More Birdies or Fewer Bogeys?

If you play golf, you probably want to play better than you do now. Much of this blog is devoted to determining how to do just that. Before getting into the Gaussian distributions of your putting dispersion (don’t worry, that’s coming!), let’s consider some more simple and straightforward metrics for designing an improvement strategy.

In my last post I talked about the concept of par as a model. It is a pretty basic and not a particularly descriptive model, but it is still a good place to start. In virtually every round we play or watch, we judge a player’s performance on a hole based on his or her score relative to par. With so much discussion in the golf world centered on the language of pars (fine), birdies (good), and bogeys (bad), I think it is fitting to consider how a score relative to par should be viewed.¹

So starting with par as a model, we may ask, “How does one shoot lower scores?” There are two straightforward answers:

1. Make more birdies (or eagles)
2. Make fewer bogeys (or double bogeys, etc.)

This may all appear trivial so far, but part of good science involves stating all of the facts upfront and making sure we take nothing for granted. It may appear that these two answers give us nothing new to consider (and of course nearly anyone trying to get better is going to try to do both simultaneously), but let’s think more critically about what these two strategies are saying. The first one basically says, “If you hit enough good shots, you can afford to make a few mistakes and still shoot a low score.” The second statement is saying “The fewer bad shots you make, the fewer heroic birdie holes you need to shoot low scores.”

In reality we have here two contradictory strategies for improvement. The first promotes a strategy in which a player aims to make lots of birdies without reducing his bogeys, while the second has a player making lots of pars, with few bogeys but few birdies as well. I will call the first strategy the “erratic” strategy, since this players hole-by-hole scores oscillate wildly, and the second strategy the “consistent” strategy, since this player will mostly make pars.

The question of consistent vs. erratic play appears in a variety of forms throughout golf: Is it better to consistently have shorter tee shots in the fairway or longer tee shots that could end up anywhere? Should you go for a well-defended par 5 in two or play safe? Many of these risk-reward questions will be considered later in this blog, but to get us started let’s ask: Is it better to make many birdies and bogeys or to just make a lot of pars?

We will try to answer this question in a few different ways. As we will see, the answer will depend on what we mean by “better” and will require finding an appropriate mathematical model.

G.H. Hardy — A Historical Detour

Godfrey Harold Hardy was one of the greatest mathematicians of the twentieth century. Hardy is what is called a “pure mathematician” – he studied math to explore the elegance and beauty of the field, not for any sense of practical application. However, he did publish one article on our subject of consistent and erratic golfers. Hardy created a simplified mathematical model, and used it to calculate which type of golfer has the advantage.

Hardy’s Model

To simplify the problem, let’s assume every hole is a par 4. Let’s also assume there are only three types of shots: normal (N), excellent (E), and bad (B). It takes four normal shots to finish a hole. If a player took four normal shots, we might represent this by writing NNNN. Excellent shots count for two normal shots. Thus, a player can finish a hole with two normal shots and one excellent shot, such as NNE. A player could make eagle by hitting two excellent shots in a row (EE), but NEE and NEN would both be a birdie, and NNNE and NNNN would both be a par. Bad shots, in contrast, don’t contribute to finishing a hole. In other words, if the first shot is bad, there still need to be four normal shots to finish the hole (BNNNN). Note this means that a hole can never finish with a B, it must finsh with an E or N.

To summarize, this model says to keep hitting shots until reaching the equivalent of four normal shots. Then count how many shots were taken – this is the score for the hole. Some examples of birdie combinations are NNE, ENN, NEN, ENE, BEE, etc. Par combinations could be NEBN, EBBE, BNEE, etc. Bogey combinations include BBNNE, NNBNN, BNEBN, etc. We can draw analogies between these shot combinations and shots on the course. For instance, NEN would be a player who hit a normal drive, stuffed an approach to 2 feet and made a normal putt, while NNBNN might be a normal drive, approach to the center of the green, but a bad lag that leads to a 3-putt.

