Bettig o Football Pools by Edward A. Beder I a pool, oe tries to guess the wiers i a set of games. For example, oe may have te matches this weeked ad oe bets o who the wiers will be. We ve put wiers i quotes because the pool may hadicap the matches so that it is expected that each side has a equal chace. For example, if the Jets are playig the Sharks ad the Sharks are a weaker team, the there will be poit spread the Jets must score some specified umber of poits more tha the Jets to be declared the wier i the pool. There a two commo types of pools. Oe is amog a group of frieds ad the perso with the most correct guesses wis all the moey that was bet. If several people have the same umber of correct guesses, the moey is divided evely amog them. We ll call this the office pool sice that s where it s ofte doe. The other type of pool is like the lotteries that are ru i may states: How much you wi depeds o how may correct guesses you have ad it is set up so that the orgaizers expect to pay out less tha they take i. We ll call this the for-profit pool. I both kids of pools, a player ormally does ot kow what wiers the other players have chose. 1 Ca You Make Moey i a Football Pool? Suppose you play the football pool may times. Ca you expect to come out ahead i the log ru or must you certaily lose? 1
Suppose the hadicappig is fair; that is, i each match, both teams have a 50% chace of beig declared the (hadicapped) wier. I that case, everyoe i the pool may as well guess radomly at the outcome of each game ad each perso has a equal chace of wiig. Sice everythig that comes i is paid out i the office pool, you should ted to break eve there i the log ru. Sice ot everythig is paid out i the for profit pool, you should ted to lose there i the log ru. This seems to be the best you ca do. Of course, you ca do better if you somehow kow that some teams are better or worse tha believed. We ll explore some cosequeces of this i the for profit pool. Amazigly, you ca do better tha break eve i the office pool! How is this possible? The idea is to eter more tha oe bet each time, say two bets. Sice each bet oly breaks eve i the log ru, how ca it help to make two bets, each of which seems to break eve i the log ru? Before we go ito this, which is a little complicated mathematically, let s look at a simpler situatio which has the same strage property. Three people are told that they will play a game as a team agaist the Mad Hatter. Here s how it works. The Mad Hatter will place either a red or a blue hat o each perso s head. He ll choose the hats radomly. Each perso ca see the other two hats, but ot his ow. They will ot be allowed to commuicate. Each perso must write o a slip of paper either a guess (red or blue) of his hat color or o guess. If everyoe writes o guess or if someoe guesses the wrog color, the team loses. Coversely, if there is at least oe guess ad all guesses are correct, the team wis. The team is told to work out a strategy ad the they will play the game. Oe perso says: 2
It s clear that a guess ca oly be right half the time sice seeig the other two hats is o iformatio at all. Thus, if oe perso guesses, we have a eve chace of wiig. If two people guess, there is oly oe chace i four that both will be right. If all of us guess, there is oly oe chace i eight that all of us will be right. Therefore, it is obvious that we should simply choose oe of us to guess ad the other two should write o guess. Everythig this perso said is correct, but the strategy is ot the best possible. I fact, there is a strategy gives a 75% chace of wiig! Here s the wiig strategy. All three people do the same thig, amely: If the other two people have the same color hat, write the opposite color; otherwise write o guess. Let s see what happes. There are eight possibilities which are listed i the followig table alog with each perso s guess ad whether it is a wi or ot. HAT guess HAT guess HAT guess wi? RED blue RED blue RED blue o RED oe RED oe BLUE blue yes RED oe BLUE blue RED oe yes RED red BLUE oe BLUE oe yes BLUE blue RED oe RED oe yes BLUE oe RED red BLUE oe yes BLUE oe BLUE oe RED red yes BLUE red BLUE red BLUE red o Why did this work? Each perso guessed half the time ad half of those guesses were wrog, but the right ad wrog guesses were distributed differetly: Whe a perso guessed correctly, the other two people did ot guess. So every correct guess led to a wi. Whe a perso guessed icorrectly, everyoe guessed icorrectly. So the wrog guesses piled up. 3
