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**THE DUCKWORTH-LEWIS METHOD**

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

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**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?

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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

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**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

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**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

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**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

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**MATHEMATICAL MODELS b) World Cup 1996**

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

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**MATHEMATICAL MODELS c) Clark Curves Too complicated**

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

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**MATHEMATICAL MODELS d) Duckworth-Lewis**

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

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**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’

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**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)

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**DUCKWORTH-LEWIS EXPRESSION**

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**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

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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

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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)]}

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**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.

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**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.

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**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.

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