There is one more piece to Hardy’s model: Hardy assumes that a player is equally likely to hit an excellent shot or a bad shot — that way every player’s average is a normal shot. He then assigns a probability $p$ to each player, which is the probability that player will hit an excellent shot (also the probability that a player hits a bad shot). Since probabilities must add up to 1, the probability of hitting a normal shot is $1-2p$. Notice the lower a player’s $p$ value is, the more consistent he is. The highest $p$ value possible is $p = 0.5$, which corresponds to half excellent shots, half bad shots, and no normal shots.

Before going any further, recall the four principles I enumerated in the last post for evaluating a mathematical model:

• Models always have assumptions built in to simplify the problem — it is important to be aware of when these assumptions are valid.
• No model is perfect, and there are certain points where the model will break down
• Using models we can gain insight into why the world behaves the way it does
• Models can help us answer more complicated questions

With these in mind, let’s analyze this model in the same way we as we have for previous models:

The Hardy Golf Model

• What it does: It assigns a probability p to each player, which is the probability excellent or bad shots, as opposed to normal shots.

• What it assumes: There are only 3 types of shots, N, E, and B. Each hole is a par 4. There is no distinction between driving, iron, play, putting, etc. Everyone has the same average ability on each shot. The shots are uncorrelated – the likelyhood of N, E, or B does not depend on what the previous shot was.
• Where it breaks down: This model has limited application because of its simplicity. When we start considering how someone’s abilities vary across shot types, or when players of different abilities compete with each other, this model does not apply.
• Insights and extensions: We could make the model more accurate by assigning different probabilities to tee shots, approaches, and putts. However, if our question is simply whether an overall consistent or erratic strategy is better, those additions may not be necessary. Generally it is good to start by solving an easy problem and then evaluate whether adding complexities will be insightful and solvable.

The benefit of the Hardy model is its simplicity, but this is also a danger – on the one hand it is a rare golf model that can be solved easily, but we must also approach its applicability to golf with healthy skepticism. However, we will see that the Hardy model provides great insight into where the advantages and disadvantages of conservative and erratic strategies arise.

Solving the Hardy Model

Hardy and others have solved this model exactly using methods of probability theory. Hardy decided to imagine a player with probability $p$ competed in match play against a player who hits only normal shots ($p = 0$, also known as “Old Man Par”). In the graph below, we see the probability of the more erratic player winning or losing a single hole, as a function of his probability p. We see that the erratic player is more likely to lose the hole than to win the hole unless his $p$ value is greater than about 0.4. However, having $p = 0.4$ is an impractically large value (only 20% of that player’s shots would be normal). Most players hit mostly normal shots, so it seems that the consistent player is almost always more likely to win than to lose. (Notice that as $p$ approaches zero on the x axis, the probabilities of both winning and losing go down. This is because the probability of tying the hole goes up.)

Why is it that the consistent player beats the erratic player, even though their average shot is the same? The difference can be seen by looking at par putts. Let’s say both players hit 3 normal shots in a row (NNN), meaning that they only need one normal shot to finish the hole. The consistent player only hits normal shots, so he does so and gets a par (NNNN). The erratic player, on the other hand, sometimes hits N, sometimes E, and sometimes B. Let’s look at these three cases. If he has an N, he gets a par, just like the consistent player (NNNN). If he hits an E, he also makes a par (NNNE). However, if he hits a B, he is not finished with the hole (NNNB), and will make a bogey or worse. This asymmetry means in this case the consistent player has an advantage, and this advantage leads to a greater chance of winning. Another way to see this is to imagine you have a 2 foot putt. A normal shot will go in, so you don’t get any bonus for hitting an excellent 2 foot putt. However, a bad shot will miss, so from 2 feet you lose more than you gain if you are inconsistent. This asymmetry also shows the danger of looking at only averages. Both of these players have the same average shot, but their results are very different. With all the recent focus on the strokes gained statistic (more on this in a later post), there has been a lot of focus on players’ averages. One must be aware that two players with the same average may not be equally matched.