2 The Office Pool Ca we use this idea somehow i the office pool? If all the etries i the pool are radom, the sometimes oe perso will wi ad other times there are ties ad the pool is shared. To do better tha average, we should try to avoid ties. We might be able to do this by aalyzig all the other bets ad the makig our choice, but we wat a simpler strategy. The key idea is to make more tha oe bet is such a way that they will seldom if ever be tied. Suppose our bets ca ever be tied. Each is still as likely to wi as a radom bet; however, if it wis it will have to share with less tha a radom bet would sice it will ever have to share with our other bets. Here s oe way to achieve this. Make two bets. Choose the first ay way at all. Choose the secod so that all its guesses for the wiers are just the reverse of the first bet. If there are N games ad the first bet gets k of them correct, the the secod will get N k correct, amely those the first bet gets wrog. There will be a tie if ad oly if k = N k. Thus N must be eve ad k = N/2. Sice someoe else is likely to guess more tha half the games correctly, it is ulikely that our two bets will have to share the pool. This may all soud good, but is it correct or is there a flaw i our reasoig? We should be suspicious sice the obvious strategy with the hats tured out to be wrog. The surest way to deal with this is to carefully calculate the result for some umber of games ad people i the pool. Suppose there are three games, we make two bets as suggested above, oly oe other perso is i the pool, ad all games have 50:50 odds. The followig table shows the eight possible wi/lose results for our two bets. For example, lwl meas the bet was correct (wo) o the secod game but was wrog (lost) o the the other two games. The bottom row shows the umber of correct guesses for the best bet. Each of the eight outcomes is equally likely because the games have 50:50 odds. 1st www wwl wlw wll lww lwl llw lll 2d lll llw lwl lww wll wlw wwl www best 3 2 2 2 2 2 2 3 4
Thus we have 1 chace i 4 of gettig all games correct ad the remaiig time we get two games correct. The other player has eight possible outcomes, all of which are equally likely. They look like the top row of the precedig table: 1 i 8 times we expect him to guess all three games correctly, 3 i 8 times we expect him to guess exactly two games correctly ad half the time we expect him to guess less tha two games correctly. We eed to pair his possibilities with ours. That is doe i the followig table. The colum headigs are his correct guesses ad the chaces of that happeig. The row headigs are our correct guesses ad the chaces of that happeig. We have made row ad colum widths proportioal to the chaces. Thus the area of each rectagle is proportioal to its chaces. The umber i each rectagle is the fractio of the pot that we wi. all 2 right less tha 2 1/8 3/8 1/2 all 1/4 1/2 all all 2 3/4 0 1/2 all Sice the area of the six rectagles totals 1, the chaces of beig i ay particular rectagle equals its area. We multiply each area by the fractio of the pot we get ad add up the results to get the fractio of the pot we expect i the log ru: ( 1 1 4 8 1 2 + 3 8 1 + 1 ) 2 1 + 3 ( 1 4 8 0 + 3 8 1 2 + 1 ) 2 1 = 3 4. If a bet costs a dollar, the we put i $2, the pot is $3 ad we get o average $3 (3/4) = $2.25. Thus we expect to make o average 25 cets per pool. Of course, this is a urealistic situatio: oly three games ad oly oe other perso i the pool. Let s see what we ca say i geeral. For simplicity, we assume that the umber of games N is odd. I this case, exactly oe of our bets will have more tha N/2 correct guesses. Why is this? For each of the N games, exactly oe of our bets will 5
be correct. Sice the two bets have N correct guesses, the average umber of correct guesses is N/2. Give a average of umbers, oe or more of the umbers is at least as large as the average ad oe or more is at least as small. Oe of our bets must be right at least N/2 times. Sice N/2 is ot a iteger, the bet must be right more tha N/2 times. Similarly the other is right less tha N/2 times. Look at the set S of all the bets that are right more tha N/2 times. Except for the fact that they re right more tha N/2 times, they are radom. (Why? Because the oly oradom feature of the bets was the fact that oe of our bets was opposite the other but oly oe of our bets is i S.) Sice the bets i S are radom, each bet is equally likely to be a wier. Suppose there are p people i the pool besides us. Thus the pool has p + 2 dollars. Suppose there are s bets i S. Sice each bet is equally likely to be a wier, i the log term each bet i S will retur 1/s of