Nearly a year ago, the mathletics blog of Dr. Wayne Winston has an interesting analysis of whether Pats coach Bellicheck should have gone for it on a critical 4th down, and put the coach's decision down to his confidence in Tom Brady. Yesterday, some thing similar (but far more serious) happened on the last day (D5) of one of the greatest test cricket matches in history.
Check out the amazing scorecard. In the 120-year history of test cricket, there have been only 12 such finishes.
The match situation
It is day 5 of the contest. About 26 hours of play time has passed and we are into the bottom of the 4th and final innings. India is batting and has lost 9 of their 10 wickets on a wicked last-day pitch with widening cracks and the ball spitting off the pitch. Many have gotten out to bouncers. Just an hour ago, India was down and out having lost 8 wickets with 92 runs still to get. A remarkable partnership between two injured players brings the match to a screaming knife edge. Australia requires 1 wicket to win. India needs 6 runs. At the crease is the injured Indian genius VVS Laxman overcoming back spasms with painkillers and with sheer grit, he is attempting to steal a miraculous win. But the one who is taking strike is P Ojha, a spin-bowler and India's no.11 player, who can't really bat. Bowling is Australian fast bowler Mitchell Johnson who can bowl past speeds of 95mph (there are others who can crank it up to 100mph).
Here's cricinfo's description of the third-from-last ball of the match (edited version here):
Johnson to Ojha, 4 runs, 90.5 mph, Lbw Shout And oh boy what we get .. Four Over throws! That looked out. Was there some wood on leather? Oh well ... What an insane little game this is! .. Steve Smith fires the throw and the ball misses the stumps and runs through the vacant covers. No Aussie fielder could back that up. But that throw was on. Had he hit - and he didn't miss by much - Ojha would have been run out. .....
(india wins 2 balls later)
Many cricket fans have criticized Steve Smith's decision to take a shy at the stumps. The Australian captain himself felt it was the right call and praised the rookie. If he had hit the wicket it would have been match over. Indeed several sports writers have called it a gutsy and worthy call. I know that if it was an Indian fielder instead, he would have been roasted by India's trigger-happy media.
Probability that australia wins = A. Probability that India wins = 1-A.
There are two scenarios:
1. If successful hit (probability p), then match over, australia win with probability 1.
2. If miss (probability 1-p) then there are two sub-outcomes:
2a. if fielder backing up, no overthrows, and australia win again with probability A (reset).
2b. if no protection, then overthrows, and australia win with probability A' < a =" p(1-A)"> A.
If (2b) is given:
p(win given hit) = p + (1-p)A'
This is statistically a good decision provided p + (1-p)A' >= A. A simple condition where this holds true would be if he were such a good fielder that statitically he hits the wicket more than 100A % of the time, i.e. if he felt that his chance of hitting the wicket was at least as good as the chance that Australia currently has of winning the game (fixing p = A would result in the LHS being > A)
Let's play with some numbers
1. Let's assume that with every run scored, the chance of success for Australia proportionally drops, so the cost of each overthrow run is A/6. For a worst-case 4 overthrows, this gives A' = A-4A/6 = A/3
2. so steve's accuracy rate had to satisfy:
p(1-A/3) >= 2A/3, or p >= 2A/(3-A)
3. If we assume that with just a single wicket to get, but only 6 runs to score, its anybody's game, and set A= 0.5. then steve's accuracy rate had to be atleast 1/2.5.
4. On the other hand, if you start with the premise that A is lower, say 25%, then steve only required an 18% accuracy to justify the throw. In other words, if you perceive that you have little chance of winning, then it is certainly a great idea to take the risk.
You can also calculate 'A' in a more sophisticated manner by looking at the competing counting process to determine P(wicket falls before 6 runs are scored). The probability that a wicket falls in some 'n' balls is a geometric distribution. Each ball is a Bernoulli trial. The distribution of runs per ball could be Poisson. India scored 6 runs in the previous 3 overs (18 balls), so purely statistically, you can extrapolate that to say there's about 18 balls to get the last wicket. All said and done, a 50-50 chance is the most practical choice for 'A' here.
If Steve Smith was generally able to hit the stumps 40% of the time (i.e. slightly less than even chance), then it would have a good call from a statistical point of view. I haven't gotten a chance to review to video to see if it was an easy versus difficult angle to hit the wicket, but on average, 40% does look like a fairly high required conversion rate.
Statistically it was not a great call especially if he knew there was nobody to back up. All the pressure previously built up on the batsman was released. But cricket is played on the field and making a match-defining split-second decision after 26 hours of exhilarating play is no easy ask. No guts, no glory. It's the Aussie way and the glorious game of cricket is better off with that choice, (especially since India won :-)