THE DUCKWORTH-LEWIS METHOD http://www.flickr.com/photos/elkinator/3624920915 (to decide a result in interrupted one-day cricket)

ONE-DAY CRICKET THE PROBLEM Match restricted to one day Fixed number, N, overs for each team Draw is unacceptable if match is not finished THE PROBLEM How can a result be decided if rain stops play?

POSSIBILITIES a) Team 1 completes Team 2 interrupted b) Team 1 completes late Team 2 left short of overs c) No of overs reduced for both teams d) Both teams interrupted

A SOLUTION DIFFICULTIES Team 1 had all N overs Suppose Team 2 interrupted after u overs Compare average runs per over Compare Team 2 total with u overs of Team 1 (First u, Last u, Best u?) Compare best u’ < u overs of each – still questions DIFFICULTIES All these solutions can cause bias. We could Use c) with Team 1’s best overs scaled

A SOLUTION VIA MATHEMATICAL MODELS Formulate and quantify Team 2’s expected score allowing for the remaining N-u overs – compare A target that Team 2 needs to win

MATHEMATICAL MODELS a) Parabola No of runs, Z(u), in u overs Z(u)=7.46 u – 0.059 u2 (1)   225 runs in 50 overs – assumed typical Allows for team getting tired Anomalous maximum at u = 63. Negative for u > 126

MATHEMATICAL MODELS b) World Cup 1996 Identical to parabola with Z(u) expressed as a percentage of 225, i.e. 100 Z(u)/225

MATHEMATICAL MODELS c) Clark Curves Too complicated Allows for different kinds of stoppage and adjusts for the number of wickets, w, fallen

MATHEMATICAL MODELS d) Duckworth-Lewis Includes explicitly the number of wickets, w, fallen. (w < 10)

DUCKWORTH-LEWIS 1) Starting point is w-independent   Z(u)= Z0[1-exp(-bu)] (2) b accounts for the team getting tired If b small Eq. (2) is essentially Eq. (1) DL call Z0 ‘asymptotic’

Z(u,w)= Z0(w){1-exp[-b(w)u]} (3) DUCKWORTH-LEWIS 2) Influence of w If many overs, N-u, and few wickets, 10-w, are left or vice versa Eq. (2) needs to be changed DL modified it to include w-dependence Z(u,w)= Z0(w){1-exp[-b(w)u]} (3)

DUCKWORTH-LEWIS EXPRESSION

DUCKWORTH-LEWIS EXPRESSION ~ 260 runs maximum for 80 overs ~ 225 runs for maximum 50 overs DL formula (3) for 0 wickets is roughly parabola or World Cup 1996 Overs Parabola DL w=0 Ratio 5 36 42 1.17 10 69 78 1.14 15 99 110 1.12 20 126 135 1.07 25 150 160 30 171 175 1.03 35 189 190 1.01 40 204 205 1.00 45 216 217 50 225

EXAMPLE APPLICATION Proportion of runs still to be scored with u overs left and w wickets down is P(u,w)=Z(u,w)/ Z(N,0) (4) Wickets lost w 2 4 9 50 100 83.8 62.4 7.6 40 90.3 77.6 59.8 30 77.1 68.2 54.9 20 58.9 54.0 46.1 10 34.1 32.5 29.8 7.5 Overs left u

EXAMPLE APPLICATION Team 1 scores S runs, Team 2 stopped at u1 overs left w wickets down, play resumes but time only for u2 overs Overs lost = u1-u2. Resource lost = P(u1,w)-P(u2,w) Score to win = S{1-[P(u1,w)-P(u2,w)]}

A REAL EXAMPLE: ENGLAND VS NEW ZEALAND 1983 50 overs expected. England batted first, scored 45 for 3 in 17.3 overs, were stopped for 27 overs and scored 43 in 5.7 overs i.e. 88 in 23 overs. New Zealand were given 23 overs to score a target of 89 to win, which they did easily.

A REAL EXAMPLE: ENGLAND VS NEW ZEALAND 1983 In the DL method England’s score is altered and the calculation gives New Zealand a target of 112 to win. England were disadvantaged by the unexpected shortening of their innings. New Zealand knew in advance that they had a maximum 23 overs and planned accordingly.   DL claim that their method avoids this.

A REAL EXAMPLE: SOUTH AFRICA VS SRI LANKA 2003 50 overs expected. Sri Lanka batted first, scored 268 for 9 South Africa were 229 for 6 when rain stopped play after 45 overs. The DL target was 229, so the game was a draw.