I could have played it safe, but that wouldn’t have been me.
—Arnold Palmer
Rivalries play a critical role in defining us as sports fans. Do you root for the Tar Heels or the Blue Devils?1 Red Sox or Yankees? Your answers may determine how loud your next bar conversation gets.
In golf, the rooting process is highly personal. We can see our heroes scowl and laugh as they respond to tragedies and triumphs. If you are a Tiger fan, it is probably because you admire his dedication and fierce competitiveness and want to see history made. If you are a Phil fan, you probably enjoy his fearlessness and joyful competitiveness and want to be thrilled.
The sport of golf was blessed with the ultimate rivalry in the 1960s when sports television came of age. In many ways, Arnold Palmer and Jack Nicklaus established the characters that Phil and Tiger played in the 2000s. Palmer was the gambler, winning and losing tournaments with late charges and collapses that turned staid, old golf into high drama. Nicklaus was the calculating perfectionist who rewrote the record book. Since Palmer established himself first, he was “the King” and everybody’s favorite. Nicklaus did not just serve as Palmer’s biggest rival, he was the usurper who dethroned the King. The public saw them as wrestling hero and villain, and large galleries rooted for Palmer with a fervor that was only partially moderated by the remarkable class and sportsmanship displayed by both men. These competitors truly wanted to beat each other in every way, but there was also respect, admiration, and even envy between them which created a special bond.2
This chapter analyzes a mathematical caricature of the Jack/Arnie and Tiger/Phil rivalries. In a variety of settings, we will explore which of two golfers of “equal” ability will have an advantage. One player is more consistent and the other is more erratic. Alternatively, you can think of them as being cautious and risky golfers, respectively.
The mathematics of this study has a surprising history. The great G. H. Hardy, a preeminent mathematician of the early 1900s, first defined and analyzed the “characters” in this study. The irony is that Hardy and his colleague J. E. Littlewood were strong proponents of pure mathematics of the most cerebral kind, yet both made contributions to the mathematics of golf (see chapter 3 for Littlewood’s analysis of lip-outs).
Hardy’s golfer is capable of exactly three types of shots: excellent, normal, and poor. An excellent shot E reduces the player’s score on a hole by one; a poor shot P increases the player’s score on a hole by one; and a normal shot N maintains the golfer on a par pace. For the sake of simplicity, every hole will be a par 4.3 Then, a shot sequence of NNNN finishes the hole in par 4. However, a sequence of NEN represents a score of 3, with the excellent shot reducing the score from a par 4 to a birdie 3. A shot sequence of NPNNN represents a bogey 5, with the poor shot increasing the score by one.
There is some ambiguity in this description, and the interpretation makes a large difference mathematically. The sequence NEE could be thought of as a birdie because there are three shots listed or an eagle because the second excellent shot might reduce the score by a second stroke. We will count strokes, so this is a 3. The second excellent shot is, in a sense, wasted because it produces the same outcome that a normal shot would. The situation is analogous to stroking a birdie putt with perfect speed and line. If the putt is from 3 feet away, the perfect putt is somewhat wasted because a normal, slightly imperfect putt would also have gone into the hole.
Each shot is assumed to be random and independent. That is, the chance of an excellent shot is the same whether the previous shot was excellent, normal, or poor. (Don’t you wish this were true in real life?) The probability of an excellent shot is some number p and the probability of a poor shot is the same number p. Different Hardy golfers may have different p-values, but if a shot is not normal, it has the same chance of being excellent as it does of being poor. In this sense, every Hardy golfer has the same skill level.
The “rivalry” for this chapter is between a Hardy golfer with a small p-value and one with a larger p-value. For clarity, I will use p = 0.1 and p = 0.2. The Hardy golfer with p = 0.1 hits on the average 1 excellent shot, 1 poor shot, and 8 normal shots for every 10 shots. This is C-golfer, our consistent, cautious golfer. The Hardy golfer with p = 0.2 hits on the average 2 excellent shots, 2 poor shots, and only 6 normal shots for every 10 shots. This is R-golfer, our erratic, risky golfer. The question is whether these golfers really are equal. Is there an advantage to being consistent, or is it better to have more frequent flashes of brilliance? In real life, Jack and Tiger won more often than Arnie and Phil. We will see how their mathematical counterparts perform.
