THE DUCKWORTH-LEWIS METHOD

(to decide a result in interrupted one-day cricket)

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ONE-DAY CRICKETMatch restricted to one day Fixed number, N, overs for each teamDraw is unacceptable if match is not finished

THE PROBLEM

How can a result be decided if rain stops play?

POSSIBILITIESa)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 SOLUTIONTeam 1 had all N oversSuppose Team 2 interrupted after u overs Compare average runs per overCompare Team 2 total with u overs of Team 1 (First u, Last u, Best u?)Compare best u < u overs of each still questionsDIFFICULTIESAll these solutions can cause bias. We couldUse c) with Team 1s best overs scaled

A SOLUTIONVIAMATHEMATICAL MODELS

Formulate and quantify Team 2s expected score allowing for the remaining N-u overs compare A target that Team 2 needs to win

MATHEMATICAL MODELSa) Parabola

No of runs, Z(u), in u overs

Z(u)=7.46 u 0.059 u2(1)225 runs in 50 overs assumed typicalAllows for team getting tired Anomalous maximum at u = 63. Negative for u > 126

MATHEMATICAL MODELSb) World Cup 1996

Identical to parabola with Z(u) expressed as a percentage of 225, i.e. 100 Z(u)/225

MATHEMATICAL MODELSc) Clark Curves

Too complicatedAllows for different kinds of stoppage and adjusts for the number of wickets, w, fallen

MATHEMATICAL MODELSd) Duckworth-Lewis

Includes explicitly the number of wickets, w, fallen. (w < 10)

DUCKWORTH-LEWIS1) Starting point is w-independentZ(u)= Z0[1-exp(-bu)](2)b accounts for the team getting tiredIf b small Eq. (2) is essentially Eq. (1) DL call Z0 asymptotic

DUCKWORTH-LEWIS2) Influence of w If many overs, N-u, and few wickets, 10-w, are left or vice versa Eq. (2) needs to be changedDL 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

OversParabolaDL w=0Ratio000536421.171069781.1415991101.12201261351.07251501601.07301711751.03351891901.01402042051.00452162171.00502252251.00

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

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

Wickets lost wOvers left u

02495010083.862.47.64090.377.659.87.63077.168.254.97.62058.954.046.17.61034.132.529.87.5

EXAMPLE APPLICATIONTeam 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 198350 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 1983In the DL method Englands 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 200350 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.