1. THE DUCKWORTH-LEWIS METHOD
(to decide a result in interrupted one-day cricket)

2. 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?

3. 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

4. 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

5. 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

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

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

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

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

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’

11. 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)

12. DUCKWORTH-LEWIS EXPRESSION

13. 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

14. 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

15. 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)]}

16. 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.

17. 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.

18. 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.