A good place to start is to compute the average score for a round. While this does not tell us the odds of either player winning a match, it gives us a useful benchmark. More importantly for a mathematician, the calculation turns out to be a very elegant result.* A detailed description of this calculation is deferred to the end of the chapter.
The result is that on a par 4 the average score for the consistent golfer (p = 0.1) is almost exactly 4.1, and the average score for the erratic golfer (p = 0.2) is almost exactly 4.2.4 The higher the p-value, the more erratic the golfer is and the higher the average score will be. In general, the average score on a hole for a Hardy golfer is approximately p over par, a wonderfully simple result. Multiplying by 18 holes, C-golfer averages 73.8 to R-golfer’s 75.6.
There is a (nearly) two-stroke advantage for the consistent golfer. The numerous calculations to follow will expand on this result, but there is an immediate lesson to be learned here. If great and horrible shots are equally likely, your average score will improve if you play it safe and hit as many normal shots as possible. In other words, go for the lowest p-value you can. It may not be as much fun to keep hitting safe, basic shots, but it does turn out to be more effective.
The average score does not fully answer the question of who will win a match. The answer to that question depends in part on what type of match is played.
In stroke play, both players go 18 holes, counting all strokes, and the lower total wins. As noted above, C-golfer averages almost 2 shots fewer than R-golfer for a round. The real question, however, is what the odds are for each player to win a single match.
A comparison of average scores does not necessarily answer the question. If R-golfer averages 75 by shooting 70, 70, 70, and 90, that might be good enough to win 3 out of 4 matches. To determine the odds of winning the match, it is necessary to compute the probability that each player shoots 70, 71, 72, and so on and compare the probabilities.
The actual probabilities are given in table 4.1. Figure 4.1 shows the probabilities graphically for each golfer.
The erratic R-golfer is more likely than C-golfer to shoot a historic 59 (or any score of 67 or less). This is balanced by the fact that R-golfer is also more likely to blow up to 77 (or any higher score). The bottom line is that C-golfer wins about 56.8% of the rounds, R-golfer wins about 37.1%, and about 6.1% are tied. To illustrate this result, I had my computer simulate some rounds using the Hardy model. The following group of 20 rounds each is representative of the other simulations, and the averages match the theoretical averages closely.
C-golfer: 71, 77, 71, 74, 74, 76, 79, 73, 73, 74, 74, 71, 74, 73, 74, 78, 73, 69, 77, 74
R-golfer: 80, 66, 79, 75, 78, 89, 68, 76, 72, 72, 80, 74, 72, 82, 78, 75, 74, 69, 77, 71
Notice that C-golfer wins 11 rounds (55%) and ties 2 (10%). R-golfer turns in some wildly erratic rounds, ranging from a hot 66 to an embarrassing 89, while winning 7 matches (35%). The consistent player has a definite advantage. (In the next chapter, we will see that the USGA handicap system is not designed to even out this match.) The lesson is that, even if the odds are 50-50 of pulling off a risky shot, the conservative shot is better in stroke play.
Table 4.1 Probabilities for shooting scores from 59 to 86, rounded to 4 decimals, for C-golfer and R-golfer
Figure 4.1 Probabilities of 18-hole scores for C-golfer and R-golfer
In match play, each hole is a separate contest, and the player who wins the most holes wins the match. The standard thinking is that match play is kinder to erratic golfers. A terrible shot leading to a “cane” (7) or “snowman” (8) can put you several strokes behind in stroke play, but only costs you one hole in match play. Meanwhile, the erratic player has a higher propensity for making birdies, which can win holes. To compare C-golfer and R-golfer in match play, you need the probabilities for each golfer to make a 2, 3, 4, and so on for a single hole. The probabilities for one hole can then be manipulated to give the probabilities for an 18-hole match. The probabilities, rounded to two decimal places, are:
The table shows that the erratic R-golfer is more likely to make eagle or birdie but is also more likely to make double bogey or worse. The consistent C-golfer pars more than half of the holes.
Comparing probabilities, it turns out that C-golfer wins 36.3% of the holes, R-golfer wins 35.8% of the holes, and 27.9% of the holes are halved. Again, there is an advantage for the consistent player, although it is much smaller this time. Extrapolating over an 18-hole match, C-golfer will win 45.5% of the matches, R-golfer will win 43.3% of the matches, and 11.2% of the matches will be draws. Match play is, indeed, kinder to the erratic player, but the consistent player still has the edge. The conclusion is that, in the standard forms of head-to-head competition, the more consistent you are, the better. If you insist on going for the impossible shot, it will cost you less in match play than in stroke play.
