Inaugural Lecture. Jan Vecer

Transcription

1 Inaugural Lecture Frankfurt School of Finance and Management March 22nd, 2012

2 Abstract In this talk we investigate the possibilities of adopting a profitable strategy in two games: betting on a roulette and playing in a lottery. In the roulette betting, no profitable strategy in a long run is possible, but one can create a reasonably well performing strategy that is extremely hard to distinguish from a legitimate investment strategy. A lottery is possible to beat in the statistical sense when the jackpot is sufficiently high. Historically this has been possible on a few occasions during the draws of transnational lottery. We determine the conditions when to enter the draw in order to achieve a positive return on the lottery ticket.

3 : Major pan-european lottery. Played in: Austria, Belgium, France, Ireland, Luxembourg, Portugal, Spain, Switzerland, United Kingdom. A ticket costs e2.00. Largest win: e185,000,000 on July 12, 2011 (single winner).

4 Play The player selects five main numbers which can be any number from 1 to 50. The player selects two different lucky star numbers from a pool of 11 numbers. Number of possible combinations: ( ) ( ) = 50! 5 2 5!45! 11! 2!9! = 116,531,800. Probability of winning the Jackpot: p = 1 116,531,800.

5 Prize Structure The size of the Jackpot is capped: e185m before July 12, e190m since July 12, Thee2 cost of the ticket is split in the following way: e1 goes to the lottery organizers, e0.32 goes to the Jackpot before the cap is reached, e0.594 goes to the lower prizes plus additional e0.32 if the Jackpot is reached, e0.086 goes to the booster fund (future Jackpots).

6 Expected Win A random ticket filed at a random time is to expect to win only e1. One can improve the return by betting at the draws with a high enough jackpot. There are two variables to consider: J - the level of the rolled over (= not won) Jackpot from the last draw. N - the number of tickets entering the current draw.

7 Let us assume for now that every ticket is random and thus it has the same expected win. This is not the case when a lottery syndicate systematically files different combinations (in extreme case files all combinations). tickets have a tendency to file some combinations repeatedly while leaving other combinations out.

8 Figure: 12,000 random points in a square. points tend to cluster in some parts of the space and leave the void in other parts.

9 Figure: 12,000 pseudorandom points in a square. People usually perceive the fact that the points fill in the space in a uniform fashion as a quality associated with randomness.

10 Probability N m illion s Figure: Probability of winning the Jackpot for random and nonrandom tickets. It takes 349,098,074 random tickets to cover 95% combinations and 536,648,771 random tickets to cover 99% combinations.

11 Prize Fund N number of tickets, J rolled over Jackpot from the last draw. Consider Jackpot cap e185m. P(Nobody wins Jackpot) = (1 p) N. Thus the expected Prize Fund is equal to min(j N,185 millions) [1 (1 p) N ] N +max(j N 185 millions,0). Compare with 2N, the total cost of N tickets.

12 Prize Fund When Prize Fund Cost? Case 1: J N 185 millions. Solving for the inequality (J N) [1 (1 p) N ] N 2N finds the minimal level J to be J min = This holds for N 28,900,906. N 1 (1 p) N 0.32 N.

13 Prize Fund Case 2: J N > 185 millions. The expected Prize Fund equals to 185,000,000 [1 (1 p) N ]+(J+0.32 N 185,000,000) N. Solving for the minimal level of the jackpot J min so that the Prize Fund exceeds the revenues 2N leads to J min = N +185,000,000 (1 p) N. This is valid in the region 28,900,907 N 94,896,649.

14 Prize Fund Jackp ot m illion s N m illion s Figure: The minimal level of the Jackpot J min as a function of number of tickets N.

15 Prize Fund Note that once the number of tickets sold exceeds 28,900,907, the minimal jackpot starts to decrease until it reaches a local minimum at N = 44,246,007. The reason is that the Jackpot Fund will exceed e185 million, and out of each additional ticket sold above N = 28,900,907,e0.914 will go directly to the Prize Pool. Thus the remaininge1.086 out of each ticket sold should be compared with the expectation of winning the e185 million Jackpot. The marginal contribution of the Jackpot win of the additional ticket sold to the expected Prize Pool: 185,000,000 (1 p) N p.

16 Prize Fund N m illion s Figure: The marginal contribution to the Prize Pool of each additional ticket sold is more than for N up to 44,246,007.

