One of the most difficult situations towards the end of a game is one down with hammer playing the next to last end (9^th or 7^th).  Statistically, the lowest probability of winning when one down with hammer is in the next to last end (34.9%).  In the last end or with two ends remaining it is over 38%.  In fact, tied with hammer with two ends to play is 67.5%, only 2.4% higher than if you are one up without!
 I have seen every type of play in this situation, from both teams keeping it clean to produce a blank (and a 5 minute end) to every rock in play.  So, using mathematics, what is the correct strategy, or at the very least, how do we approach this scenario to be better prepared when it happens?
 Recall statistical outcomes for the final end:
 

Expected Results (ER) with 9(7) ends remaining:
Odds of winning if tied with hammer (x) = 60.3% (60.7%)                      
Odds of winning if one down with hammer (y) = 43% (42.7%)
Odds of winning if two down with hammer (z) = 27.1% (26%)
Odds of winning if three down with hammer (m) = 14.7% (15%)
 

A nearly identical outcome.  If we expect our chance of allowing a steal in the first end less than 30%, when playing Aggressive, we increase our chances further.  For example, in Option 1 (Aggressive) if s=.2 and t=.4 we now win 62%.
 Given the early stages of a game, I suspect the differences in actual results between teams would be significant.  For example, Kevin Martin will win higher than 60% if tied with hammer after 1 end and, I suspect, the lower teams on the WCT rankings would be the opposite.  The more ends remaining, we would expect a greater discrepancy between teams.  But as a comparison, the analysis shows us being conservative is not the correct strategy as believed, it is actually the same or slightly worse than being aggressive.
           
The outcome for 8 end games is, oddly enough, nearly identical.  
           
Now, what about when we do not have last rock?  This decision appears to be obvious by answering this simple question:
 When tied without hammer, starting the 3^rd end of a ten end game, what strategy would we use?
 I expect nearly every team would choose to place a centre guard, and attempt to play for a steal.  Mathematically, the 3^rd end of a 10 end game is identical to the 1^st end of an 8 end game.  I’ll leave any further estimation to be done by the reader, as I am already convinced based on this fact alone.  There is no doubt, risk of a deuce or three is high in the first end and puts us into a poor position.  However, it doesn’t get any better as the game progresses and the more ends remaining to come back, the better our chances.  A team without hammer should play the first end to steal or force opposition to one.  In my opinion, the best way to achieve this is by aggressive play early (centre guard, draws to middle of house).
 It is perhaps simplistic to say Aggressive vs. Conservative play.  Commonly a team may move from one strategy to the other, and back, during a single end.  However, the common misconception that supports conservative play in early ends does not appear correct.
  Myth #2: A deuce early is not significant
 In the CCA document, it is suggested that, when starting without hammer:
A general objective is to limit the opposition to scoring a single point when you do not have last rock. Even a two-ender is acceptable.
 The author then goes on to suggest placing the first rock of the game in the four foot!  Per Myth 1, this is not the approach I would recommend.  So, is surrendering a deuce acceptable?  From Myth 1, we now know down two after the first end puts us at a 26% or 27% chance of winning (based on 8 or 10 end game).  This doesn’t sound acceptable to me.  We may need to surrender a deuce because that is how the end develops, but we certainly should not play the end to make a deuce acceptable. 
 Let’s take a scenario where two rocks remain in the first end and we do not have last rock.  Our team has missed a few shots and we are in trouble.  Our opposition, holding hammer, lays two.  One rock is open in the twelve foot; the other is partially buried in the top four foot. We can hit the open rock and surrender a deuce or try a corner freeze on shot rock to force the opposition to 1.  What percentage do we need to succeed to make the draw the correct call?  We’ll use the statistics from 8 end games for this analysis.
 Hit
W (if down two) = z = 26%
 Draw
 Assumptions
The team with hammer either scores three or is held to one.  The chance for a steal or deuce is considered negligible.
 Variables
Successful draw (hold team to one) = d
 W = dy + (1-d)m
 Setting W to 26%
 

.26 = d(.43) + (1-d)(.26)
 

Solving for d = .395
 

We need to make the draw greater than 40% of the time for it to be correct call.  Given skills of top skips, I would suggest the correct play is the draw and “accepting a deuce” is not the correct decision.  Better to risk a three at the possibility of forcing opposition to a single.
 

Myth #3: Play conservative against a stronger team
It is a commonly held belief that when playing a team that is stronger than yours; play conservative, keep the game close and hope to win a close battle at the end.  The U.S. team in the 2007 World Championship appeared to use this strategy to perfection against Team Canada during the round robin.  I disagree with this approach and will attempt to explain why….in another article.

The following three articles are taken from the 2007/08 Black Book of Curling. 