Even though an erratic player is likely to lose to a more consistent player one-on-one, perhaps two erratic players working together can beat two consistent players in “best ball.” In this type of match, each player plays each hole to completion. The team’s score on the hole is the better of the scores of the two players on the team. So, the lesser-used term “better ball” is a more accurate name.
The most extreme example I have experienced of the oddities of team play came in a best ball championship match. My partner and I both started the round playing horribly. Each of our individual front-nine scores would have been well into the 40s; however, on each hole one or the other of us would make a putt, so our best ball score was even par.5 Our opponents each shot par individually but missed numerous short birdie putts and were only 1-up at the turn. Aside from casting some confused and increasingly irritated glances at us, they kept their cool and were rewarded on the back nine. My partner and I each had better individual scores on the back nine than did our opponents, but on this nine they made all of the clutch putts and closed us out. On both nines, the team scores were very different from the individual scores.
The mathematical question is how the erratic R-golfer fares in best ball competition. In individual play, his relatively high chance of making a hole-winning birdie is more than offset by numerous hole-losing bogeys and doubles. With a partner to save him on his bad holes, perhaps he will finally be able to beat the consistent C-golfer.
One possible match is between two erratic players and two consistent players. To analyze this match, the probabilities of the two R-golfers making a 2, 3, 4, and so on are combined to find the probabilities for the team’s score. Details on how this can be done are at the end of the chapter. The calculation is repeated for the team of two C-golfers. The results are shown below.
Team R is much more likely to make birdie or eagle, but is this enough to offset the 37% birdies and 54% pars of Team C? It turns out that, with these probabilities, Team R wins 38% of the holes, Team C wins 26%, and 36% of the holes are halved. Score one for the risk-takers.
Another possibility is to pair an erratic player with a consistent player. The logic is that the consistent player removes the chance of the team making bogey or worse, while the erratic player makes enough birdies to win the match. How does this play mathematically? The team breakdown is as follows:
In a match against Team C (two C-golfers), Team RC wins 33% of the holes to Team C’s 27%, with 40% halved. In a match against Team R (two R-golfers), Team RC wins 29% of the holes to Team R’s 35%, with 36% halved.
It seems that, in best ball, the more erratic players you have, the better. Said differently, if you can balance your great shots and bad shots, it pays in best ball to be aggressive and take chances. Chris Conklin approached the issue of team play as a college golfer. In a college team match, five golfers play for each team, with the team score being the sum of the best four 18-hole scores. Conklin’s hypothesis was that supplementing three consistent players with two erratic players might produce a better team score than using five consistent players. Using players with p = 0.1 and p = 0.05, his simulations showed that the team with the most consistent players always had an advantage.6
The ultimate game for aggressive golfers is “skins.” In this game, all four players record their scores on a hole and the lowest score wins. If two or more players tie with the lowest score, then the group moves on to the next hole. The phrase “two tie, all tie” means that everybody is eligible to win the next hole, regardless of how they played on the previous hole. Many golfers love to wager, and there are several ways to bet on skins. The most common is to “carry over” all bets. If the bet is $1 per hole and the first hole is tied, then everybody plays the second hole for $2. If the second hole is also tied, then everybody plays the third hole for $3. An erratic golfer can have a string of bad holes and still be in line to win all of the money if the other three are cooperating and tying every hole.
Who has the advantage if the foursome consists of three C-golfers (p = 0.1) and one R-golfer (p = 0.2)? The calculations show that, on a given hole, each of the C-golfers has a 10.8% chance of winning the hole, the R-golfer has a 17.7% chance of winning the hole, and 49.9% of the holes are tied. The erratic golfer has a clear advantage here.
If the foursome consists of two C-golfers and two R-golfers, the percentages change only slightly. On a given hole, each of the C-golfers has a 10% chance of winning the hole, each of the R-golfers has a 16% chance of winning the hole, and 48% of the holes are tied. Again, about half of the skins are carried over.
The number of holes halved does not vary much with the composition of the foursome, at least for foursomes composed of C-golfers and R-golfers. If all four golfers are C-golfers, then 53% of the holes are tied. If all four golfers are R-golfers, the percentage of ties drops, but only to 45%.
