Stroke and Distance, Part Two

My last post discussed when the potential for a stroke-and-distance penalty (out of bounds or lost ball) is worth the risk. I derived a criterium for attempting the risky shot that depended on the odds of going out of bounds (OB) and the expected gain from succeeding. This calculation assumed that if the risky strategy was worthwhile on the first shot, it was still worthwhile regardless of the number of failed attempts.

I defended this assumption by arguing that if the risky strategy predicted a lower expected score before an OB, it should still be the lower expected score after the OB. After all, the only thing that has changed is that your score is 2 strokes higher; you face the same shot with the same obstacles.

While this argument is logically sound, it does miss one difference between the situations before and after the OB shot: the mishit provides new information about the likelihood of a second failed attempt. We have previously discussed the problems with overreacting to a small sample size, and certainly one tee shot out of bounds is not enough information to support a drastic change of strategy. On the other hand, after several out-of-bounds shots in a row, it seems plausible that the chance of a stroke-and-distance penalty is higher than originally expected.

This post attempts to recalculate the optimal strategy when a stroke-and-distance penalty is possible – this time accounting for the inherent, real-world uncertainties in estimating potential outcomes and adjusting those estimates based on new information. As I discuss below, the best strategy depends on the how confident the estimates are and how much risk is considered acceptable.

The discussion below requires some in-depth probability theory, which has been really fun for me to explore! Those only interested in the (surprisingly simple and elegant) end result, however, should jump ahead to the Applying It section below.

An intuitive picture

Before we consider the rigorous argument, let us begin with some simple examples to highlight how one might intuitively determine when the optimal strategy would change.

Case 1: Lots of data

Suppose a player has played this hole and holes like it hundreds of times, and 90% of the time his driver is in bounds. How much information does a single out-of-bounds shot provide? Not a lot. For instance, if he has attempted 200 similar shots, and 20 have been out of bounds, hitting 21 of 201 out of bounds barely changes the statistics.

What if he hits 2 shots out of bounds in a row? 3? 4? 5? Since this person has a lot of data about his past performance, a streak of bad luck is more likely than a miscalculation. Thus we expect this person to stick to his aggressive strategy for a long time.

Case 2: A new scenario

Suppose a player is faced with a long iron shot from fescue over water. If she attempts the shot and misses, she will have to drop a ball in the fescue, effectively a stroke-and-distance penalty. She does not have much experience hitting from fescue, but on her home course she regularly pulls off shots from thick rough 80% of the time from this distance. Based on this probability of success, she decides the risk is worthwhile. What should she do if she hits 1,2,3 shots into the water?

Even if this player has hit hundreds of similar shots from the rough, there is some uncertainty as to how well this data will transfer to shots from fescue. It seems very possible that her success rate could be 50%, 60% or 70% (it is also possible, though probably less likely, that she performs better out of fescue than tall rough). With more uncertainty, this player’s predicted success rate should be more sensitive to one or more failed attempts. Thus we expect that this player will quickly abandon her risky strategy after just a few mishits.

Case 3: Loss aversion

Suppose a player is playing a hole with out of bounds down one side. He has some data about his abilities, but it is limited. He is very confident that the chance of keeping the ball in play with a driver is between 70% and 90%, but he thinks that any value in that range has an equal chance of being his true probability. He computes that the driver is the best choice if his chances of putting it in play are 78% or better. Should he attempt the risky strategy or layup?

This question does not have a right answer; it depends on how much risk this player wants to take. The player does not know exactly what his probability of success is, and since all the values between 70% and 90% are equally likely, there is a better than 1 in 2 chance (it is actually 3 in 5) that his success rate is higher than 78%.¹ In other words, if the player chooses to take the risky strategy, there is a 3 in 5 chance he made the right choice.

This player should decide how certain he needs to be about the efficacy of the risky strategy in order to attempt it. A risk-averse player may decide not to take the risk unless he is 90% certain it is the right choice. A risk-taker may decide to attempt the risk even when there is only a 30% chance it is the best option. A reasonable middle ground could be to attempt the risky strategy when the odds of it being the right choice are more than 50%.

These three examples highlight intuitive features we expect the proper strategy to have:

1. With lots of good data, the ideal strategy should not change much even with new information.
2. When there is a lot of uncertainty about the probability of success, the ideal strategy should be highly sensitive to new information.
3. A risk-averse player is less willing to mistakenly choose the aggressive strategy.

