Blackbook of Curling 2007/08 - Reprints

1.To go for two… or not? Masters of Curling final: Howard vs Ferbey
Ontario’s Glenn Howard squad had a spectacular Seasons of Champions run in 2006-07, going undefeated throughout the Ontario playdowns, and losing only a single round-robin game at the Tim Horton’s Brier and the Ford World Men’s Championship. Earlier in the season, they defeated Edmonton’s Randy Ferbey in the finals of The Masters, one of the four Grand Slam events. An intriguing situation occurred during that final, showing that even the best teams in the world are not always certain of the “correct” strategic decision… even when faced with a common scenario.

 

It is the sixth end of the new Grand Slam format – eight-end games – and there are only two ends remaining. Howard has the hammer, the teams are tied at 3-3 and Randy Ferbey, throwing third stones, has just placed his first stone behind a centre guard. It is hanging on the back side of the button. These are the only two rocks in play.

The first indication from Glenn is for third Richard Hart to throw his lefty out-turn (in-turn for a right hander) and draw to the face of the Ferbey stone, to possibly sit shot-stone. While sitting in the hack, Richard sees another option – a soft hit on the Ferbey stone situated out of the rings with his in-turn, and roll to the open.

After much debate, they agree to hit the Ferbey stone. The decision centers on whether the preference is to aggressively play for a deuce – at a risk of a steal or possible single point scored – or to play conservatively, to increase the possibility of a blank end, in order to keep last-rock advantage in the seventh end. The outcome, at this stage, is unimportant, the question is: what is the correct call?

The dilemma Team Howard faced is very common. With three ends remaining, tied with hammer, what position would I prefer to have going into the second-last end, and what risk should I take in order to get to that position? It is quite fascinating to note that at the highest levels of the sport, much of this decision is made on instinct or “situational analysis”… because, in fact, we can examine the decision using mathematics.

Let’s attempt to analyse the situation based on available statistics and mathematical in-game analysis. To examine the decision, 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 (ER) with 2 ends remaining:
Odds of winning if tied with hammer (x) = 68.3%
Odds of winning if one down with hammer (y) = 35.2%
Odds of winning if two down with hammer (z) = 14.8%
Odds of winning if three down with hammer (m) = 4.5%

These Expected Results are based on available statistics for four-rock FGZ games played in the Brier, European and World Championships, Grand Slams and other WCT events, Provincial Championships, Olympic Games and Olympic Trials. In using these statistics, we assume skill is relatively equal, conditions are similar and shotmaking ability is similar. These assumptions are clearly debatable, but we disregard any difference in skill level amongst specific teams or events.

Assumptions:
· If Howard attempts to draw a blank will not occur
· Howard will not score 3
· The most Ferbey will steal is a single point
· Richard Hart will not have a complete miss. He will either:
   o Draw and place his stone in the rings (counting first or second)
   o Hit and remove Ferbey stone or remove the guard
· Richard attempts a soft hit and will not roll his shooter out if he removes the Ferbey stone.

Option 1: Draw
With his opponent on the back of the button, a draw around the guard is clearly the aggressive play. Assuming Richard will not miss, two outcomes exist – Howard will sit one or Ferbey will sit one. For each outcome, we can assess the possible outcomes of the ends and assign probabilities to each:

Variables:
Howard sits shot stone:
Score two (a) = 30%
Score one (b) = 60%
Steal (c) = 10%

Howard sits second shot:
Score one = (d) 60%
Steal (e) = 40%
We assume Glenn Howard will have a similar chance at scoring one, regardless of the shot’s outcome. We also assume a deuce is not likely if Richard does not achieve shot-stone position. Both of these assumptions are, as always, debatable, but for our analysis we consider such possibilities negligible.

This appeared to be a difficult shot. Let’s assume Hart will achieve shot-stone position 50% of the time. He might argue this point, but we’ll save that for later analysis.

Win (W) = (.5)(a(1-z)+b(1-y)+cy) + (.5)(d(1-y)+ey))

W = 60.5%

Option 2: Hit and roll
Variables:
Howard sits shot:
Blank (a) = 75%
Score two (b) = 5%
Score one (c) = 15%
Steal (d) = 5%

Howard removes guard, Ferbey sits shot:
Blank (e) = 40%
Score one (f) = 40%
Steal (g) = 20%

We’ll assume Richard will remove the shot stone ¾ of the time (75%).