Now let’s place several C-golfers and R-golfers into a tournament setting. Recall that, for a single round, C-golfers average almost 2 strokes better than R-golfers. Nevertheless, R-golfers have a greater chance of scoring 67 or lower. Generally, C-golfers will perform better, but our question here is whether there will always be at least one “hot” R-golfer to take over the tournament. This analysis is done for a 140-player tournament with 90 C-golfers and 50 R-golfers.
In a one-round tournament, an R-golfer wins 56% of the time, a C-golfer wins 29% of the time, and R-golfers and C-golfers tie for first 15% of the time. Even though the R-golfers are outnumbered, the odds are high that one of them will go low and win the tournament. Of course, many of the R-golfers will blow up and be at the bottom of the leaderboard.
This represents an interesting statistical “paradox.” For an individual golfer, the lower your p-value is, the better your average will be. However, for a one-round tournament, the absolute best score is likely to come from a golfer with a higher p-value. If the goal is to improve your chances of winning without worrying about your overall placement, try some risky shots. If you would like a nice top-20 finish, conservative plays are best.
The situation changes for a four-round tournament. While a gambling R-golfer might steal the lead for one round, the odds that the gambles continue to pay off decrease as the tournament progresses. In a four-round tournament, a C-golfer will win 68% of the time, while R-golfers win 29% of the time. For the pros, then, a long-term plan is important, and conservative play is often rewarded.
This points out an important difference between an event like the U.S. Open and an event like the Ryder Cup. Long-term strategies that lead to good finishes in the Open do not always translate to victories in the one-round sprints that comprise the Ryder Cup. This is true mathematically, as we have seen with our C-golfers and R-golfers, and psychologically, as the Ryder Cup teams show every two years.
To illustrate the above results, I simulated a four-round tournament consisting of 90 C-golfers and 50 R-golfers. I call it the Hardy Open, since it consists of Hardy golfers but shows some of the oddities often found in Open championships.
The first-round lead is grabbed by two unknown R-golfers who shoot 65, R21 and R43 (the 21st and 43rd R-golfers in the field). Third place at 66 and fourth place at 67 are also taken by R-golfers.
R21 |
65 |
R43 |
65 |
R7 |
66 |
R27 |
67 |
C18 |
68 |
C77 |
68 |
R8 |
68 |
R48 |
68 |
The pressure of the overnight lead gets to R43, who blows up to an 85 and misses the cut. R21 fires a second-round 70 to grab the halfway lead at 135. The cut of the top 70 plus ties falls at 148. Of the 90 C-golfers, 56 make the cut. Only 20 of the 50 R-golfers make the cut. A course-record 61 is turned in by R12. A trio of R-golfers sit at 139, with four C-golfers at 140 and 13 of the next 15 spots taken by C-golfers.
R21 |
135 |
C14 |
137 |
R12 |
137 |
R49 |
137 |
C49 |
138 |
R7 |
139 |
R24 |
139 |
R27 |
139 |
On “moving day” (third round), R21 and C14 move out of contention with an 80 and 81, respectively. R12 is unable to duplicate the magic of his 61, falling to a 78. He will, however, rebound in the final round with a 68 to finish tied for 6th. C73 grabs a share of the lead with a 67, tying the unusually consistent R49’s 69-68-71.
C73 |
208 |
R49 |
208 |
C53 |
209 |
R24 |
209 |
C51 |
210 |
C85 |
210 |
R37 |
210 |
C77 |
212 |
In the final round, C77’s strong 67 wins the tournament. He receives the traditional laurel and a Hardy handshake for the winner. Third-round coleader R49 cannot handle the pressure of being in the final group and shoots 83. Meanwhile, second-round leader R21 recovers from a third-round 80 to nearly steal the tournament with a final round 65. In the broadcast booth, Johnny Miller notes that the 63 he shot at Oakmont would have won the tournament.
C77 |
279 |
R21 |
280 |
C73 |
281 |
R24 |
281 |
R26 |
281 |
C25 |
283 |
R12 |
283 |
C59 |
284 |
R37 |
284 |
Of the 12 players in the top-10-plus ties, half are C-golfers and half are R-golfers. The next 12 players are all C-golfers. Of the 54 players in the top-50-plus ties, 40 are C-golfers. The conclusion is that playing risky golf gives you a shot at the top spot, but cautious play leads to made cuts and good money.
