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

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Presentation on theme: "THE DUCKWORTH-LEWIS METHOD (to decide a result in interrupted one-day cricket)"— Presentation transcript:

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

2 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?

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

5 A SOLUTION VIA MATHEMATICAL MODELS Formulate and quantify a)Team 2s expected score allowing for the remaining N-u overs – compare b)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 – u 2 (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)= Z 0 [1-exp(-bu)](2) b accounts for the team getting tired If b small Eq. (2) is essentially Eq. (1) DL call Z 0 asymptotic

11 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)= Z 0 (w){1-exp[-b(w)u]} (3)


13 ~ 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 OversParabolaDL w=0Ratio

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 Overs left u

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

16 A REAL EXAMPLE: ENGLAND VS NEW ZEALAND 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 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.

18 A REAL EXAMPLE: SOUTH AFRICA VS SRI LANKA 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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