Hardy, Littlewood and Golf

In the course of reading several papers about the mathematics of golf, I was surprised to …nd J.E. Littlewood cited as one of the …rst to solve a particular problem. This surprise turned to astonishment two days later when I found a reference to a completely di¤erent problem in a 1945 paper called "A Math- ematical Theorem about Golf" by G.H. Hardy. My reaction was due to the reputations of Hardy and Littlewood as pure mathematicians par excellence. G.H. Hardy (1877-1947) and J.E. Littlewood (1885-1977) are high on any serious list of the most important mathematicians of all time. Individually, they made deep and numerous contributions to analysis and number theory, publish- ing hundreds of papers each. Together, they formed what many consider to be the greatest mathematical collaboration ever. The Hardy-Littlewood conjectures in number theory contain important insights into the distribution of prime numbers. They revitalized the study of analysis in England, and the rigor that they brought to their work elevated all of British mathematics. A saying, at- tributed to Harald Bohr as a friend’s"joke" but expressing a common belief in the mathematics community, was that in the early 1900’sthere were only three great English mathematicians: Hardy, Littlewood and Hardy-Littlewood. Hardy and Littlewood were also well known for their collaboration with the fantastic Srinivasa Ramanujan, whose mysterious conjectures are still being investigated. After discovering the Hardy paper, I was concerned that I would next read that a page of one of Ramanujan’s notebooks contains a formula for the perfect golf swing. At that point, I feared, Rod Serling would materialize and explain how it was that I had become embedded ... in the twilight zone. Hardy was quite outspoken about the beauty and purity of mathematics. In his book A Mathematician’sApology, he wrote, "The mathematician’spatterns, like those of the painter’s or the poet’s, must be beautiful, the ideas, like the colours or the words, must …t together in a harmonious way. There is no per- manent place in the world for ugly mathematics." He also wrote, "I have never done anything ’useful’. No discovery of mine has made, or is likely to make, directly or indirectly, for good or ill, the least di¤erence to the amenity of the world." This statement is often quoted and, undoubtedly, often misunderstood. I believe that his primary intent was to a¢ rm the purity of his motives, that he worked on problems for the beauty of the mathematics and not for personal pro…t or an illusion of saving the world. Given this background, Hardy and Littlewood are extremely unlikely can- didates for contributors to the mathematics of golf. However, I did not know at …rst that both men followed cricket avidly. Hardy scoured the newspapers for cricket statistics and often characterized other mathematicians’abilities by comparing them to famous cricketers. In modern golf terms, he might call Isaac Newton "in the Tiger Woods class" and Brook Taylor "in the Rocco Mediate class." In addition, Hardy regularly played real tennis, the game from which our modern tennis is descended. In this light, perhaps an interest in golf is not shocking. For completeness, a brief description of Littlewood’s golf problem follows.

1 The balance of this article concerns Hardy’sgolf problem. Littlewood’sGolf Problem

One chapter of Littlewood’s book A Mathematician’s Miscellany discusses problems from the Mathematical Tripos examination used to rank mathematics graduates of Cambridge University. Littlewood was himself a Senior Wrangler in 1905, this being the term for the candidate with the highest score on the Tripos. Hardy had been fourth wrangler in 1898. One of the problems involves a spinning sphere bumped from a perch on top of a horizontal cylinder. The student is to prove that the sphere follows the path of a helix as long as it remains in contact with the cylinder. In the …gure on the left, a ball sits on top of a cylinder. If it is pushed, it will start to roll down the cylinder. The …gure on the right shows a helix (spiral) superimposed onto the cylinder. Obviously, the ball would not stick to the outside of the cylinder and follow the helix all the way around. The challenge is to show that the sphere will follow the helix until it loses contact with the cylinder and falls o¤.

Ball on a Cylinder Cylinder with Helix

Littlewood then asked what would happen if the cylinder is turned upright and the ball rolls along the inside of the cylinder. This could represent a ball rolling around the edge of a golf cup. What is its path? The most reasonable guess is that the ball will spiral down to the bottom of the hole. You might argue about whether the path is a perfect helix or whether gravity stretches out

2 the helix. However, almost nobody guesses the actual path.

Ball in Hole Helical Path?

The mathematical assumptions are that gravity is the only force and that the ball stays in contact with the cylinder. This is an obvious situation for using cylindrical coordinates. In the two dimensions seen from above the hole, the motion is circular and is easily described in polar coordinates. The motion is periodic. The third dimension can be left as a spatial variable representing, for example, the height z of the ball above the bottom of the cup.