Although I described the tournament as if players were responding well or poorly to pressure, remember that these were all results that were randomly generated on the computer. This is not to say that the golfers we watch in tournaments are not choking, but it is instructive to notice that random golfers show some of the same scoring characteristics. In particular, the phenomenon of the unknown first-round leader was present here. The fast exit of R43 was not due to pressure but to some bad random numbers. As well, R27 followed a first-round 67 with rounds of 72-80-76. Notice that his scores in the last three rounds averaged 76, which is the average score for an R-golfer. This is called regression to the mean, where an unusually good (lucky) round is followed by a more typical, not-so-good round. I suspect that this explains the collapse of many first-round leaders.
Hardy’s model gives us some insight into the question of how advisable risky shots can be. In this section, actual data from PGA tournaments gives us a different way to approach the same question. The mathematical tool that we will use is the expected value, a weighted average that can give us information about whether cautious strategies pay off in the long run.
To illustrate the expected value, let’s start on the green. Suppose that we one-putt 30% of the time and two-putt the other 70% of the time. This cannot literally hold for an 18-hole round, because 30% of 18 is 0.3 * 18 = 5.4, and one-putting 5.4 times is not possible. However, over a large number of rounds it would be possible to have these averages, in which case we would average 5.4 one-putts and 12.6 two-putts per round, for an average (expected value) of 5.4(1) + 12.6(2) = 30.6 putts per round.
We will now apply expected values to the question of whether to unleash the driver on a short par 4. Let’s say that our hole is 380 yards long, and our options are to bomb a driver 320 yards or smooth a 240-yard hybrid into the fairway. In addition, assume that we hit the fairway 90% of the time with the hybrid and 40% of the time with the driver. Figure 4.2 shows average scores on par 4s on the 2008 PGA Tour, when playing from different distances from the fairway compared to the rough.
An interesting fact illustrated in figure 4.2 is that missing the fairway costs the pros about one-fourth of a shot on a par 4. That is, for a given distance from the hole, the average score is about 0.25 strokes higher from the rough than it is from the fairway.
Using the numbers behind figure 4.2, we can evaluate strategies. Taking the safe hybrid from the tee leaves us 140 yards out. We are in the fairway 90% of the time, from where we expect to average 3.883 strokes on the hole. We are in the rough 10% of the time, from where we average 4.180 strokes. The expected value for the safe play is then 0.9(3.883) + 0.1(4.180) = 3.9127. Taking the driver from the tee, we leave ourselves 60 yards from the hole. In the fairway, we average 3.685 strokes, and in the rough we average 3.918 strokes. The expected value if we hit 40% of the fairways is 0.4(3.685) + 0.6(3.918) = 3.8248. Comparing the expected values, we see that the driver gives us a lower average score, saving us about 0.1 strokes on the average.
Figure 4.2 Average scores on 2008 PGA Tour on par 4s with second shot from the given distance and position
Of course, this analysis is only as good as the data we are given in figure 4.2. If the particular hole that we are on has a lake covering most of the last 100 yards to the green, then the averages in figure 4.2 may not apply, and we might prefer the cautious tee shot.
What if we do not hit the fairway with our driver 40% of the time? Is the driver still the better play if we only hit 30% of the fairways? Or 20%? This is a case in which the mathematician’s use of variables can save a substantial amount of work. Instead of repeating the above calculation for many different percentages, we can answer all of the questions at once by naming the percentage of fairways hit: I’ll call it p. The calculations are easier if we have p represent the proportion of fairways hit (so p = 0.2 instead of p = 20%).
The hybrid still has an expected value of 3.9127. To compute the expected value for the driver, we use the same format as above, just replacing the proportions 0.4 and 0.6. Notice that these proportions add up to 1, or 100%. If our fairway proportion is 0.4, the rough proportion is 1 − 0.4 = 0.6. In general, if the fairway proportion is p, the rough proportion is 1 − p. Then the expected value is p(3.685) + (1 − p)(3.918) = 3.918 − 0.223p. For the driver to be the better play, we need 3.918 − 0.223p < 3.9127. Some algebra shows that this is true as long as p > 0.0227. In other words, if we can hit the fairway more than 2% of the time, we should use the driver!
