1. Isle of Wight Maths Society Probability and Gambling
  2. 2. IntroductionWhere did the ideas of probabilityoriginate?What is probability?Elementary probability theoryExamples from the world ofgambling
  3. 3. A little background to the developmentof probability theory"A gamblers dispute in 1654 led to the creation of amathematical theory of probability by two famous Frenchmathematicians, Blaise Pascal and Pierre de Fermat. Because of the inherent appeal of games of chance,probability theory soon became popular, and the subjectdeveloped rapidly during the 18th century. The majorcontributors during this period were Jakob Bernoulli(1654-1705) and Abraham de Moivre (1667-1754).In 1812 Pierre de Laplace (1749-1827) introduced a host ofnew ideas and mathematical techniques in his book, ThéorieAnalytique des Probabilités. Laplace applied probabilisticideas to many scientific and practical problems. The theoryof errors, actuarial mathematics, and statistical mechanicsare examples of some of the important applications ofprobability theory developed in the l9th century.
  4. 4. What is probability and how do we measure it?We envisage a situation in which an outcome or event is in doubt -the face of uncertainty!Probability may be regarded as the measurement of uncertainty of theevent and the theory developed axiomatically as an exercise inmeasure theory. However for practical purposes, statisticians havechosen the follow possible definitions:Equally likely outcomesIf an event E happens when one of a proportion p of equally likely outcomesactually occurs, we say that the probability of E is p and write: P(E) = pThis definition is frequently adopted when calculating probabilities in games ofchance e.g. roulette, card games, dice games, coin tossing etc.Relative frequencyIf , in the long run, E happens 100*p% of the time then we say: P(E) = pIf we cannot reasonably assume equally likely outcomes, then this defintionprovides a practical way of estimating p through sampling.Degree of beliefIf you personally regard that the odds that E occurs as opposed to nothappening, to bep/(1-p) then we say: P(E) = p
  5. 5. Axioms of probability TheoryMutually exclusive events cannot occur together. If all events in asample space are ME then we say they are mutually exclusive andexhaustive (MEE).
  6. 6. NOTE: The probabilty that E does NOT occur is: If P(E) = 1 => E certain to occur If P(E) = 0 => E certain not to occur Outcome spaces (sets)This is a description of the outcomes of interest in the calculation ofprobabilities.For gambling situation this could be simply:S = { win, lose} and P(win) = 0.4 and P(lose) = 0.6The events win and lose would be associated with the outcome of the gamble(or trial).Thus, if we were to toss two (fair) coins and gamble on the occurrence of twoheads occuring, then: S = {[H,H], [H,T], [T,H], [T,T]} assuming equally likely outcomes. P(win) = p([H,H]) = 1/4 and P(lose) = 3/4If we bet on the faces being different i.e. the COMBINATION {H,T} occurs then P({H,T}) =or put simply 2 cases out of four[ Note:In this trial we could toss a single coin twice instead of tossing two coinsand we assume P(coin lands on edge) = 0]
  7. 7. Combinations, Permutations and the Binomial Discrete Probability Distribution In the calculation of probabilities where equally likely outcomes are assumed, many problemsrequire us to calculate the number of ways of selecting r objects (or events) from n objects (orevents), without regard to order. Each selection of r objects is called a COMBINATION. Acombination is just a sub-set. A permutation is a selection in a particular order.e.g. toss a coin four times (or toss four coins at once). How many of the possible outcomes wouldshow 1, 2, 3 or 4 heads?There are 16 possible ways (permutations) the coins can be arranged i.e. S = {HHHH, HHHT, HHTH, .. ,TTTT} each one is a specific outcome ,so our outcome space in this sense would have 16 elements.However, suppose we are only interested in the number of heads that show. The outcome spacewould be {0,1,2,3,4} and if we were gambling on the possibilty of 3 heads our outcome spacecould be simply {3heads, not 3 heads}!But how do we find how many elements of S (all equally likely - why?) would have 0,1,2,3 and 4heads?We could count and then draw up a table thus: Random variables N(H) 0 1 2 3 4 Frequency 1 4 6 4 1Thus for 3 heads the chances are 4/16 = 1/4 i.e. P(3 heads) = 0.25In terms of probabilities we have: E 0 1 2 3 4 P(E) 1/16 4/16 6/16 4/16 1/16 Probability distributionThis is an example of a discrete probability distribution.