17 Let us analyze the case when the lottery syndicate chooses all possible S = 1 p = 116,531,800 combinations, and thus the win of the Jackpot is guaranteed. The main question in this case is splitting the Jackpot in the case of multiple winners. Let us assume that there are additional N random bets entering a given draw. The probability that there are exactly k winners of the Jackpot in this pool is approximately Poisson, so P(k Jackpot wins out of N tickets) e N p (N p)k. k! When there are k other winners, the syndicate will receive the fraction of the Jackpot. 1 k+1

18 Case 1: Jackpot Pool J (N +S) is belowe185 million so that the Jackpot cap is not reached in the current draw. This holds for N 28,900,906. E [ST] = E[Winnings of Lower Prizes] + E[Share of the Jackpot] J (N + S) = p P(k Jackpot wins out of N tickets) k + 1 k=0 J (N + S) p e N p (N p)k k + 1 k! k=0 J (N + S) = e N p (N p)k+1 N (k + 1)! k=0 J (N + S) = e N p (N p)k N k! k=1 J (N + S) = e N p (N p)k e N p N k! k=0 J (N + S) ( = e N p). N

19 Solving for J min such that E[ST] = 2 leads to J min N 0.32 (N +116,531,800). 1 e N p The expected fraction of a Jackpot won for the syndicate as a function of N: E[Fraction of the Jackpot] = = 1 P(k Jackpot wins out of N tickets) k + 1 k=0 1 k + 1 e N p (N p) k = 1 e N p. k! N p k=0

20 Exp ected Fraction of J N m illion s Figure: The expected Jackpot share for the lottery syndicate as a function of the number of the random tickets N.

21 Jackpot Share P(k Jackpot wins out of N) J k k+1 30M 60M 90M 120M 0 J J/ J/ J/ J/ E[ k+1 ]

22 Case 2: When J (N +S) exceedse185 million, the Jackpot will reach its cap and the remaining amount of J 185,000, (N +S) will be split among the winners of the next highest prize. Since the lottery syndicate bets all combinations, the next highest prize is guaranteed to be 5+1 (5 correct numbers out of 5 drawn from 50, and 1 correct number out of 2 drawn from 11). The number of such winning combinations is 18 (2 choices for the right number, and 9 choices for the wrong number). The probability of a random ticket winning the combination 5+1 is thus q = ,531, ,473,989.

23 The expected value of the syndicate ticket E [ST] = E[Winnings of Lower Prizes] + E[Share of the Jackpot Pool] M 1 e N p N J 185M (N + S) + S k=0 18 k + 18 e N q k (N q). k! Solving for J min such that E[ST] = 2 leads to J min 185M 0.32 (N + 116,531,800) + ( M ( k=0 1 e N p ) N 18 k+18 e N q (N q)k k! ).

24 Jackp ot m illion s N m illion s Figure: The minimal level of the Jackpot J min as a function of number of tickets N for the lottery syndicate. Once N reaches 96,002,622, the lottery is not profitable for the lottery syndicate as the J min would have to be over its cape185 million.

25 e185m cap e190m cap N for which J min reaches cap 28,900,907 35,415,550 Marginal effect becomes negative 44,246,007 47,305,606 Max profitable N (random ticket) 94,896, ,117,434 Max profitable N (syndicate) 96,002, ,244,221

26 EuroMillions draws in July 2011: Date J N P(No J) E[RT] E[ST] ST: P&L 05/07/ M 62.0 M M 08/07/ M 97.1 M M 12/07/ M 72.8 M M The loss distribution for a lottery syndicate on July 12, 2011: k l M M M M M P(L = l) P(L l) VaR 95% = M, VaR 99% = M.

27 How many times n does the syndicate need to bet to ensure that it makes a profit with probability 1 α? Using the Central Limit Theorem, the average win for the syndicate X(n) has approximately normal distribution N(µ, σ n ), where µ M, and σ M. The probability that the syndicate makes the profit is approximately equal to P( X(n) > 0) = 1 P( X(n) 0) ( X(n) µ = 1 P n σ ( 1 Φ n µ ). σ n µ ) σ

28 We want to find n such that ( Φ n µ ) = α. σ If we denote the α quantile by q α (defined as Φ(q α ) = α), we get n µ σ = q α, or ( ) qα σ 2 n =. µ

29 In particular, when α = 0.05, q , and n 57. Thus the lottery syndicate would have to bet all tickets in at least 57 such favorable draws to ensure that it would end up with profit with 95% or higher probability. When α = 0.01, q , and n 114. Thus it would take at least 114 draws to ensure profit with 99% or higher probability.

30 Profit and loss from betting on Red has distribution: { +1 if Red, with probability 18 X 1 = 37, 1 if Black or Zero, with probability 19 37, betting on 13 has distribution: { +35 if 13, with probability 1 X 2 = 37, 1 if not 13, with probability

31 Note that µ = E[X 1 ] = E[X 2 ] = 1 37, 1368 σ 1 = , σ 2 = How long does it take before 95% scenarios end up being below zero? ( ) q0.95 σ 2 1 n 1 = = 3,701. µ ( ) q0.95 σ 2 2 n 2 = = 126,230. µ

32 How long does it take before 99% scenarios end up being below zero? ( ) q0.99 σ 2 1 n 1 = = 7,403. µ ( ) q0.99 σ 2 2 n 2 = = 252,497. µ In the second case, it would take 1000 years(!!!) (assuming one bet on a roulette is made in one trading day) and there would still be 1% probability that the betting strategy is positive.

33 Figure: A simulated wealth evolution when betting on 13 in the roulette together with the expected trend and the confidence interval. The last positive value is at the bet number 116,279.