Cylindrical coordinates

The result is counterintuitive: the height z is also described by a periodic function. In a helix or stretched helix, the height z would be a strictly decreasing function. Because z is periodic, the ball does not roll to the bottom of the hole. It spirals down some, and then spirals back up! (And then back down and then back up and so on.) A full solution is given in "Golfer’sdilemma" by Gualtieri,

3 Tokieda, Advis-Gaete, Carry, Re¤et and Guthmann in the June 2006 American Journal of Physics. They show that the ratio of the period of the vertical motion to the period of the circular motion is 7=2. Mathematically, then, a ball entering the hole could roll down into the hole p and then roll up and back out! A possible path is shown below.

A Mathematical Lip-Out

This does not seem reasonable to most of us. The reality is that the assump- tion that the ball stays in contact with the cylinder is not often met on a golf course, especially since the metal cup does not extend all the way to the top of the hole. If the ball enters the hole near the edge rolling along the dirt portion of the hole, a collision with the metal cup will bounce it free. As well, a ball entering the hole is more likely to bounce o¤ the lip or drop down into the hole than it is to roll along the edge. Unless the ball is rolling, this mathematical analysis is not valid. However, any experienced golfer can tell you stories about putts that did go into the hole and came back out. The greatest golfer of our time, Tiger Woods, had a putt in the 2007 PGA Championship to shoot a 62, which would have been the lowest score ever shot in a major championship. The putt went most of the way in, did a quick 180 turn and came back out. The mathematics and physics of a rolling ball show that this can happen. In his Miscellany, Littlewood comments that, "Golfers are not so unlucky as they think." Somehow, I doubt that this makes Tiger Woods feels any better. Hardy’sGolf Problem The problem proposed by G.H. Hardy is motivated by a hypothetical match between two golfers of "equal" ability. If one golfer is much more consistent than the other, which golfer has the advantage? Another way to state the problem is

4 in terms of strategy, known to golfers as "course management." Golfers are often faced with a choice of attempting a risky shot or playing safely. For instance, they may choose to try to hit over a lake directly at the hole, or play safely around the lake but farther from the hole. Is it better to use a cautious strategy or a risky strategy? Hardy approached this question by imagining a golfer who can hit three types of shots: excellent (E), normal (N) or poor (P). The golfer plays a hole with a par (target score) of four. A player hitting four normal shots (NNNN) …nishes the hole with a score of 4. A poor shot adds one to the required number of shots. Therefore, a player who hits three normal shots and then a poor shot (NNNP) has not …nished the hole. Another normal shot (making the shot sequence NNNPN) …nishes the hole with a score of 5. However, another poor shot extends the hole to at least six shots.,The sequence NNNPPN …nishes the hole with a score of 6. By contrast, an excellent shot reduces the required number of shots by one. Thus, a shot sequence of NNE …nishes the hole with a score of 3. Poor shots and excellent shots are essentially inverses of each other. The sequence NPEN …nishes the hole with a score of 4. More examples follow.

Shot Sequence Score NPNNN 5 PNNE 4 ENPPN 5 NEPPPN 6 ENN 3

Notice that a sequence can never end in P. A poor shot always adds one to the sequence. However, a sequence can end in E. A careful understanding of how an ending E a¤ects the score will be the key to our analysis. As shown above, the sequence ENN …nishes the hole with a score of 3. However, what score is represented by the sequence ENE? The answer is 3: the score equals the length of the sequence. In a sense, the golfer is "cheated" out of a possible bene…t of an excellent shot: ENN and ENE receive the same score. The gol…ng analogy is that the …rst two shots (EN) leave the ball close to the hole. A decent (normal) putt from this distance will go in. An excellent putt that goes into the exact center of the hole is enjoyable to watch, but the golfer gets no extra credit for perfection; this putt still counts as the third stroke. The most important assumptions involve the distribution of shots. All shots are independent. One poor shot does not a¤ect the probability that the next shot is excellent. (All golfers wish that this was realistic.) The probability of 1 a poor shot is some number p with 0 p < 2 . The probability of an excellent shot is the same number p. This leaves a probability of 1 2p for a normal shot. The only di¤erence between one such golfer (I will use the phrase "Hardy golfer" for a golfer playing with these constraints) and another is the value for p. At …rst glance, all Hardy golfers appear to have "equal" ability, since excellent shots and poor shots have equal probability and a sequence starting with PE is equivalent to a sequence starting with NN.