  8. 8. Counting the PermutationsThe four coins could be re-arranged amongst themselves in 24 different ways(permuations) viz:1st coin choice = 4 ways, followed by 2nd coin choice = 3 ways, and so on giving 4*3*2*1 =24 ways or 4! (four factorial)In general n objects could be re-arranged in n! different ways. Now consider where r of thecoins show H and (n-r) show T. Clearly the r lots of H could be re-arranged amongstthemselves in r! ways without it looking a different and similarly the (n-r) tails in (n-r)! ways.Hence the number of different arrangements of four coins with r heads and (n-r) tails is: 1.0and denoted asIt can easily be shown that this can be written as : 1.1So, if we had 10 coins the number of ways that 4 heads (or 6 tails) could appear is : 10*9*8*7 = 210 4*3*2*1NOTE: Equation 1.1 is also used to calculate the number of COMBINATIONS of any robjects out of n objects. This will be explained in the talk. Thus, given 5 objects, I can select3 objects inWhat about the probabilities?To answer this we need to look at joint events/outcomes
  9. 9. Joint Probabilities and Independent OutcomesIf I toss a fair coin, then P(H) = P(T) = 1/2This is true if I repeat the trial. This is because the two trials or experiments have independentoutcomes.The joint occurrence of H followed by H is P(H) * P(H) = 1/2 * 1/2 = 1/4This could have been arrived at by looking at all the joint outcomes and assuming equally likelyprobabilities for each 2.0In general for n INDEPENDENT events,This in tossing 5 coins the probability of getting exactly 3 heads = (number of ways 3 heads could occur) * P(H) *P(H) * P(H) * P(T) * P(T) = 10 * (1/2) 5 = 10/32 = 5/16 = 0.3125General resultIf in n independent trials the probability of an outcome or event is is p and the probabilty itdoesnt occur is (1-p) = q, the probability of exactly r successes is given by:This is called the binomial probability distribution function. It should be noted that the value r isthe realisation of a random variable. If we denote by X the random variable, it is customary towrite: 3.0 P(X = r) = P(r) =
  10. 10. Non-independent outcomes - conditional probability and Bayes TheoremGiven two outcomes or events A and B say which are not independendent i.e. P(A and B) = P(A) * P(B)then we define the joint occurence as a conditional probability viz: P(A and B) = P(A given B) * P(B) = P(A|B) also, P(A and B) = P(A|B) * P(B) = P(B|A) * P(A)This leads to a form of Bayes theorem: 4.0 P(A|B) = P(B|A) * P(A) P(B)The result can be generalised: 4.1 (See video example))This is an extremely important result in probability theory and has been used as a basis forreasoning under uncertainty in expert systems, a branch if artificail intelligence. Example: from a standard pack of 52 cards I select two cards. What is the probabilitythey are both aces? Given one of the cards is an ace (say the first), this has a probability of 4/52 = 1/13 ofbeing selected. It is wrong to infer the probability of two aces is therefor 1/13 * 1/13 because the selection ofthe first ace has altered the probability of selecting the second ace on the same basis. P(A1) = 4/52 P(A2 |A1) = 3/51. hence P(A1 and A2) = P(A1) * P(A2 |A1) = 4/52 * 3/51 We could have deduced this result from combinatorial analysis (left as an exercise!)
  11. 11. Examples:The National LotteryProbability of JackpotSelect 5 numbers from 49Total number of possible combinations = = 49 * 48 * 47 * 46 *45 * 44 6*5*4*3*2*1 = 13,983,816 P(jackpot) = 0.0000000715 or 13,983,815:1 againstImagine putting nearly 14 million raffle tickets into a box and selecting just 1 which is thewinner.Most people gamble on this extremely unlikely event because the prize is large. However, theprize is rarely £13983816 on a single draw. Anything less and the bet is technically unfair. Weshall look at EXPECTATIONS shortly.We can calculate this probability another way.We have 1 chance in 49 of selecting the first number = 1/49.1/48 of selecting the second corectly,1/47 the third etc.So to get all six numbers in the order of the draw (a permutation) the probability is: 1/49 * 1/48 * ...*1/44 (conditonal probabilities)But these 6 numbers can be arranged in any order in 6! ways = 720 so the probabilty is: 720 * 1/49 * 1/48 * .... * 1/44 = 0.0000000715 as before.Probability of three numbers correct out of the six chosenAs before,P(first number matches) = 6/49 Hence, P(selection as given)P(second matches) = 5/48 = 6/49 * 5/48 * 4/47 * 43/46 * 42/45 * 41/44 = 0.00088 We need to scale this by the number of ways three correctP(third matches) = 3/47 and three incorrect could occur in the six selectionP(fourth doesnt match) = 43/46 = (6 *5 *4)/(3 * 2* 1) = 20 waysP(fifth doesnt match) = 42/45 So the probability = 20 * 0.00088 = 0.0177P(sixth doesnt match) = 41/44 or 0.923 : 0.0177 = 55:1 roughly The fair odds should be 10:1.
