Let’s start by repeating the background from that article:
It’s the 2005 Brier Final. Ferbey, appearing in the final game for a 5th straight year, is facing what appears to be a simple decision. It is the 9th end, and the game is tied. A miss by the opposing skip, Shawn Adams of Nova Scotia, has left the Alberta rink lying one with no other rocks in play. It is commonly accepted curling strategy to play for two (or more) when you have the hammer and hold your opponent to one (or steal) without last rock. In this case, Ferbey would be expected to draw to the open side of the rings and likely have a final shot at two or perhaps three if Adams attempts a freeze and makes a poor shot. David Nedohin, who throws the final rocks for the Ferbey team, discusses the options with Randy. Much to the surprise of fans and commentators, they decide to remove their own stone and play out the end for a blank in order to keep the hammer in the 10th end. The final outcome eliminated any need for second guessing the decision, as Ferbey won - David making a draw to the open four foot in the 10th end for the victory.
Specifically we used math and statistics to analyse the decision by Ferbey to attempt a blank end in 9 in order to secure last rock coming home. An assumption made from that article was that Adams will always attempt to freeze to the Alberta stone, if Ferbey chooses to draw and lie two.
Our intention in this article is to:
1. Determine the correct choice for Adam’s shot, if Ferbey attempts a multiple score.
2. Re-examine the Ferbey decision based on a weighting factor of Adam’s decision.
First, we need to determine all the Expected Results (ER) and Variables (V) to be included and state which are being omitted. ERs are numbers, in percentages, taken from existing statistics. Variables are numbers, in percentages, which are estimated during a game situation. Variables are based on the chance of making a single shot or the expected outcome following a succession of shots during an end. The ERs, Vs and related assumptions I used are listed below. I have indicated in each case which letter is used to represent each ER or V.
Expected Results in the 10th end
Odds of winning if tied with hammer (x) = 75.7%
Odds of winning if one down with hammer (y) = 39.5%
Odds of winning if two down with hammer (z) = 11.7%
Odds of winning if 3 down with hammer (m) = 0.4%
These Expected Results are based on available statistics for 4-rock games played in the Brier, European and World Championships, Grand Slam and other WCT events, Provincial Championships, Olympics and Olympic Trials. In using these statistics, we assume skill is relatively equal, conditions are similar and shot making ability is similar.
Situation 1. Adams Decision
Game Situation Assumptions:
- Alberta chooses to draw and lie two. Nedohin’s first stone lands on the tee-line in the eight foot, higher than shot stone, leaving a possible freeze, double or hit and roll.
- If Adams attempts a hit, he will not miss both rocks. Alberta cannot score greater than two.
- When going for two in 9th, it is assumed that at worse Alberta will take 1. Odds of a steal for Nova Scotia are considered negligible.
- If we miss the double, Alberta will always score two. The chance they will miss is considered negligible.
x = Odds of winning tied with hammer = .757
y = Odds of winning if one down with hammer = .395
z = Odds of winning if two down with hammer = .117
m = Odds of winning if three down with hammer = .004
a = Odds of forcing Alberta to 1 with a freeze = .75
b = Odds of Alberta scoring 2 = .2
c = Odds of Alberta scoring 3 = .05
d = Odds of double = .75
These numbers are estimates based on evaluation of where the rocks may have been positioned. We have assumed Ferbey places the second stone on the tee-line, higher than shot stone, making a double possible. Specific location will impact our estimate of the ability to make a specific shot. If we evaluate our chances of making a double the same as the freeze (75%):
W = ay + bz + cm = 32%
W = d(1-x) + (1-d)z = 21%
Based on the analysis, the freeze is the correct call. If we are 100% sure of making a double, we are still only 24% likely to win. If we estimate our chance of the freeze as 50%, W= 24.5% and the decision is much closer if the double is 100%
Situation 2: Weighing the options
In the Ferbey decision, we assumed that Nova Scotia will always attempt a freeze. Now that we have analysed their options, let’s use a weighting factor to determine and re-examine Alberta’s options.
We are Alberta, faced with an option to blank or attempt a multiple score in the 9th end. If we expect to make a perfect shot 50% of the time and leave little option for a hit and roll, then Adam’s will have no choice but to choose a freeze. But, perhaps 50% of the time our shot leaves an easy double. Let’s estimate that when this happens, 10% of the time Adam’s will freeze, 40% attempt a double.
If Alberta also estimates Adams is successful 75% of the time with his freeze, we score a deuce 20% and three points 5%, resulting in a win percentage of 68%. Using results from Situation 1 above:
W = .5(.68) + .1(.68) + .4(.79) = .72%
What is very clear is our desire for Adams to choose to hit. This increases our statistical probability of winning. What also comes into mind is the possibility of enticing Nova Scotia to attempt the double, by intentionally placing the rock in a better position for them to choose this option.
This leads us into another application of statistical analysis, where we choose shots which may appear to be the incorrect call, but in fact are intended to lead the opposition into choosing an incorrect shot call in order to increase our chance of winning. For now I will leave it to the reader to ponder similar examples.
Extra-End: Martin vs. Koe - Provincial Finals
There were very dramatic events in the recent final game to determine Alberta’s 2007 Men’s Provincial Champions. Kevin Martin took the game in heart wrenching fashion when Blake MacDonald, skip for the Koe rink, missed a draw in 10 and another in the extra end to hand him the title.
An interesting moment is Martin’s decision in the 9th end. Kevin was down 2 with hammer, 7-5, and Koe had a single rock in the twelve foot. With their last stone, the Martin team opted to draw for a single point rather than blank and keep the hammer in the 10th. Feedback I’ve read on Curlingzone forums is that Martin felt taking two on the rather straight ice would be more difficult than stealing a single. What is the math in this situation?
Odds of winning if two down with hammer = .117
Odds of winning if one down without hammer = .103
The mathematical difference is very small, less than 1.5%. This makes 2 down with or 1 down without as nearly a coin-flip decision. In this case, Martin weighed in the impact of ice conditions and determined a steal was more likely. The numbers indicate that when this situation occurs, a team should not quickly assume a blank the correct call.