5 Suppose that golfer C is a Hardy golfer with p-value p1 and golfer R is a Hardy golfer with p-value p2 > p1. Golfer C has a higher probability of hitting a normal shot than golfer R, so golfer C is the more consistent golfer (equivalently, the more cautious golfer). Golfer R has a higher probability of hitting either an excellent shot or a poor shot than golfer C, so golfer R is the more erratic (or risky) golfer. The problem is to determine which golfer, C or R, is more likely to win a match? Hardy’sAnalysis

Hardy’s "A Mathematical Theorem about Golf" presents the case where p1 = 0, so that golfer C always makes a par 4. Golfer R has a probability x of hitting an E shot (which Hardy calls a supershot) and probability x of hitting a P shot (which Hardy calls a subshot). Hardy computes the probability of golfer R winning a hole as w(x) = 3x 9x2 + 10x3. This follows from golfer R making a 2 (EE) with probability x2 and making a 3 (ENN or NEN or NNE or ENE or NEE or EPE or PEE) with probability 3x(1 2x)2 + 2x2(1 2x) + 2x3. He then computes the probability that golfer R loses the hole as l(x) = 4x 18x2 + 40x3 35x4. This is most easily obtained by computing 1 p where p is the probability that golfer R makes 2, 3, or 4. (Since golfer R could lose with a score of any positive integer greater than 4, the direct computation would be an in…nite series.) A score of 4 can result from four N’swith probability (1 2x)4; the sequence NNNE with probability (1 2x)3x; two P’s and two E’s (in 3 possible orders) with probability 3x4; a starting sequence of PEN (in 6 possible orders) followed by either an E or an N with probability 6x2(1 2x)(1 x); or one P and two N’s(in 3 possible orders) followed by an E. Summing these gives the probability of a score of 4 as 1 7x + 27x2 50x3 + 35x4, from which we get p = 1 4x + 18x2 40x3 + 35x4 and Hardy’sresult follows. In his pre-computer era, Hardy compares the probabilities by graphing (ac- tually, having a colleague plot numerous points) the di¤erence of the two prob-

6 abilities. We can simply graph the two functions w(x) and l(x).

Probabilities of winning, losing

For all values of x below approximately 0:37, player R is more likely to lose than win. The maximum di¤erence between the two curves occurs at approximately 0:09. This is a much more realistic value of x than 0:37. If x = 0:09, then 9% of player R’sshots are Excellent, 9% are Poor and the remaining 82% are Normal. At x = 0:37, only 26% of the player’sshots are normal. For realistic values of x, then, the consistent player is more likely to win the hole than the erratic player. Hardy says that this is at odds with the standard gol…ng wisdom that an erratic player is better o¤ at match play (counting each hole as a separate contest) than at medal play (where all strokes are counted for all 18 holes). As we will see, the standard wisdom is correct, because the consistent player has an even larger advantage in medal play. A Mean Calculation To start to analyze the results of a medal match between two Hardy golfers, we can calculate the mean score on a single hole. A reasonable guess is that the mean should be four, since E and P shots are equally likely. This would be true if it were not the case that some E shots are "wasted." Recall that both sequences ENN and ENE produce scores of 3; the second E in the sequence ENE does not improve the golfer’sscore. For this reason, the mean score for a Hardy golfer with p > 0 will be larger than 4. One strategy for computing the mean is to take our above calculations for the probability of making a 2, 3, 4 and so on, generalize to the probability P (k) 1 of making a score of k and then compute the mean kP (k). The patterns k=2 in P (k), however, eluded me. As is often the case inX mathematics, a slightly

7 di¤erent way of looking at the problem reveals some very nice patterns that are easy to manipulate. The key is to follow up on the explanation of why the mean is not 4: some E shots are wasted. If w equals the probability that a Hardy golfer’sshot sequence has reached a point where an E shot could be wasted, then the probability that one is wasted is wp. Each wasted E shot raises the golfer’s score by 1, so the average score is raised by wp. That is, the mean is 4 + wp. The next step is to characterize the set W of sequences that make it possible to have a wasted E. Take any complete shot sequence that ends in a normal shot N, and the partial sequence with the ending N removed has potential for a wasted E. For example, the sequence EN can be completed with either N or E; the E would be wasted. The sequence NNN can be completed with either N or E; so can the sequence EPN. The sequence NNPN can be completed with either N or E; so can the sequence PEPN. The challenge is to identify the pattern here and accurately compute probabilities. There are two sequences of length 2 in W: EN and NE. The probability of getting such a sequence is 2p(1 2p). There are two types of sequences of length 3 in W. The maximum number of N’sis 3, illustrated by the sequence NNN. This has probability (1 2p)3. Two of the N’scan be swapped out for a PE combination. All six of the permutations of EPN are in W. The probability of getting one of them is 6p2(1 2p): There are two types of sequences of length 4 in W. The maximum number of N’s is again 3, illustrated by the sequence NNPN. Given the four possible locations for the P, the probability is 4p(1 2p)3. Two of the N’scan swapped out for a PE combination, as in PEPN. There are 4 3 locations for the E and the N, so the probability of getting this type of sequence is 4 3p3(1 2p). Generalizing, there are two types of sequences of length k in W for k 3. The maximum number of N’s is 3, with the remaining k 3 shots being P’s. k k(k 1)(k 2) There are = distinct orders for the N’s and P’s. The 3 6 k(k 1)(k 2) k 3 3 probability of getting a sequence of this type is p (1 2p) . 6 Two of the N’scould be swapped out for a PE combination. The sequence has one N and one E, with k(k 1) distinct orders possible. The probability is k 1 k(k 1)p (1 2p). Combining all possibilities together, the probability of getting a sequence in W is