  12. 12. Random Variables and ExpectationGiven a (discrete) outcome space {E1 , E2 , ....., En) and a 1:1 mapping of each Eto a real valued number n.i.e. f(E) -> ni then n is a random variable which has a real value. We have alreadymet this in the case of tossing coins.A probability distribution (pd) may be defined as a function f in which the domainconsists of the possible values that a random variable (rv) can take on, and therange is composed of the probabilities associated with those values.The probability that a random variable X can take on the value x is f(x) and usuallydenoted as P(X = x).Similar definitions can be given for continuous real-valued random variables from areal domain.Thus in tossing two coins then if X = number of heads, P(X = 2) = 1/4Thus, a pd as a set of rv values and their associated probabilities.The expectation of X denoted by E(X) is defined asIt can be interpreted as the arithmetic mean (called the 1st moment about the origin)of the distribution.Example: (from coin tossing distribution x 0 1 2 3 4 P(x) 1/16 4/16 6/16 4/16 1/16 E(X) = 0 + 1/4 + 3/4 + 3/4 + 1/4 = 2Thus the expected number of heads (mean number) is 2 heads. This can be verifiedexperimentally by performing large numbers of tosses of 2 coins and averaging thenumber of heads.
  13. 13. Expectation in gambling (expected loss or gain)We shall use pdf to mean probability distribution function. Suppose we have a pdf as follows: Outcome WIN LOSE Amount(x) £10 -£2 P(x) 0.2 0.8The expectation is : 10 * 0.2 + (-2) * 0.8 = 2 - 1.6 = £0.4I.e. in the long run you will win 40p (nice if you can find it!!)If E(X) = 0 then we say the bet is FAIR, in that in the long run neither side gains norloses.If E(X) < 0 then in the long run you will always LOSE All games of chance with a negative expected gain means that in the long run you will lose money Roulette is a classical exampleStatistically odds(E) is defined as P(E)/(1-P(E)) = P(win)/P(lose) Bookmakers tend to quote odds the other way up as P(lose)/p(win). Thus if abookmakers odds were interpreted as personal probabilities then 2:1 againstimplies P(lose)/P(win) = (1- P(win))/P(win) = 2 so P(win) = 1/3 and P(lose) = 2/3.Such probabilities do not necessarily reflect the true values whatever these mightbe. Games of pure chance such as roulette and the lottery allow the calculation ofprobabilities based on equally likely outcomes. However, the payout is not basedon these odds since the house or betting organiser must take a percentage. Thebet in these cases is not fair and hence E(X) < 0Horse racing and certain card games are different in that they allow an element ofskill or judgment in deciding the bet.
  14. 14. RouletteOn the right is a standard roulette wheel having 18 redslots, 18 black slots and one green zero (American wheelshave two green zeros, 0 and 00).The payput odds are based solely on probability. Whether you bet on European or American tables the payout is simply: where n = number of squares bet on
  15. 15. Expectations and the House EdgeAssume a single zero wheel and bet £1 on 24Assume equally likely outcomes Total number of outcomes = 37 P(ball in any given slot) = 1/37 P(24) = 1/37 = P(win) and P(lose) = 36/37If you lose, then you lose your £1 bet i.e. outcome = -£1 with probability 36/37The house offers 35:1 for the return on a single slot bet i.e on any number notzero. Thus your return in £35 (profit) if you win with probability 1/37Note that the fair payout odds should be 37:1Expectation on the bet = -1 * 36/37 + 35 * 1/37 = 1/37(35 -36) = 1 /37 = £ -0.027For a bet on red or black (also odd or even) the house offers even odds i.e. 1:1P(lose) = 19/37 and P(win) = 18/37E(bet ) = -1 * 19/37 + 1 * 18/37 = -1/37 = -£0.027 as before.The single or double green zero is effectively the house edge because theirpayout odds do not include it. i.e if you kept the house odds and removed the zero(s) from the wheel you would have a fair bet.
  16. 16. PokerThe probabilities of getting the various hands at poker (e.g. four of a kind, pair,two pair etc) can be found using the theoretical results we have already lookedat. However, calculating these probabilites can be quite tricky!Hand Probability DistributionNo pair 1,303,560 in 2,598,960 50 .16%One pair 1,098,240 in 2,598,960 42 .26%Two pair 123,552 in 2,598,960 4 .75%Three of a kind 54,912 in 2,598,960 2 .11%Straight 9,180 in 2,598,960 0 .353%Flush 5,112 in 2,598,960 0 .197%Full house 3,744 in 2,598,960 0 .144%Four of a kind 624 in 2,598,960 0 .0240%Straight flush (excluding Royal flush) 32 in 2,598,960 0 .00123%Royal straight flush 4 in 2,598,960 0 .000154%TOTAL 2,598,960 in 2,598,960 100The total number of five card hands = 52C5 = 2,598,960 different handsExample calculation:Probability of three of a kind (e.g. three kings and any two others [not pairs])For convenience letFor a given rank of four cards (e.g. Kh,Kd,Ks,Kc) we can select any three in 4C3= 4 ways.We have 13 ranks so we can create 13 * 4 = 52 different triples (three-of-a-kinds.)Now consider those cards not in the rank of the triple. We have 12 ranks in thefour suits and and we can select any pair that are not the same rank. We have12 ranks to choose from. Ignoring suits for the present we can select 12C2 pairsamong any rank (so no pairs) and each number in the pair could come from foursuits, so the total number of different pairs is 12C2 x 42 = 1056
  17. 17. Total number of three-of-a-kinds = 52 * 1056 = 54,912P(three-of-a-kind) = 54,912/2,598,960 = 0.02113 BUT - having a winning hand in poker doesnt mean you will win. Poker is not just a game of chance but a game of bluff also.