Stroke Play

In stroke play the consistent player has an advantage for the same reason as in match play. The occassional poor “short putts” lead to an average score for the erratic player which is over par. Specifically, a player with probability $p$ of hitting an excellent or bad shot will have an average score of:²

\[ (\text{score}) = 4+p\left(1-\frac{p^4}{(1-p)^4}\right) \]

This number is always at least 4, and grows larger with large values of $p$. Thus we see that in both match play and stroke play the consistent player is better, and it appears to be because the erratic player is not rewarded for some of his excellent shots (“short putts”) but is always penalized for bad shots.

Team Events

I discovered this problem in college, when the mathematician Roland Minton published an article on the topic (see Resources). As a college player, I wondered how this work applied to college team events. In college tournaments, a team plays five players and counts the lowest four scores; that way if one player has a really poor day, his score doesn’t hurt the team (in high school the top 4 of 6 players count for score). Thus it seems that the format favors having a couple of streaky players so that if they go low, the team has a good score, but if one of them struggles it doesn’t matter. On my team we had recently needed to decide who the fourth and fifth players would be at our conference tournament. We had three players who all had shown potential but each had his weaknesses. I thought the Hardy Model could answer whether we should have taken the consistent players or the erratic ones.

I created a computer simulation (what is called a Monte Carlo Simulation) of the Hardy Model for golf teams. One benefit of computer simulations is that they are often easy to set up and can reveal insight by being able to run situations millions of times to see what the most likely outcome will be. My simulations found that, contrary to what I expected, having 5 consistent players was better than having 4 consistent and 1 erratic, which was better than 3 consistent and 2 erratic, etc. I believe this is because even consistent players have bad days, and having an erratic player means the chances of having two players play poorly goes up. My work along with Professor Minton’s analysis can be found in his book, Golf by the Numbers.

Extending the Hardy model

One of the great benefits of mathematical modeling is that, once we think we understand why we get a particular behavior, we can try modifying the model to see if there is anything else to learn. For instance, in the Hardy model it seems clear that the reason the consistent player has an edge is because when a normal shot will finish the hole, the erratic player isn’t rewarded for an excellent shot and is still punished by a bad shot. If we remove this advantage, does the consistent player still win?

I created what I call the Smarty Hardy Model, in which a player has two $p$ values, a conservative $p$ value and an erratic $p$ value. The player uses a conservative shot when a normal shot is good enough to finish the hole, and uses the erratic shot for everything else. How will this player do against someone who is conservative on every shot?

First, let’s look at match play. I ran another Monte Carlo simulation in which the Smart Hardy player competed against the standard conservative player in 1 million holes of match play (for simplicity I again made the conservative player Old Man Par, $p = 0$, and I also made the Smart Hardy player’s conservative game at $p=0$ as well). The figure below plots the fraction of times the Smart Hardy player won and lost the hole across a variety of $p$ values for his erratic play. Amazingly, we see that with the asymmetry on “short putts” removed, the erratic Smart Hardy player wins more often than he loses – the opposite of the original conclusion!

Why does the Smart Hardy player now have an advantage? The answer lies in his ability to recover from bad shots. To see this more simply, let’s imagine a hole that is only a par 2. Old Man Par scores 2 every time (NN). The Smart Hardy player hits his first shot, which is either excellent (E), normal (N), or bad (B). If it is an E, his hole is over and he wins. If it is an N, he only needs another N to finish the hole, so he switches to his conservative strategy, hits another N, and halves the hole with Old Man Par (NN). If his first shot is a B, he hits the next shot with his same erratic strategy, since he still needs 2 normal shots to finish the hole. If his second shot is an E, he still halves the hole with Old Man Par (BE). An N or another B will mean he loses the hole, as his score would be at best a bogey 3.

To summarize above, on a par 2 the Smart Hardy player:

• Wins the hole every time his first shot is an E.
• Ties the hole every time his first shot is an N.
• Sometimes ties the hole and sometimes loses when his first shot is a B.