1 k(k 1)(k 2) k 3 3 1 k 1 w = 2p(1 2p) + p (1 2p) + k(k 1)p (1 2p) 6 k=3 k=3 X X 1 k(k 1) k 2 3 1 k(k 1)(k 2) k 3 = 2p(1 2p) p + (1 2p) p 2 6 k=2 k=3 X X All that remains is to evaluate the two series. They can both be evaluated using basic power series rules from …rst year calculus! Recall the geometric

8 series result 1 1 pk = for 1 < p < 1 1 p k=0 X 1 For us, 0 p < 2 , so our p-values are within the interval of convergence of the series. We can then take derivatives and get

1 k 1 1 kp = for 1 < p < 1 (1 p)2 k=1 X 1 k 2 2 k(k 1)p = for 1 < p < 1 (1 p)3 k=2 X 1 k 3 6 k(k 1)(k 2)p = for 1 < p < 1 (1 p)4 k=3 X As a side note, we have started a proof that for 1 < p < 1, the series 1 k 1 pk m converges to for any positive integer m. Sub- m (1 p)m+1 k=m stitutingX these values for the series into the previous expression for w, we get 1 1 w = 2p(1 2p) + (1 2p)3 (1 p)3 (1 p)4 p 4 = 1 1 p where the last expression follows from …nding a common denominator and ex- panding terms. The mean score for a Hardy golfer is thus

p 4 mean = 4 + p 1 1 p " # which for reasonable values of p is very close to 4 + p. For example, if p = 0:1 the mean is about 4:09998 and if p = 0:2 the mean is about 4:1992. Assuming that all holes are played independently, the mean score for a hole can be extrapolated to the mean score for a whole round. If p = 0 and all 18 holes are par 4, the mean score is 72. If p = 0:1, the mean increases to about 73:8. If p = 0:2, the mean increases to about 75:6. The more erratic the golfer is, the higher the mean score is. Di¤erent Strokes

An important point to make is that having a lower mean score does not necessarily imply that you will win a majority of your matches. The type of match being played has a strong in‡uence on who wins.

9 In stroke play (also called medal play), each player counts all strokes over 18 holes and the lower total wins. The probabilities of players making di¤erent scores on a given hole can be combined to compute the probabilities of a player having a particular score for the entire round. Comparing these probabilities, a Hardy golfer with p = 0 will beat a Hardy golfer with p = 0:1 in stroke play 63% of the time. The two will tie 10% of the time, and the more erratic (p = 0:1) golfer wins 27% of the time. The percentages are similar but not quite as one-sided in a match between a Hardy golfer with p = 0:1 and a Hardy golfer with p = 0:2. The more consistent (p = 0:1) golfer wins 57% of the time, 6% are tied, and the more erratic golfer wins 37%. This is illustrated below, where the results of 20 simulated rounds are shown.

p = 0:1: 72,72,69,78,73,76,75,73,76,74,71,75,72,72,75,78,73,75,70,74 p = 0:2: 79,71,67,81,83,74,78,77,80,71,75,90,72,71,79,73,82,74,74,76

Notice that the more erratic golfer (p = 0:2) is quite erratic, recording the best score (67) and the worst score (90). The more consistent golfer has the lower score 12 times (60%), the higher score 7 times (35%) and there is one tie. The Match Game

In match play, each hole is a separate contest, and the player who wins the most holes wins the match. This is the competition that Hardy analyzed. Continuing with a comparison between a consistent golfer with p = 0:1 and an erratic golfer with p = 0:2, the consistent golfer wins 36:3% of the holes, the erratic golfer wins 35:8% of the holes, and 27:9% of the holes are halved. Over an 18-hole match, the consistent golfer wins 46% of the matches, the erratic golfer wins 43% and 11% are tied. As Hardy commented, the edge goes to the consistent player. However, the edge is relatively small. The erratic golfer is de…nitely better o¤ playing match play than playing stroke play. Thus, the common golf folklore is correct. The A-Team