Win (W) = (.75)(ax + b(1-z) + cy + c(1-y)) + (.25)(ex + fy + g(1-y))

W = 61.6%

It would appear the correct call is the hit – but it is very close. If Howard believes Hart can achieve shot-stone position greater than 50% of the time, the draw would appear to be correct. If in Option 1, for example, Howard “gets shot” 60% W = 62% and at 75% W = 64.2%.

We are using many variables in this analysis (5 and 7 respectively), which can lead to greater variance in our final numbers. For example, with Option 1, if we set d=40% and e=60%, W drops from 60.5% to 57.5%. And because we are examining a situation in which several (in fact seven) rocks have yet to be thrown, this analysis becomes very difficult.

What is perhaps more important than the analysis of this specific situation is the common situation of a close game with two ends remaining. An interesting trend was uncovered when we started digging into available game statistics. Virtually every scenario (tied with hammer, two down without, three down without, one up with, etc.) shows a steady increase or decrease in probability of a Win as you near the final end. For example, tied with hammer reads as follows:

Your chance of winning with hammer increases from being 61% with five ends remaining to 75% in the last or extra end.

However, the scenario of one up without hammer is shown in the next table:

The results show a 3-4% difference both up and down, depending on end completed. The most interesting point we see is the increased chance of winning when one up with hammer with two ends remaining, versus three or one end remaining. 65% versus 62%. Now, commonly a team tied with three ends to play may risk a steal in an attempt to score two. The next table shows the outcome based on two down with hammer:

 

If the team with hammer in the eight end (or sixth of an eight-end game) scores the deuce, their opponent still has a 15% chance of winning and, surprisingly, statistics show a 4% chance when three down with two ends to play.

Yet another interesting indication is that two down with hammer is difficult. A team will win only 24% in this situation after playing the second end in an eight-end game. Useful for a future discussion on the danger of eight end games, but let’s digress…

Statistics would indicate a team with hammer should ensure a blank or score to be in control with two ends remaining, rather than risk a possible steal. Let’s choose variables which represent likely outcomes, based on our approach to strategy in the third-to-last end. For example, if we play aggressively, our chance of a blank is estimated to be 0% and a steal is 30%.

Aggressive starting the 8th (or 6th) end
Blank (b) = 0%
Take 1 (t) = 40%
Take 2 (u) = 25%
Take 3 (v) = 5%
Steal = (s) 30%

Win (W) = bx + t(1-y)+u(1-z)+v(1-m)+sy

W = 62.5%

 

Conservative starting the 8th (or 6th) end
Blank = 20%
Take 1 = 60%
Take 2 = 10%
Take 3 = 0%
Steal = 10%

 

W = 64.6

 

Based on my estimated variables, it appears the conservative approach, with three ends to play, is the preferred strategy. I’d be interested to see what percentages others might estimate for these variables. Once we have enough shot data collected, we may eventually be able to change these numbers from estimates to actual statistics.

 

Going back to our original game situation, Team Howard appeared to be unsure of their desired outcome for the end. If top teams use statistics to analyse both in game shots and late game situations, they can be better prepared to make these decisions while “on the clock”.

 

In case you don’t remember or didn’t see the outcome, Howard took one in the sixth, Ferbey took one in the seventh and Glenn made a great shot in the final end to win the game 5-4.

 

2. Go hard and go home 2007 Brier Final: Howard vs. Gushue
The most interesting – ie. debated – shot call of the 2006/07 curling season was that of Newfoundland’s Brad Gushue during the seventh end of the Brier final versus Glenn Howard. The Gushue team, despite missing the grizzled veteran of their gold medal win at the 2006 Olympics – Glenn’s brother Russ – were on the verge of adding a Canadian championship to their resume. After getting off to a rocky start at 1-3, the young foursome from the St. John’s Curling Club in Newfoundland had reeled off eight straight wins and was now in control, tied with the hammer, with four ends left to play. The critical seventh end came down to a decision of whether to attempt a difficult double-takeout for a possible score of three, or take a single point via the much easier shot of hitting the open stone belonging to the Howard team. The stakes were high as the Ontario team currently sat two, and any miss by Gushue could be disastrous. The reward of a three-shot lead and near certain victory, it seemed, was too great and the decision was made to play the double take-out. Ultimately, crashing on a guard led to a steal of two for Ontario and eventually a Brier and World Championship for the Howard team.