Now you can see why the modern game emphasizes length over accuracy.
The mathematical field of game theory gives us another way to assess risk and reward situations. Game theory is often applied to competitions between two players, each of whom must choose among several strategies. In golf, the competition is usually between golfer and course, or golfer versus nature, as in the case of playing in a gusty wind.
On a par 3, the golfer must decide whether to aim for the pin or play it safe. This time, I will make up some average scores. Playing it safe, let’s assume that the golfer averages 2.9 if the wind does not blow and 3.1 if the wind does kick up. Aiming for the pin, the golfer averages 2.4 if the wind does not blow and 3.4 if it does. These are summarized in a payoff matrix, shown in figure 4.3.
Figure 4.3 Average scores if the wind is gusty (G) or not (N) and if the golfer aims for the pin (P) or plays it safe (S)
One way to analyze this game is to assume that the wind is an active player in the game. This requires a certain amount of paranoia, but I believe most golfers will be able to relate. Regardless of which way the golfer plays it, the wind can force a higher score by choosing strategy G (gusting). Betting on the wind playing this strategy, the golfer would need to play it safe by using strategy S. The golfer accepts an average score of 3.1 and moves on to the next hole.
Assuming that the wind is more of a random phenomenon and less of an evil adversary, we can simply assume that the wind blows with probability p and use expected values as before. If the golfer is going for the pin (strategy P), the expected value is 3.4p+2.4(1−p). This simplifies to 2.4 +p. If the golfer is playing it safe (strategy S), the expected value is 3.1p + 2.9(1 −p), which simplifies to 2.9 + 0.2p. The break-even point is where these two expected values are equal, so that 2.4 +p = 2.9 + 0.2p. Solving this equation, we get p = 
. This tells us that, if p > (that is, we expect the wind to kick up more than 5 out of 8 times), we should play it safe. If p < , we should go for the pin.
In this case, “we should go for the pin” means that the long-run average score for playing this hole under the same conditions numerous times will be better if we aim for the pin each time than if we play it safe each time. The fact is that, as we stand on a tee box, we will play the hole only once. A good long-run average is of little consolation if the wind kicks up this one time and blows our shot into a lake.
A story related in Kevin Cook’s Driven illustrates this dilemma perfectly. Isabelle Lendl, then eleven years old, was given an assignment by her dad, Ivan Lendl (the tennis star). She was to estimate her average score if she tried to hit a long 3-wood over a lake versus laying up in front of the lake. Figuring that she could carry the lake 20 times out of 100, going for it produced an average score of 5.2 compared to an average lay-up score of 4.8. As fate would have it, that weekend Isabelle faced that exact choice. She went for the green and dumped her shot in the lake. Afterwards, Dad said, “Isabelle, what were you thinking? You told me you could hit that shot only twenty times out of a hundred.” Isabelle’s response epitomizes the R-golfer in us all: “I thought it was gonna be one of those twenty times.”7
To close the chapter, I fill in some of the details left out of the Hardy discussion. If the mathematics here gets a little heavy, feel free to “pick up” and move on to the next chapter.
To analyze each of the games between C-golfers and R-golfers, you need to be able to take probabilities for individual golfers and compute probabilities for the best score in a group. This is an example of what is called an order statistic. For example, given four golfers on a particular hole, what will be the minimum score posted by any of the four? In many cases, we are especially interested in which golfer has the best score.
To simplify the calculations, suppose that our golfers are capable of five scores on a hole, with probabilities as shown.
If two of these golfers are paired together for a best ball match, one question is how to compute the probabilities for the team score. Start with two C-golfers. Since neither player is capable of a 2, the team has 0 probability of a 2. For the team to score a 3, either both players make 3s or one makes a 3 and the other a higher score. The probability that they both make a 3 is the product 0.2 × 0.2 = 0.04. Next, consider the possibility that the first C-golfer makes a 3 (probability 0.2), and the second C-golfer makes a higher score. The second player could make a 4 or 5, with probability 0.6 + 0.2 = 0.8. So, the probability that the first C-golfer makes 3 and the second C-golfer makes higher than 3 is 0.2 × 0.8 = 0.16. The other way for the team to make a 3 is for the first player to make higher than 3 (probability 0.6 + 0.2 = 0.8) and the second player to make a 3 (probability 0.2), which again has probability 0.8 × 0.2 = 0.16. Putting this together, the probability of a team 3 is
Pr(3) = 0.2 × 0.2 + 2 × 0.2 × (0.6 + 0.2) = 0.36.
