THE DUCKWORTH-LEWIS METHOD

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

ONE-DAY CRICKET

• 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 interruptedb)Team 1 completes late Team 2 left short of oversc) No of overs reduced for both teamsd)Both teams interrupted

A SOLUTION• 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

DIFFICULTIESa)All these solutions can cause bias. We could

a) Use c) with Team 1’s best overs scaled

A SOLUTIONVIA

MATHEMATICAL MODELS

• Formulate and quantify a) Team 2’s expected score allowing for the

remaining N-u overs – compare b) 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’

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 oversDL formula (3) for 0 wickets is roughly parabola or World Cup 1996

Overs Parabola DL w=0 Ratio

0 0 0

5 36 42 1.17

10 69 78 1.14

15 99 110 1.12

20 126 135 1.07

25 150 160 1.07

30 171 175 1.03

35 189 190 1.01

40 204 205 1.00

45 216 217 1.00

50 225 225 1.00

EXAMPLE APPLICATION

Proportion of runs still to be scored with u oversleft and w wickets down is

P(u,w)=Z(u,w)/ Z(N,0) (4)

0 2 4 9

50 100 83.8 62.4 7.6

40 90.3 77.6 59.8 7.6

30 77.1 68.2 54.9 7.6

20 58.9 54.0 46.1 7.6

10 34.1 32.5 29.8 7.5

Wickets lost w

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.