11/23/24
August 19, 2005
Tiger's Big Gamble: Tiger took a chance; but how big a gamble was it, exactly?
Last week, Tiger Woods elected to go home rather than wait out the delayed ending of the 2005 PGA Championship. Woods was the leader in the clubhouse and trailed the leader on the course (Phil Mickelson) by just two strokes. When play resumed on Monday, Mickelson had five holes to complete. There were eleven other golfers on the course, five of whom were either tied with or ahead of Woods at the time.
Had Woods finished in a tie for the lead, he would have forfeited the required playoff and somewhere in the neighbourhood of $500,000 (the difference between first- and second-place money).
Tiger, in defending his decision, stated that there was little chance that all of the golfers ahead of him would have 'come back to' him. It's clear from his comments that he looked at the number of players ahead of him, who they were, and the holes that were remaining, and intuitively assessed that his odds of finishing tied for the lead were very very long.
So, how long were they, exactly? Tiger probably didn't bust out his laptop and fire up MATLAB, but then, I'm no Tiger Woods.
Let's assume that the 12 players still on the course played the last five holes just like the average golfer at the 2005 PGA Championship. We then can determine the hole-by-hole probability of a given score from the official course Statistics. Table 1 shows the probability (%) of making an eagle, birdie, etc. on holes 14-18, based on the whole tournament.
| Hole | Hole Score Relative to Par | ||||
|---|---|---|---|---|---|
| -2 | -1 | 0 | +1 | +2 | |
| 14 | 0.0% | 19.4 | 66.2 | 13.4 | 1.1 |
| 15 | 0.0 | 11.9 | 61.1 | 23.4 | 3.6 |
| 16 | 0.0 | 6.8 | 67.7 | 24.0 | 1.5 |
| 17 | 0.2 | 18.5 | 60.6 | 18.7 | 1.9 |
| 18 | 4.3 | 47.5 | 41.1 | 6.0 | 1.3 |
Note that I am making a conservative assumption here. Even neglecting the possibility that the overnight rain made the course easier to play, these were the last twelve golfers in a major tournament, and probably better than average. That means that my final calculation will probably be an overestimate of Tiger's chances.
To make life easy, I'll also ignore the fact that some holes had already been started when play was halted, and I'll ignore triple bogeys and beyond as being extrelely unlikely outcomes.
Now it is straightforward (although tedious) to calculate the probability for any given combined score. It involves multiplying together probabilities from different holes, and then adding together all the probabilities for combinations resulting in the same total score. For example, one way to score -1 for the last five holes is to score (0, 0, 0, 0, -1). The probability of this sequence, from Table 1, is (0.662 * 0.611 * 0.677 * 0.606 * 0.475) = 0.079. There are, of course, lots of other ways to score -1 over five holes, so we just need to find them all, calculate the probability as above, and then add them together.
This is why computers exist.
Table 2 shows the results. Each entry is the probability (%) of achieving a particular score (-4 through +5) based on the number of holes remaining to play. (The probability of doing better then -4 or worse than +5 is essentially zero, so I didn't include it in the table.)
| Score Change | Holes Remaining | ||||
|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | |
| -4 | 0.0% | 0.0% | 0.1% | 0.2% | 0.5% |
| -3 | 0.0 | 0.9 | 1.4 | 2.1 | 3.4 |
| -2 | 4.3 | 11.5 | 10.5 | 10.3 | 12.0 |
| -1 | 47.4 | 37.2 | 30.3 | 25.0 | 23.8 |
| 0 | 41.1 | 35.0 | 33.6 | 30.1 | 27.3 |
| +1 | 6.0 | 12.4 | 17.6 | 20.3 | 19.4 |
| +2 | 1.3 | 2.7 | 5.3 | 8.7 | 9.3 |
| +3 | 0.0 | 0.4 | 1.1 | 2.6 | 3.2 |
| +4 | 0.0 | 0.0 | 0.1 | 0.5 | 0.8 |
| +5 | 0.0 | 0.0 | 0.0 | 0.1 | 0.2 |
The next thing we need to do is to combine this with the player scores on the course, to determine a given player's probability of finishing at or above -2 (Tiger's score). So for Phil Mickelson, who was -4 with five holes remaining, we want to know the probability that he would score +2 or worse over 5 holes. That's just the probability of +2, plus the probability of +3, plus the probability of +4, plus the probability of +5, which we can take from Table 2: (0.093 + 0.032 + 0.008 + 0.002) = 0.135. Table 3 extends this same calculation for each of the other eleven players, and also does the calculation for a final score of -1 or worse.
| Player | Holes to play | Starting Score | Probability of finishing -2 or worse | Probability of finishing -1 or worse |
|---|---|---|---|---|
| Mickelson | 5 | -4 | 13.5% | 4.2% |
| Love | 5 | -2 | 60.2 | 32.9 |
| Bjorn | 4 | -3 | 32.2 | 11.9 |
| Perez | 4 | +0 | 97.7 | 87.3 |
| Elkington | 3 | -3 | 24.1 | 6.6 |
| Appleby | 3 | +1 | 99.9 | 98.5 |
| Singh | 3 | -2 | 57.8 | 24.1 |
| Curtis | 3 | +4 | 100.0 | 100.0 |
| Goosen | 2 | -1 | 87.6 | 50.5 |
| Bohn | 2 | +4 | 100.0 | 100.0 |
| Owen | 1 | +7 | 100.0 | 100.0 |
| Westwood | 1 | +3 | 100.0 | 100.0 |
So the probability that Mickelson would finish tied or behind Woods was only 13.5%. Not good odds for Tiger, but probably worth sticking around New Jersey. However, what Tiger really needed was for everybody to finish at -2 or worse. The chance of that is just the product of the twelve probabilities, which works out to 0.31%. That is, Woods had about 1 chance in 320 that he would finish in either a tie for the lead, or in the outright lead. The probability that he would finish in the outright lead is much less even than this (about 1 in 90,000), so we can neglect that possibility. That means that Tiger, sitting in his hotel room on Sunday night, had about one chance in 320 (probably less, as noted above) at a chance at half a million dollars. Now if it was me, I would take those odds, but half a million dollars is worth quite a lot more to me than it is to him.
posted by Amateur to commentary at 01:45 PM - 0 comments