All three of these features will appear in the detailed statistical analysis of the next section.

A Bayesian approach to the optimal strategy

As discussed in a previous post, Bayesian inference is a statistical technique by which probabilities are updated to account for new information. It is an ideal approach for evaluating how our strategy might change after one or more stroke-and-distance penalties.

In order to simplify the problem, I will make a few assumptions. First, I will assume there is some true probability of a successful shot, $q_0$, that is unknown. I will assume that $q_0$ is fixed for this specific hole.²

For each value of $q$ between 0 and 1, we can assign a probability $p(q)$ that $q=q_0$.³ As noted before, there is no perfect way to determine these probabilities initially. I will make the assumption that the prior probability is a Beta distribution, $B(q;\alpha,\beta)$. I chose this distribution for two reasons. First, this choice is a conjugate prior, meaning that each updated probability distribution will be a Beta distribution as well. Second, if a player hits $s$ successful shots and $f$ failed shots and the initial prior distribution was uniform (all outcomes equally likely), the posterior probability distribution is $B(q;s+1,f+1)$. This seems like an appropriate prior for a player who approaches a hole with some historical data about his success and failures on similar previous holes.

These assumptions make the incorporation of new information very easy. If a player is considering a risky shot and has succeeded $s$ times and failed $f$ times, her initial probability that any particular value of $q$ is equal to the true success rate $q_0$ is $p(q; s,f) = B(q; s+1, f+1)$. After $k$ failed attempts in a row, the new probability will be $p(q; s, f+k) = B(q; s+1, f+k+1)$.

For example, suppose a player attempts a tough hole and has succeeded at 9 similar shots and has failed only once. What do we think his probability of success $q_0$ is? A good guess is probably $q_0 =0.9$, since he has succeed in 9 of his previous 10 shots. But it is also possible that his odds are 8 in 10 ($q_0=0.8$) and he got a little lucky in the past. It is also possible that his odds are 19 in 20 ($q_0=0.95$) and he got a little unlucky to miss one in ten. It seems pretty unlikely $q_0$ is less than 0.5, since it would require a large streak of luck to succeed 9 out of 10 times. Thus intuitively we expect the most likely scenario is $q_0=0.9$, but it seems reasonable that the true probability of success could be a little higher or lower than this value.

The figure below shows the prior distribution I described above, using $s=9$ prior successes and $f=1$ prior fails. This chart describes the chances that various values of $q$ are the true probability of success $q_0$. As expected, the most likely scenario is if $q_0=0.9$. However, the $q$ values around 0.9 are also fairly likely. The size of the spread highlights the uncertainty we have in estimating the true probability of success. With only 10 previous shots to inform the prediction, the chances of random luck affecting our prediction are high.

Now what happens to this probability distribution after the player misses 1, 2, 3, 4 shots out of bounds? We expect the chances of a high $q_0$ to go down, and the chances of a low $q_0$ to go up. With 9 out of 10 successes, it seemed pretty unlikely that the probability of success was really 50%. However, after 4 fails in a row, the player has succeeded in only 9 out of 14 attempts (64%). The chances of $q_0$ equaling 0.5 now seem a lot more plausible than when we only knew about the 9 successes and 1 fail.

The chart below shows how the probability distribution in this example changes after each miss. With each failed attempt as new information, we can update the relative likelihood of each $q$ value being the true probability of success (lower values are more likely, higher values are less likely). There is still a large spread in each distribution – after all, the sample size is still small. Nevertheless, this method accounts for the fact that our confidence in keeping the ball in play decreases with each miss.

As noted above, more data should provide more confidence in knowing the true probability value. There should be more precision in our initial prediction, and the prediction should not change much with new information. The chart below illustrates this fact. In this example, we assume the player has succeeded 90 times and failed 10 times. As with the above example, the most likely probability of success given this information is $q_0 =0.9$, but because there is a lot more data here, the spread of possible values is a lot smaller, so that even the odds of $q_0$ being 0.8 are very small.⁴

When to take the risk

The probability of going out of bounds itself is not the end goal – we want to know when a player should attempt the risky shot. The last post determined a criteria for when the risk is worthwhile: the risk should be attempted if

\[ q_0 > \frac{2}{ISG+2}, \]

where ISG is the Ideal Strokes Gained by the successful risk over the conservative play. In other words, the attempt is worth the risk if the true probability of success is greater than a critical value ,$q_0>q_c$, where

\[ q_c = \frac{2}{ISG+2}. \]