The erratic players have their revenge in competitions involving more players. In best ball (also called foursomes) matches, two teams of two players each compete. The score for a given team on a hole is the minimum of the two scores posted by the players on that team. This competition encourages risk-taking, because even if a gamble does not pay o¤ for one player, the other player on the team could still have the best score on a hole. A team of two erratic Hardy golfers, each with p = 0:2, has the advantage over two consistent Hardy golfers, each with p = 0:1. The pair of erratic players wins 38% of the holes, loses 26% of the holes and ties 36% of the holes. This team of two erratic players also has the advantage over a team consisting of one erratic p = 0:2 Hardy golfer and one consistent p = 0:1 Hardy golfer. The analysis ignores any psychological bene…t the erratic golfer may receive from

10 having a consistent partner always hitting good shots. The "mixed" team loses 35% of the holes, wins 29% and ties 36%. It seems that, mathematically, in the best ball format the more erratic players a team has, the better. Tournament

The …nal setting in which I will compare Hardy golfers is a large tournament. An interesting statistical "paradox" is present, that players with the highest average scores are most likely to win the tournament. To see why this happens, look back at the simulated scores listed previously. This time, imagine that they represent twenty di¤erent consistent golfers (p = 0:1) and twenty di¤erent erratic golfers (p = 0:2). The consistent golfers have a mean of 73:75, close to the theoretical mean of 73:8. The erratic golfers have a mean of 76:4, slightly higher than the theoretical mean of 75:6. And yet, one of the erratic golfers has the best score of 67. Further, four players tie for second with 71’sand three of them are erratic golfers. Of the top …ve, four are golfers from the group with the larger expected value! This is not terribly surprising if you also notice that the seven highest scores are all erratic golfers. The much larger variance of the erratic golfers gives them a higher potential for both low scores and high scores. In the context of a tournament with many such players, the odds are high that one or more of the erratic golfers will have a good round, and post a lower score than the consistent players are likely to achieve. As an example of this, consider a tournament with 90 consistent (p = 0:1) Hardy golfers and 50 erratic (p = 0:2) Hardy golfers. In a one-round tournament, one of the erratic Hardy golfers will have the best score 56% of the time. One of the consistent golfers will have the best score a mere 29% of the time. You might expect that the consistent players will shine in a four- round tournament. Clearly, the longer the tournament is, the less likely it is that an erratic golfer will continue to have a run of good luck. This is correct, as a consistent golfer will win 68% of four-round tournaments to 29% for the erratic golfers. (Of course, the tournament was seeded with almost twice as many consistent golfers.) Real Golf

In A Mathematician’sApology, G.H. Hardy puts forth his argument that the mathematicians’work is more real than the physicists’work. Physicists must work with mathematical models that provide, at best, only slightly simpli…ed descriptions of a perceived reality. Given the inaccessibility of stars and quarks, the version of the real world with which the physicist works is incomplete and often changed by new experiments and discoveries. A theory that "works" today may be overturned tomorrow. By contrast, the world in which the mathematician works is precisely de…ned. Theorems proved are indisputable and eternal. The mathematical truth that is real today will also be real tomorrow. The gol…ng model presented here is not real. The shots hit by real golfers fall into many more than three categories. A "poor" shot does not always cost

11 exactly one stroke. In fact, a shot that goes out of bounds is essentially a two- stroke penalty. Excellent shots and poor shots are not independent occurrences that have equal probabilities. Nevertheless, Hardy’s model provides a simple basis for a comparison of di¤erent strategies in di¤erent competitions. Cautious plays are rewarded in head-to-head competitions, especially stroke play, and in longer tournaments. The calculation of mean score does not turn out to be useful in making these comparisons, but it is a surprisingly elegant result. I believe that Hardy would be pleased with that trade-o¤. Further Reading

Both Hardy’s A Mathematician’s Apology and Littlewood’s Miscellany are readily available. They both open a small window into the mind of a great mathematician. Hardy and LIttlewood come across as exceedingly clear and level-headed thinkers, with strong philosophical bents and biting wits. Hardy’soriginal paper appeared in the December 1945 issue of Mathematical Gazette. "On a mathematical theorem on golf by G.H. Hardy" by HHH appears in the March 2006 Mathematical Gazette. In this article, the analysis is taken in a di¤erent direction than what is presented here.