the pool. Thus we expect to average p+2. s To complete this we eed to kow the chaces that s has a particular value ad add up all the possibilities. We do t have the mathematics to do this i geeral see [1] but we ca do particular cases. If there is oe other perso i the pool, the the chaces are 50:50 that he will guess more tha half the games correctly. Thus Half the time s = 1 (ours is the oly bet i S) ad we get $3. Half the time s = 2 ad we average $3/2. Thus our average retur is 3 1 2 + 3 2 1 2 = 2.25, a average gai of 25 cets per pool. Notice what happeed here: Our earlier calculatios were for three games ad oe other bettor. To work out the aswer we eeded some tables. Istead of just plugig ahead with the calculatios, we first removed the oe oradom situatio our paired opposite bets. The the calculatios were easier ad we did more: Still just oe other bettor, but ow ay umber of games (as log as it s odd). 6
We ve doe p = 1 ad could do other values of p the same way. For example, whe p = 2 we have s = 1 o average 1/4 of the time, s = 2 o average 1/2 the time, ad s = 3 o average 1/4 of the time. Istead, we ll simply quote the geeral formula from [1]. If the umber of games is odd, each game has a 50:50 chace, there are p other bettors ad each of them guess radomly, the our average profit per pool is 2 p + 1 ( 1 p + 2 2 p+1 ) dollars. (1) 3 Various Averages ad a Partial Result We said we lacked the tools to derive (1), but ca we at least show that our double bet strategy wis moey for us. It may seem clear that it should from the earlier discussio, but we d better be careful about trustig ituitio after playig with the Mad Hatter. Ca we fid a proof? It all depeds o kowig somethig about averages. As we observed, if S has s people, we expect to wi o average p+2 dollars. To compute our s expected wiigs, we simply eed to compute the average of 1/s over the various outcomes ad multiply by p + 2. We do t have the tools to do this. O the other had, we ca compute the average value of s. Every bet has a 50:50 chace of beig i S. Thus, o average, there are 1 (p + 2) bets 2 i S. I other words, the average value of s is p+2. Of course, that s ot the 2 average we wated, so what good does it do us? We eed to digress for a discussio of averages. Give positive umbers a 1,a 2,...,a. Oe possible average is the arithmetic mea A, defied by A = a 1 + a 2 + + a. Aother average that is sometimes used is the geometric mea G, defied by l G = l(a 1) + l(a 2 ) + + l(a ), 7
Expoetiatig this equatio ad doig a bit of algebra, we ca rewrite the geometric mea i a more commo form: G = (a 1 a 2 a ) 1/. Aother average is the harmoic mea H, defied by (1/H) = (1/a 1) + (1/a 2 ) + + (1/a ). For example, suppose = 2, a 1 = 2 ad a 2 = 8. The A = (2 + 8)/2 = 5 G = 2 8 = 16 = 4 1 H = (1/2 + 1/8)/2 = 2 5/8 = 16 5 = 3.2, three differet umbers. Which is correct? There is o correct aswer. Which average we wat depeds o what we are doig. Notice that H < G < A i this example. The followig fact is proved i Appedix A: Uless all the a i are the same, H < A. (2) Back to our office pool. I that case, we wat the average value of p+2 s, which is p+2, times the average value of 1/s. The average value of 1/s is the reciprocal of the harmoic mea of the s values. Thus, our average retur is p + 2 harmoic mea of s values > p + 2 arithmetic mea of s values, by (2). We already oticed that the arithmetic mea is p+2 2. Thus which proves that we wi. average retur > p + 2 (p + 2)/2 = 2, 8
4 The For-Profit Pool The typical for-profit pool differs i three importat respects from the typical office pool. There are usually may more bettors. Oly a part of the moey take i is paid out, the rest beig kept by those ruig the pool. Istead of the people or persos with the most correct guesses dividig the prize, the for-profit pool usually has prizes for o errors i the bettor s guesses (first prize), a lesser prize (secod prize) for oe error ad perhaps other lesser prizes. With assumptios as i the office pool (50:50 wiig chaces i a game ad bettors guessig radomly), will the office pool strategy work? No. The umber p of bettors i (1) will be large so that the expected gai per bet will be very small, assumig all moey is retured as a prize. Ufortuately, the moey take by those ruig the pool will be much larger tha the small gai i (1). Is there ay way to wi? Not with our preset assumptios; however, if we kow 1 the outcomes of some of the games, the the situatio is differet. I the office-pool case, we would make two bets. They would agree o the games where we kew what the outcome would be. This situatio could be aalyzed, but we ll tur to a differet sceario. Suppose there are N games ad we kow what the outcome of k of them will be. We wat to place bets so that, assumig our kowledge is correct we are certai to wi either a first or secod prize. How ca we do this? Because we wat to guaratee a wi regardless of what happes, we do t eed to make ay assumptios about how the other bettors behave or the odds i the games we re ucertai about. 