Similarly, the team makes 4 if both players make 4s or if one player makes 4 and the other makes higher. The probability is
Pr(4) = 0.6 × 0.6 + 2 × 0.6 × 0.2 = 0.6.
Finally, the team makes 5 only if both players make 5, which happens with probability
Pr(5) = 0.2 × 0.2 = 0.04.
Notice that the three probabilities add up to 1, or 100%.
The probabilities for a team of two R-golfers are computed as
Pr(2) = 0.1 × 0.1 + 2 × 0.1 × (0.2 + 0.3 + 0.3 + 0.1) = 0.19
Pr(3) = 0.2 × 0.2 + 2 × 0.2 × (0.3 + 0.3 + 0.1) = 0.32
Pr(4) = 0.3 × 0.3 + 2 × 0.3 × (0.3 + 0.1) = 0.33
Pr(5) = 0.3 × 0.3 + 2 × 0.3 × 0.1 = 0.15
Pr(6) = 0.1 × 0.1 = 0.01.
For a match between Team C and Team R, then, the team probabilities are
The next question might be to compute the probability that Team C wins a hole. There are three possibilities for this outcome. Team C could make 3 while Team R makes 4 or higher. The probabilities are 0.36 and 0.33 + 0.15 + 0.01 = 0.49, respectively. The probability of both occurring is the product 0.36 × 0.49 = 0.176 4. Team C could make 4 while Team R makes 5 or higher. The probability is 0.6 × (0.15 + 0.01) = 0.096. Finally, Team C could win with a 5 if Team R makes 6. This happens with probability 0.04 × 0.01 = 0.0004. The total Probability of Team C winning is 0.1764 + 0.096 + 0.0004 = 0.2728, which is slightly larger than 27%.
Team R can win a hole in three ways, as shown.
Adding the probabilities in the right-hand column together, the probability that Team R wins is 0.408, or about 41%.
Finally, the teams can tie in three ways, as shown.
The probabilities add up to 0.3192, for a 32% chance of halving the hole.
Other statistics can be computed in similar ways. The first step is to identify all of the different ways that the event of interest (such as Team C getting a 3 or Team C winning with a 3) can happen. For each possibility, identify the relevant probabilities. Add the Probabilities when any of the options are valid (such as Team R making a 4 or a 5). Multiply the Probabilities when all options have to occur (such as Team C making a 3 and Team R making a 3). Then add up all of the component probabilities. When there are more than four or five possibilities, it is very nice to have a computer to do the arithmetic for you. The principles remain the same even when the details get more complicated.
Having indicated some of the details behind the probability calculations, I now want to return to the calculation of the mean for a Hardy golfer on a par 4. The key is to realize that some excellent shots are wasted. For example, the sequences ENN and ENE both represent birdies. The partial sequence EN creates the possibility of a wasted E, because a second E does not reduce the score at all.
These wasted Es raise the mean above 4 to 4 + u, where u is the probability of a wasted E. To compute u, rewrite it in the form u = pw, where w is the probability of having a partial sequence like EN that creates the possibility of a wasted E. If the next shot is an E (probability p), then the E is wasted. The probability of being in a position to have a wasted E equals
The logic is that the partial sequence can have either one normal shot N (e.g., EN) or three normal shots (e.g., NNN).
The first sum describes all of the possibilities for having one N. For example, a final score of 3 can follow partial sequences of EN or NE, occurring with probability 2p (1 − 2p). A score of 4 can follow partial sequences of EPN, PNE, … (6 possible orders), occurring with probability 2p (1 − 2p)3p. A score of 5 can follow partial sequences of NPEP, PEPN, … (12 possible orders), occurring with probability 2p (1 − 2p)6p2.
The second sum describes all of the possibilities for having three Ns. For example, a score of 4 can follow NNN, occurring with probability (1 − 2p)3. A score of 5 can follow NNNP, NPNN, … (4 possible orders), occurring with probability (1 − 2p)34p.
Some calculus shows that the sums are 
and 
, respectively. The resulting expression for w simplifies to
,
and therefore the mean for a par 4 is given by
.
And that sums up our situation nicely!