  18. 18. The Martingale Betting Strategy in RouletteThere are various forms of Martingale betting but the simplest is as follows:Bet either on red/black or odd/even (payout odds are 1:1). If you lose thendouble your bet and play again. Repeat until you win then restart your betsequence again. Foolproof??? After each win you only win back your initial bet. For example suppose you bet £1 initially and double -up on each loss for the next bet. The final net gain is: Bet(£) 1 2 4 8 Total returned = £16 Outcome L L L W Winnings = £16 - £8(stake) - £7(total outlay) = £1 Net Gain -1 -3 -7 1Assume the gambler is only prepared to bet on up to six losing spins in a row(e.g. he/she may be bankrupt after this).In reality, the odds of a streak of 6 losses in a row are much higher than themany people intuitively believe. Psychological studies have shown that sincepeople know that the odds of losing 6 times in a row out of 6 plays are low, theyincorrectly assume that in a longer string of plays the odds are also very low.When people are asked to invent data representing 200 coin tosses, they oftendo not add streaks of more than 5 because they believe that these streaks arevery unlikely. Assuming a US roulette wheel then the odds of losing a single spin at roulette are q = 20/38 = 52.6316%. If you play a total of 6 spins, the odds of losing 6 times are q6 = 2.1256%. However if you play more and more spins, the odds of losing 6 times in a row begin to increase rapidly. In 73 spins, there is a 50.3% chance that you will at some point have lost at least 6 spins in a row. (The chance of still being solvent after the first six spins is 0.978744, and the chance of becoming bankrupt at each subsequent spin is (1-0.526316)x0.021256 = 0.010069, where the first term is the chance that you won the (n-6)th spin - if you had lost the (n-6)th spin, you would have become bankrupt on the (n-1)th spin. Thus over 73 spins the probability of remaining solvent is 0.978744 x (1-0.010069)^67 = 0.49683, and thus the chance of becoming bankrupt is 1-0.49683 = 50.3%.) Similarly, in 150 spins, there is a 77.2% chance that you will lose at least 6 spins in a row at some point. And in 250 spins, there is a 91.1% chance that you will lose at least 6 spins in a row at some point.
  19. 19. Betting on the GGsBookmakers payout odds on horses winning may not be coherent if consideredas probabilities. The probability of a horse winning is unknown but might beguessed in relation to the other horses.So, how much are you prepared to bet on a horse winning? A lot depends onyour knowledge of horse racing and betting, as well as how much you cancomfortably lose I expect but let us look at a possible way you could win.If a bookmaker sets odds on horses winning, which if considered as probabilitiesare not coherent i.e. do not sum to 1, then it is possible for the bookmaker toALWAYS win if the gambler bets on every horse optimally according to the odds.If he/she doesnt bet this way they will lose more.The total implied probability is 1.05 !! (not coherent with probability theory).The following example is taken from WikipaediaHorse number Offered odds Implied probability Bet Price Bookie Pays if Horse Wins 1 Even 0.5 £100 £100 stake + £100 2 3 to 1 against 0.25 £50 £50 stake + £150 3 4 to 1 against 0.2 £40 £40 stake + £160 4 9 to 1 against 0.1 £20 £20 stake + £180 Total: 1.05 Total: £210 Always: £200The bookmaker is guaranteed to make a £10 profit for this type of bet. It is calleda Dutch Book or lock.Now consider if a horse 4 is scratched and the bookmaker does not adjust theodds. The better can guarantee to win £10 by betting on each horse as before,since the bookmaker only makes £190 and must pay out £200 on each bet.With competitive fixed-odds gambling being offered electronically, gamblers cansometimes create a Dutch book by selecting the best odds from differentbookmakers, in effect by undertaking an arbitrage operation.
  20. 20. Two interesting Problems in Probability1. The Monty Hall Problem (see video)2. The Birthday Problem
  21. 21. Solutions:1. Change your selection2. The birthday probabilities