2 1 We might kow that some games are fixed or have iside iformatio about a player ijury. O the other had, we might just be usig ituitio. Regardless, we ll treat this kowledge as if it is a certaity. 2 We would eed such assumptios if we wated to compute how much moey we could expect to get back from our bet a importat questio sice we could still be losig 9
Sice we kow how we ll bet o the k games we re sure of, we ll just look at how we should bet o the other N k games. Call this umber. A bet cosists of a list of the wiers. We ca reprset this by a list of zeros ad oes: A oe meas the first-amed team i a cotest wis ad a zero meas that team loses. We wat a collectio of -log lists of zeros ad oes (the bets) so that, give ay list A of zeros ad oes (the outcomes), there is a list B i our collectio that either agrees exactly with A or differs from it i oly oe positio. For example, whe = 3, the two bets (0, 0, 0) ad (1, 1, 1) work. Ca we say aythig about how may bets we must make? Yes. There are 1 + possible outcome lists for which a bet will wi: a first prize if the outcome list agrees with it ad a secod prize if the outcom list differs from the bet i just oe of the games. Thus b bets ca cover at most (1 + )b outcome lists. (It may be less because two bets might cover the same outcome list.) Sice there are 2 possible outcome lists ad we wat to cover every outcome list, we must have (1 + )b 2. I other words, We must make at least 2 + 1 bets. (3) For example, whe = 4 we must make at least 2 4 /5 = 16/5 = 3 1 5 bets. Sice the umber of bets must be a iteger, this meas we must make at least four bets. This problem belogs to a area called coverig problems ad is related to error correctig codes, which we look at ext. For more iformatio o the for-profit pool, see [2]. 5 Error Correctig Codes It is coveiet to itroduce the otio of distace betwee two -log strigs of zeros ad oes: d(a 1,...,a ; b 1,...,b ) is the umber of i for which a i b i. moey. I this lecture, we ll just look at how we eed to place our bets to guaratee a prize, ot whether it s expected to be fiacially rewardig. 10
For example, d(0, 0, 0; 1, 1, 1) = 3. The previous football pool problem we were just lookig at ca be phrased as follows: Fid a set B of -log strigs of zeros ad oes such that, give ay -log strig s of zeros ad oes, there is at least oe strig b i B such that d(b;s) < 2. Also, make B as small as possible. (4) Of course, if we just wat to make sure we oe of the first k prizes, we ca replace d(b;s) < 2 with d(b;s) < k. Error correctig codes are used i digital commuicatio. The idea is that we have certai -log strigs of zeros ad oes that are allowed to be set ad, if a few of the zeros ad oes are garbled, we wat to be able to guess what was set. Examples iclude commuicatios betwee (a) cell phoes ad towers, (b) satellites ad groud statios, ad (c) computers. The idea is, if the strig s is received we wat to choose the allowed strig b that is as close as possible. Naturally, we wat as may code words allowed -log strigs as possible sice this meas we ca sed iformatio faster. Also, we do t wat more tha oe allowed strig to be close. This ca be phrased as follows: Fid a set B of -log strigs of zeros ad oes such that, give ay -log strig s of zeros ad oes, there is at most oe strig b i B such that d(b;s) < k. Also, make B as large as possible. Except for the replacemet of 2 with k, there are oly two differeces betwee (4) ad (5), prited i bold. Whe k = 2, we ca rewrite (3) as We ca have at most 2 + 1 (5) code words. (6) Of course, sice this must be a iteger, we ca roud it dow just as we were able to roud (3) up. Whe 2 is a iteger, it might be possible to +1 fid a set B that works for both the football pool ad error correctig code problem. This ca be doe. We ll discuss it ow. Sice 2 is a power of 2, + 1 must also be a power of 2 to divide it evely. So let = 2 l 1. Whe l = 2, we have = 3 ad we foud the solutio with the 23 = 2 words 0,0,0 ad 1,1,1. 3+1 The ext case is l = 3. We have = 2 3 1 = 7 ad 27 = 16. Sice the 7+1 problem is gettig larger, it would be ice to have a more systematic way of 11