Can you see the asymmetry above that leads to the Smart Hardy player’s advantage? His first shot is just as likely to be an E as a B, but with an E he wins every time, while he can still sometimes get a tie. Therefore the Smart Hardy player will win more holes than he loses.³ Of course this analysis is simplified with the hole being a par 2, but the logic remains the same if we increase the par.

What about stroke play? It turns out stroke play is a bit different because there is the possiblity of the erratic player making a big number by hitting many bad shots in a row. Through a relatively straightforward calculation, one can show that amazingly this Smart Hardy strategy leads to an average score of exactly par, regardless of the p value!

There are so many ways to extend this problem. What would happen if a player abandoned the erratic strategy after hitting 2 bad shots in a row? What if only the first shot was erratic? What if the shots are correlated, so that your are more likely to hit two good shots (or two bad shots) in a row? If you are interested in this problem and know a little Python, I posted my code on github here. If you find anything new and interesting, let me know!

Summary

So where does this investigation leave us? The question we asked at the beginning was whether it is better to have lots of good and bad shots or to have a consistent game. As was stated before, the simplifying assumptions in this model mean we have to be careful about how much faith we can have in the results. We cannot say that it is always better to play it safe than to play aggressively, and of course the situation in the match will dictate the proper play. But generally the Hardy model and its extension highlight two assymetries between consistent and erratic play:

1. The consistent player gets an advantage over the erratic player when a normal shot is good enough to finsih a hole, since the erratic player gets no benefit for an excellent shot and occasionally hits a bad one.
2. The erratic player gets a match play advantage over the consistent player because she can recover from a bad shot. In stroke play, this advantage is negated by the possiblity of many bad shots leading to a really high score.

We should ask, how well does this model compare with reality? This is a very hard question to ask, since we want to compare players with the same skills over a long period of time. One method would be to look at two players with similar strokes gained statistics, but one whose individual shots have more highs and lows than the other. Unfortunately this level of detail is not publicly available from the PGA Tour’s Shotlink data, but perhaps one day we will have experimental data to verify these findings. For now, let’s consider where these consequences affect strategy in a real game of golf.

Applying it: When excellent shots don’t pay off

The first insight says that its best to be consistent when a normal shot is good enough. The key is to identify situations where a successful risky shot isn’t much better than a normal shot. To me there is one situation that fits this description particularly well: tee shots on long par 5s.

Let’s say there is a par 5 that you are certain you cannot reach in two, either because it is too long or the green is too protected. Do you still hit driver off the tee? Let’s consider the options: hitting driver is inherently more risky than a wood or long iron, since nearly everyone is more consistent with a shorter club. But the benefits of hitting a driver on a long par 5 are marginal at best, since with a driver you might layup with a 6 iron, while with a 3 wood you layup with a 4 iron. Unless your layup requires a lot of accuracy, it seems the driver introduces a lot of risk without much of any reward. It is on these shots, Hardy says, that the consistent player gains her edge over the erratic one.

Applying it: Take smart risks in match play

The second insight is that after removing the imbalance from needless risk, an erratic player has an edge over a consistent player in match play because he may be able to recover from a bad shot. Here is an example of where the Smart Hardy strategy would apply:

Suppose you are playing a match and you find your ball in a deep greenside bunker, like the pot bunkers of links golf. You can easily get out by going sideways, where you are confident you will get up and down for your bogey. If you try to go at the pin, you might be able to get it close and make par, but it is just as likely that the ball will stay in the bunker and you will have the same difficult shot. Your opponent also is in trouble, and you are pretty sure he will make bogey. What should you do? According to the Hardy model, it is worth trying to hit the tough shot. If you succeed, you will make par and win the hole. If you leave it in the bunker on your first try, you may be able to recover by getting your second shot close and still halve the hole. And since it is match play, even if you spend all afternoon trying to get out of the bunker, you only lose one hole. The Smart Hardy model shows that it is worth the risk.⁴

In summary, this simple model of excellent, normal, and bad shots reveals some fundamental information about strategy: take risks in match play, but don’t ever take risks when just an average shot will do.