 

Mathematically, how do we analyse this situation, and what was the correct call?

Expected Results (ER) with 3 ends remaining:
1. Odds of winning if tied with hammer (x) = 64.6%
2. Odds of winning if one down with hammer (y) = 38.2%
3. Odds of winning if two down with hammer (z) = 19.8%
4. Odds of winning if 3 down with hammer (m) = 6.9%

 

Assumptions

• Gushue will be successful 100% of the time if he chooses to hit the open stone. Newfoundland will always score 1 point.

Option 1: Hit the open stone:
No variables, assuming 100% success, results in:

Win = (W) = 1-y = 61.8%

Option 2: Double attempt:
Many outcomes appear possible with the double attempt. Choosing the variables and estimating each of them may not be easy. Let’s try to choose most likely outcomes:

 

Successful double and stay, Newfoundland scores three points (a)
Successful double but the third Gushue stone is removed from play and Newfoundland scores two points (b)
Double is missed; Newfoundland scores a single point (c)
Double is missed; Ontario steals a single point (d)
Double is missed; Ontario steals two points (e)

 

Win (W) = a(1-m) + b(1-z) + c(1-y) + dy + ez

 

Recalling the situation, the target stone belongs to Team Gushue. They are attempting to jam it onto a Howard stone (removing it), roll to another Howard stone (also removing it) and keep the shot stone, the target stone and a third Gushue rock at the back of the house all in the rings. In discussions following the game, opinions were varied. Many people believed the double for three may not have been possible, that Gushue at best could score two (possibly spinning the target stone out of play). The outcome made this evaluation more difficult. By wrecking on the front guard, fans were unable to see what would happen if Gushue were to hit the target stone. In future, technology may allow us to study the behaviour of stones and, by mapping location, use a computer to determine the outcome of a shot. This type of analysis is likely possible now… however, the work required would be significant and I am doubtful the various curling governing bodies, would approve. Can you imagine if Gushue were to call a time-out and have his coach, sitting in the stands with a computer, advise them the shot was not possible based on a software analysis?

 

We will attempt to assign some estimates to all the variables and determine a possible outcome.

 

Variables:
Newfoundland scores three points (a) = 10%
Newfoundland scores two points (b) = 20%
Newfoundland scores a single point (c) = 10%
Ontario steals a single point (d) = 50%
Ontario steals two points (e) = 10%

 

Win (W) = a(1-m) + b(1-z) + c(1-y) + dy + ez

W = 52.6%

 

If we believe missing the guard will result in no worse than a steal of one, this analysis gives us a good start. Unfortunately, it is well below the known Expected Result for Option 1, 61.8%.

 

Let’s keep a and e constant at 10% and build a table of possible estimates for further analysis:

Interesting results. For Estimate #3, if Gushue can score 80% of the time, even if he only scores a single point 50%, the call is nearly identical to Option 1. In Estimate #7, if we assume a score of one is not possible but Gushue will score half the time, the shot call will result in two or three points scored (40% and 10%) or a steal of one or two surrendered (40%, 10%) and Gushue wins 58.7%, below Option 1. However, increase the result of a deuce by 10% in Estimate #10 (steal of one becomes 30%) and we are above Option 1 at nearly 63%.

Two other likely extremes:

• Set a, c and e to zero and assume outcome of either a deuce or a steal of one. If Gushue succeeds 60% of the time, W = 63.3%.
• Set b, c, and e to zero and assume either a score of three or a steal of one. Gushue needs to be successful more than 43% of the time to make this the correct call.

The numbers might start to confuse us and the final decision, which at first appeared incorrect, now appears fairly close. The ultimate answer depends on your assessment of how rocks will collide and move around the house, and the ability of Brad Gushue to hit what appears to be a small target on his front stone.

In this scenario, we are less concerned with assessing Brad’s abilities. The final outcome, wrecking on the front guard, was unexpected and likely a low probability. If he attempted this shot repeatedly, under the same conditions, I suspect he would hit the target stone at least 9 times out of 10. What we are more interested in is our estimate of the action on the stones, and who will score what once the dust has cleared. Unfortunately, this will never be known. If someone set the stones up exactly as they were situated we could, perhaps, try to recreate the outcome. If one of our readers decides to attempt this experiment, or if you already have, please let us know. In all seriousness!