statig the aswer. To do this, we eed to discuss how to add together two strigs of zeros ad oes. We do that elemet by elemet. We re ot allowed to have twos, so we defie 1 + 1 to equal 0. For example, 0,1,1,0 plus 1,1,0,0 equals 1,0,1,0. Here are sixtee bets (or code words) for l = 3: The strig 0,0,0,0,0,0,0. The four strigs b 1 = 1, 0, 0, 0, 0, 1, 1 b 2 = 0, 1, 0, 0, 1, 0, 1 b 3 = 0, 0, 1, 0, 1, 1, 0 b 4 = 0, 0, 0, 1, 1, 1, 1. All possible sums of two or more of the strigs b 1 b 4. For example b 1 + b 2 + b 3 = 1, 1, 1, 0, 0, 0, 0. (There are eleve such sums.) There is a method for systematically costructig solutios for ay value of l. To explai the method ad prove that it works would take us too log because we would eed some backgroud i liear algebra ad fiite fields. Most texts for a applied algebra course like Math 103 discuss error correctig codes. There are also books devoted to the subject. For example, the book [3] ca be read after takig a liear algebra course such as Math 20F. A Iequalities for Meas Let s see how we ca prove (2). Before doig so, we eed a theorem about cocave ad covex fuctios: Theorem 1 Suppose a 1,...,a lie i some iterval I ad are ot all equal. Let A be their arithmetic mea. If f is a fuctio for which f (x) > 0 for all x I, the f(a) is less tha the arithmetic mea of f(a 1 ),...,f(a ). If f is a fuctio for which f (x) < 0 for all x I, the f(a) is greater tha the arithmetic mea of f(a 1 ),...,f(a ). Proof: If you wish, you ca just assume the theorem ad skip the proof. Usig the Fudametal Theorem of Calculus ad the itegratig by part with u = f (t) ad v = t b, we have b b b f(b) f(a) = f (t) dt = (t b)f (t) b a (t b)f (t) dt = 0 (a b)f (a)+ (b t)f (t) dt. a 12 a a
With a little rearragig: f(b) = f(a)+(b a)f (a)+e(b), where E(b) = b a (b t)f (t) dt. (7) Suppose f (t) is either always positive or always egative betwee a ad b. If a < b, the b t > 0 for a < t < b ad so E(b) ad f (t) have the same sig. If a > b, the b a = a b ad you should be able to show that, agai, E(b) ad f (t) have the same sig. We ow retur to the statemet of the theorem. I (7), let a = A, let b = a i ad sum over i: f(a i ) = f(a) + (a i A)f (A) + E, wheree = i=1 i=1 E(a i ). (8) i=1 Note that E has the same sig as f ad that (a i A)f (A) = i=1 Usig this i (8) ad dividig by, we obtai ( ) a i A f (A) = (A A)f (A) = 0. i=1 i=1 1 f(a i ) = f(a) + E i, which gives the theorem. This completes the proof. We ca ow retur our attetio to (2). It s atural to work with just H ad A, but it s possible to do somethig more geeral. You may have guessed that there are lots of possible averages take your favorite ivertible fuctio f ad defie the f-mea F by f(f) = f(a 1) + f(a 2 ) + + f(a ). (9) To isure that f is ivertible (so that we ca compute F), we ll require that f ever be zero. To compare F with A, we eed oe techical fact, which we wo t prove: 13
For the harmoic mea H, f(x) = 1/x ad so f (x) = 2/x 3 > 0. By Theorem 1, f(a) is less tha the arithmetic mea of f(a 1 ),...,f(a ). By defiitio, the latter mea is f(h). Thus f(a) < f(h). Sice f is a decreasig fuctio, A > H. You should be able to use the theorem to prove that G < A. Ca we compare the harmoic ad geometric meas? Yes. I fact, let s see how to compare a f-mea ad a g-mea. Let F ad G be the f-mea ad g-mea of a 1,...,a. Let h(x) = f(g 1 (x)) ad look at the h mea of b 1 = g(a 1 ),..., b = g(a ). Call this umber H. Also, let A be the arithmetic mea of b 1,...,b. By the defiitio (9), h(h) = h(b 1) + + h(b ) = h(g(a 1)) + + h(g(a )) = f(g 1 ((g(a 1 )))) + + f(g 1 ((g(a )))) = f(a 1) + + f(a ) = f(f). Also, h(a) = h ( ) g(a1 ) + + g(a ) = h(g(g)) = f(g 1 (g(g))) = f(g). We ve just show that h(a) = f(g) ad h(h) = f(f). Now use Theorem 1 to compare h(h) ad h(a): (a) If h > 0, the h(a) < h(h). I other words, f(g) < f(f). (b) If h < 0, the f(g) > f(f). For the harmoic ad geometric meas, let f(x) = 1/x ad g(x) = lx. The F is the harmoic mea ad G is the geometric mea. Also, g 1 (x) = e x ad h(x) = 1/g 1 (x) = 1/e x = e x. Sice h (x) = e x > 0, (a) tells us that f(g) < f(f); that is, 1/G < 1/F, ad so G > F the geometric mea is greater tha the harmoic mea. 14
Refereces [1] Joseph DeStefao, Peter Doyle ad J. Laurie Sell, The evil twi strategy for a football pool, The America Mathematical Mothly, 100 (1993) 341 343. [2] Heikki Hämäläie, Iiro Hokala, Simo Litsy ad Patric Östergård, Football pools A game for mathematicias, The America Mathematical Mothly, 102 (1995) 579 588. [3] Vera Pless, Itroductio to the Theory of Error Correctig Codes, 2d ed., J. Wiley ad Sos (1990). 15