In the last post we assumed that $q_0$ was known, while here $q_0$ could be a range of values. We cannot say for certain that $q_0$ is greater than $q_c$, but instead we can ask what the probability is that $q_0$ is greater than $q_c$, $Pr(q_0>q_c)$. This is simply the sum of all the probabilities $p(q; s,f+k)$ for which $q > q_c$. In calculus, an integral is just a fancy type of sum:

\[ Pr(q_0>q_c) = \int_{q_c}^1 p(q; s,f+k)dq. \]

Note that because the distribution $p(q; s,f+k)$ shifts lower with each miss, $Pr(q_0>q_c)$ will decrease as $k$ increases. The rate at which the $Pr(q_0>q_c)$ decreases depends on the size of $s$ and $f$ – with fewer previous attempts, the likelihood that $q_0$ is above $q_c$ will be very sensitive to new data.

As discussed before, the decision of whether to attempt the aggressive shot depends on the player’s personal willingness to risk choosing the wrong strategy. A reasonable starting point would be to take the risk if there is a 1 in 2 chance that $q_0$ is above the threshold $q_c$ (i.e. $Pr(q_0>q_c) > 0.5$).

With this choice, we can now determine when to abandon the aggressive strategy; we do so when the probability of $q_0>q_c$ shifts below 50%. The number of misses it takes to reach this point depends on the prior distribution; it will be very sensitive when the initial prediction is not very precise.

Applying it: Decision Criteria

The discussion above describes how our estimates of the likelihood of success shift with each failed attempt. Given the Ideal Strokes Gained (ISG) of a particular shot, there is a threshold probability of success above which the risk is worthwhile. A player’s choice of when to attempt the risky shot depends on how much confidence she needs in the probability of success being above this threshold. A reasonable choice could be to attempt the risk when the there is a 50% chance of being above the threshold value.

Despite all the complicated math and probability theory above, the end result turns out to be unbelievably simple. If we use the 50% threshold for attempting the risk, as described above, the number of allowable failed attempts (including previous attempts) can be found by this simple formula:

\[ (\text{Failed attempts on this and similar shots})\leq \frac{(\text{Number of previous successes})\times(\text{ISG})}{2} \]

Since the left side of the inequality should be a whole number, the right hand side should be rounded to the nearest whole number.

In other words, if you multiply the number of previous successes by the strokes you gain by pulling off the shot, you should attempt the shot if the number of previous failures is no larger than half of this quantity (rounding if necessary). If after one failed attempt your total failures is still less than this number, you should continue with the risky strategy.

For example, suppose I play a hole where keeping a driver in bounds gains me 0.3 strokes over my layup. Suppose I have kept the ball in play 25 times on similar holes. I compute $25\times0.3/2 = 3.75$, which rounds up to 4. This means I should attempt the driver as long as I have not hit my drive out of bounds more than 4 times previously. Furthermore, if I have previously hit my drive out of bounds twice in similar scenarios, I am allowed a maximum of 2 attempts off the tee. If they both go out of bounds, I should switch to the conservative play.

We anticipated that a player with more data should stay with the aggressive strategy longer if they initially compute that it is the correct play. This appears in the formula. Suppose instead of 25 successes and 2 failures, I had 250 successes and 20 failures. In both cases my success rate is 92.5%, but the first comes from a lot less data and should be influenced more by new out-of-bounds shots. Indeed, while in the first case I should hit only 2 shots out of bounds before abandoning the strategy, in the second case I can hit up to 17! With more data I am more confident in my probability of keeping the drive in play, so it will take more evidence for me to change strategy.

The simplicity of this formula relies on the fact that we chose to attempt the shot when there is at least a 50% chance it is the correct strategy – the symmetry of 50% simplifies some of the calculations. The formula would be different if we chose to be more risk tolerant or risk averse – the allowable number of failures would still be proportional to the product of successes and ideal strokes gained, but the proportionality factor would be a function of the risk aversion and the number of successful shots. Furthermore, there are certain scenarios when rounding to the nearest whole number does not give the correct result, but these differences are generally too small and rare to be worth accounting.⁵

Despite all the complexities involved in accounting for new information, a few simplifying assumptions yield an amazingly straightforward formula for choosing the ideal strategy. With this in hand, you should not only be confident in knowing when to take a risk, you will also know when new information is enough to justify a change of plans.