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:
 

With all due respect to Linda Moore and Ray Turnbull, the recent game of Howard versus Stoughton during the round robin game of the 2007 Tim Horton’s Brier showed how even some of the most knowledgeable don’t understand statistics as it relates to curling.
 Howard is one up without hammer in 9^th end.  Two rocks remain.  There is a corner guard and a Stoughton rock on the opposite side of the sheet in the back twelve foot.  Richard Hart, the Ontario third, and Glen Howard discuss the options.  Glen prefers a freeze to the back stone, but eventually Richard convinces him to play a draw around the corner guard.  Rich’s argument is their uncertainty of the ice on the freeze attempt – Glen’s concern is the amount of curl he will get with his come-around attempt.
 

This article continues the discussion from “Curling with Math”, included in the 2006/2007 Black Book of Curling, available from Curlingzone at www.blackbookofcurling.com.  Also reprinted in Dec 2006 article.
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 9^th 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 10^th 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 10^th end

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.
   

Here’s a scenario that was described to me by a friend of mine. 

One down playing the ninth end and our hero, throwing Red stones, has the hammer.  His opponent is throwing his final stone, Blue.  There are no rocks in play except for one Red stone at the back of the eight foot.  The Blue skip is about to throw his last rock.  We would expect him to hit and stick, hoping our hero noses for one or, at worse, blanks and remains one down coming home.

Instead, the Blue skip chooses to attempt a freeze.  This shot call appears to conflict with common curling strategy, but, using math, what is our analysis of this play?

Expected Results in the 10^th end

1. Odds of winning if tied with hammer (x) = 75.7%
   
2. Odds of winning if one down with hammer (y) = 39.5%
   
3. Odds of winning if two down with hammer (z) = 11.7%
   

Option 1: Hit
Let’s assume the odds of Blue missing the hit as negligible.  Either he rolls out and a blank is assured or, Red rolls out most of the time for a blank.  To simplify, let’s state the odds of a blank are 95%.
 

W        = (1-y)(.95)+x(.05)
            = .6126
 

Option 2: Freeze
Possible outcomes after Blue’s shot:
 

1. If Blue makes the perfect freeze, our hero is forced to draw for one. 
   
2. If the freeze is off slightly (call it a “near-freeze”), the deuce is not likely but Red will be able to blank by blasting both rocks and rolling out as well. 
   
3. If the freeze attempt is too heavy or light or fails to cover part of the Red stone, a chance for two becomes possible.
   

Blue needs to initially estimate his chances of making the freeze.  Let’s estimate this and other Variables:
 

Blue’s shots:
f = Odds of the freeze, Red must draw for 1 = 25%
n = Odds of near-freeze = 50%
m = Odds of a miss = 25%
 

Assumptions:
Initially, let’s try to simplify our analysis as much as possible…
·        If Blue freezes, Red will always score 1
·        If Blue misses, Red will always make the shot for two. 
 

Further math is then required for the “near-freeze”.  Let’s start by estimating that, after a near-freeze, Red will always attempt a hit. Variables as follows:
 

b = blank = 70%
t = take 1 = 20%
s = steal = 10%
 

W        = fx + my + n(b(1-y) + tx + s(1-z))
            = ..6196
 

Very close to our original result.  This would imply that, based on our estimates, either decision is correct.
 

Using some other values for b, t, s
 

Not very drastic changes, even if Blue steals 20% and forces Red to one 30%, we only increase our chances 3% over hitting.  Given these numbers are very close, let’s use 62% (our original estimate), and make changes to the original variables on Blue’s shot.
 

This article is a reprint of the 2006/07 Black Book Curling 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 9^th 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 10^th 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.The question I immediately asked myself was, is that the right decision?  Not in terms of gut feelings of the moment or using hindsight, but if the game was played 1,000 times or 10,000 or infinite times, what would be the correct play based on mathematics and statistical analysis?  Could this be done and, if so, how valid are the results?Now, let’s attempt to analyse the situation based on available statistics and mathematical in-game analysis. To examine the 9^th end 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.

It’s down to the final two rocks of the end.  You are Team A, and do not have last rock.  You have a 1 point lead and the house is open except for your opponent’s stone in the eight foot at the tee-line and a center guard.  What should you do and, more importantly, what analysis can you do to determine the correct decision?
 

Assumptions:
1.      The guard is halfway from the 12 foot to the tee line.  It is your rock.
2.      Ice conditions are perfect or near perfect.
3.      If we attempt a hit we will not miss the rock completely.
 

Expected Results (ER) in the 10^th 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%
 These Expected Results are based on available statistics for mens 4-rock games played in the Brier, European and World Championships, Grand Slam and other WCT events, Provincial Championships, Olympics and Olympic Trials.
 We have three options:
1.      Hit and stick.
Variables
b = Blank = 95%
 We estimate that if we attempt to hit and stay, the chance we roll out or the chance our opponent hits and rolls out successfully to be 95%.
 W (Win) = b(1-y) + (1-b)x = 61%