Bringing some additional “non-math” analysis into the equation, if you will, is the consideration for the enormity of the moment, and the event. It is the Brier final and the position of being one up without hammer with three ends to play is, while positive, still only 62% headed to victory. Brad was no doubt drawn to visions of three points and a virtual lock on the Brier title (93% in fact). Even with a deuce the Newfoundland rink would be heading to the world championship 80% of the time. Looking back at all of our estimates, we still have Newfoundland winning greater than 50% of the time this shot is attempted. Perhaps it is worth the attempt at the big shot now – in the fourth-to-last end – when the opportunity presents itself, and take pressure off your team in the later ends of a tight game.

Just as in golf, some curlers choose to “grip-it-and-rip it” with an aggressive style, whereas others are tacticians who plod about the course and minimize both risks and mistakes. Personally, I trust the mathematics and believe the shot call Gushue attempted was indeed too risky. Specifically due to doubts about the shot being make-able for three; not because his ability is in question. One point for Howard or Gushue appeared the most likely outcome, with some possibility of a deuce (no better than estimate #2 in table above). It is, unquestionably, up to the skip to come to a final decision. As always, if you use math, it might make the decision clearer.

3. Hitting the post Canadian Olympic Trials semi-final: Morris vs. Stoughton
The use of mathematics to analyse game situations is useful in the later stages of a game and also for critical shot calls during the early ends. The John Morris call during the semi-finals of the 2005 Canadian Olympic Trials is a prime example of an early game decision that proved costly.

After stealing a single in the second end, the Jeff Stoughton team had scored a deuce in the third end to go one up without the hammer. Morris then found himself in deep trouble, facing three Stoughton counters and several guards. The two apparent options available for the Calgarians were to either a draw to force Stoughton to a steal of one point, or attempt a difficult hit between two guards and remove shot stone to count a single.

The port appeared small, but Morris was certain he could thread the needle and succeed in putting the score back to even. The shot wrecked on the front guards and Stoughton was handed a critical steal of three points. Despite a valiant comeback attempt throughout the remaining ends, Stoughton prevailed and advanced to the final, where his Winnipeg foursome eventually lost to Brad Gushue in the finals.

What mathematics can help us determine is: how certain did Morris have to be that he could make the called shot in order for it to be the correct decision?

If John makes the shot 100% of the time, it is the best call. He would much rather be tied without hammer than down two. Using available statistics, we can determine how assured he must be in order for the shot to be the correct decision.

Expected Results (ER) with 6 ends remaining:

1. Odds of winning if tied with hammer (x) = 61.8%
2. Odds of winning if one down with hammer (y) = 42.7%
3. Odds of winning if two down with hammer (z) = 24.3%
4. Odds of winning if 3 down with hammer (m) = 12.7%
5. Odds of winning if 4 down with hammer (n) = 5.4%

 

Assumptions

• The draw is successful 100% of the time. Stoughton will always be held to a single point. The chance of John being unable to draw the full eight-foot rings is considered negligible.
• The hit either results in a single for Morris or a steal of three for Stoughton. The chance of rolling out is negligible.

 Draw:

 No variables to examine, we assume 100% success. This results in:

 

Win (W) = z = 24.3%

  

Hit:

We will choose S as the single variable, indicating a successful shot by Morris.

 

W = S(1-x) + (1-S)n

 Setting W to the result for a draw, .243…

 .243 = S(1-x) + (1-S)n

 We now solve for S

 S = .577

 

The math shows us that if Morris expects to make his shot through the port more than 58% of the time, it is the correct call.

 

What at first appeared to be a foolish call now looks like a reasonable decision, based on statistical analysis. You are welcome to debate the shot call, and outcome, but the ultimate decision rests with John Morris himself, to assess if he can make the shot more than six times in ten attempts.

 

John may have felt his chances were much higher; others would counter it was lower, but the numbers tell us how he can best make the final decision.

 

Given an assessment of the Stoughton team and their abilities, Morris may determine that coming back from two down will be more difficult than 24.3%. Also, a draw is never 100% certain, therefore he may be more inclined to attempt the hit, even if his chances are as little as 50%.