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Reinventing the DLS method: Part 3a

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Roar Rookie
18th November, 2020
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With Part 2 demonstrating that there’s a lot wrong with the Duckworth-Lewis-Stern method in its operation, including target calculations, this Part 3a is all about presenting how my McWarehouse method works logically.

This series, finding something better than DLS, will be increased from the originally intended six parts to ten in order to halve the length of each individual publication on The Roar. However, they will be numbered 3a, 3b, 4a, 4b, 5a, 5b, 6a and 6b.

The McWarehouse mathematical model is based on the notion that a team chasing a target has two different run rates they need to adequately maintain: (i) chasing an example target of 300 in a maximum of 50 overs, they must firstly maintain a run rate of six per over across the whole of the 50 overs available. However, they must also (ii) maintain a run rate of 30 per partnership across the whole of the innings.

Simply, each five-over period of play must produce for the chasing side an outcome of no worse than 1 for 30.

To generate a complete table, the McWarehouse method begins with a ‘par diagonal of equilibrium’. The top row represents, in descending order, the number of wickets still to fall, while the column on the far left represents the number of available overs already used up. This par diagonal, at all points, equates to precisely those two different aforementioned run rate requirements.

Table 1 reflects a run-chase target of 300 from 50 overs.

50max 10 9 8 7 6 5 4 3 2 1
5 30
10 60
15 90
20 120
25 150
30 180
35 210
40 240
45 270
50 300

If cricketing reality says that any team’s innings will not always be precisely in that state of equilibrium demonstrated by the previous table, then they at least need to have maintained a credible balance between overs remaining and wickets still in hand. Keep in mind, this is in proportion to the runs still needed to be scored for victory at or by any given point the match might be prematurely terminated.

As wickets are preserved, the current run rate per partnership increases above the overall innings requirement, and this allows leeway for a temporary drop in the current run rate per over to slightly below the whole-innings requirement.

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However, if a disproportionately higher number of wickets fall and the chasing team moves into the isosceles triangle region, to the right of the par diagonal of equilibrium, then there is far less scope to fall behind in the run rate per partnership requirement. Namely, because on an average day, ignoring momentarily how quickly runs might be scored, the lower half of a batting line-up does not contribute more to the eventual team run total than the top and middle-order.

Mitchell Starc of Australia celebrates dismissing Kane Williamson

(Photo by Cameron Spencer/Getty Images)

For example, based on mathematical probability, a team that is already seven wickets down after 30 overs will more than likely be bowled out around the 40-over mark, failing to maximise their team’s potential 50-over run total.

Following this example, a side seven wickets down at the 30-over mark chasing 300 could conservatively be assumed to make no more than about 70 runs for the final three wickets, and further still, would need a virtual ‘timeless’ scenario in order to achieve this.

While at 5-180 after 30 overs and chasing 300, the run chase can be considered to be in equilibrium. Yet if the side was seven down instead, they have lost a disproportionate number of wickets and therefore, cannot afford to take the same risks as when five down – if they are to maintain the initial run rate per over requirement simply because the runs per partnership requirement would have risen from the original 30 to 40.

On the contrary, reducing the runs per partnership requirement from the original 30 to about 23 would seem attainable for a batting group consisting chiefly of tail-enders and perhaps one remaining specialist batsman at the crease.

This gives rise to a series of runs per wickets calculation tables that make it straightforward to recalculate run targets if an innings is reduced.

50max 10 9 8 7 6 5 4 3 2 1
15 66 78 90 115 140 165 190 215 240 270
20 92 101 110 120 145 170 195 220 245 270
25 120 127 134 142 150 175 200 225 250 275
30 -150 156 162 168 174 180 205 230 255 280
35 182 187 192 197 202 206 210 235 260 285
40 215 219 223 227 231 234 237 240 265 290
45 252 254 256 258 260 262 264 267 270 295
50 300 300 300 300 300 300 300 300 300 300
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Each necessary target, at the various points on the table above, is converted into a decimal fraction and then expanded to cover every single ball of the maximum overs set down for the match according to the playing situation. This is demonstrated in Table 3 below, which represents a small portion of the overall handbook comprising more than 100 such tables collating every possible number of maximum overs from 50 to six:

50max 10 9 8 7 6 5 4 3 2 1
29.3 2.0498 1.9672 1.8905 1.8184 1.7516 1.6712 1.4673 1.307 1.1786 1.0732
29.4 2.0332 1.9525 1.8776 1.8075 1.7424 1.6697 1.466 1.3061 1.1779 1.0726
29.5 2.0166 1.9378 1.8647 1.7966 1.7532 1.6682 1.4647 1.3052 1.1772 1.072
30 2 1.9231 1.8519 1.7857 1.7241 1.6667 1.4634 1.3043 1.1765 1.0714
30.1 1.9883 1.9125 1.8423 1.7769 1.7161 1.6597 1.4622 1.3034 1.1757 1.0708
30.2 1.9766 1.9019 1.8327 1.7681 1.7081 1.6527 1.461 1.3025 1.1749 1.0702
30.3 1.9649 1.8913 1.8231 1.7593 1.7001 1.6457 1.4598 1.3016 1.1741 1.0696
30.4 1.9532 1.8807 1.8135 1.7505 1.6921 1.6387 1.4586 1.3007 1.1733 1.069

The umpire and players simply need to divide the target score by the decimal figure from the box corresponding to the number of wickets still standing and the number of overs and balls already bowled.

For instance, a team is chasing 239 off 50 overs, and the innings is prematurely terminated after 29.3 overs. They have three wickets still standing, so the calculation will be – from Table 3 – 239 divided by 1.307. Evidently, to be awarded the victory, they would need to have 183 runs on the board.

The columns in bold represent the par scores to that point of an innings, and this par column shifts one position to the right for every ten per cent of the elapsed innings – a maximum of five overs for a 50 over match.

Therefore, if it became known before the innings that time remaining necessitated that the chasing side’s batting stint be reduced to 30 overs, then the calculation for that same 50-over target of 239 would be: 239 divided by 1.6667, which corresponds to 5-143 – a par score to this point. Subsequently, this guide target would be presented to both team’s captains in the following form:

Par Target in Runs: 143 VBO: 6 (‘Virtually Bowled Out’ principle from Part II of this series)

Par Winning Score (at required net run rate per over for whole of reduced innings): 5 for 143

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Tim Paine speaks to umpire Aleem Dar

(Ryan Pierse/Getty Images)

If the captains wish to know alternative par targets, then it is their own responsibility to make these calculations. All teams, as well as umpires in any association competition, will possess their own copy of the handbook containing all necessary tables vis-a-vis delayed or reduced matches. In this particular example, above par targets would be as follows: 0 for 120, 1 for 124, 2 for 129, 3 for 134 and 4 for 139. Below par scores would be as follows: 6 for 163, 7 for 183, 8 for 203 and 9 for 223.

With nine wickets down, for example, the assumption is that with the last pair at the crease, we would expect them to gather 16 runs for victory, provided they had so much time to do it that run rate per over was superfluous. That is not to say that they should scratch and poke around for 20 overs, rather that there is no way they can be backed to get their team home if they are required to score at the overall innings run rate requirement of 4.78 per over.

There is another factor that always makes the nine wickets down scenario extremely significant in a mathematical sense, and this will be explored in detail later. By the same token, in chasing 239 to win off 50 overs, if the openers were still together at the 30-over point, we would back the team to double their innings total with all ten wickets still in hand.

Importantly, under the McWarehouse system, the match does not end simply because the chasing side reaches the par score – 143 runs in the above example. If they reach 143 with only five wickets down by the end of the 25th over, they must continue batting until either a) the 30 overs their innings has been reduced to have been bowled in full, b) they get bowled out before the end of those 30 overs, in a wickets-sense rather than VBO, or c) they pass the actual 50-over target – 239 here.

In the case of c), this provision applies whether the 50-over target is one run more than the opposition actually scored with full access to their maximum overs, or a projected 50-over score for the team batting first in the event that their overs were also reduced after their innings began.

For example, let’s imagine that this team chasing 239 from 50 overs has their innings reduced to 30 overs, producing a par guideline score of 5 for 143. Let’s also imagine that this is precisely the score they have reached off 25 overs, with another five to be bowled. If they then lose another wicket off those remaining five overs, then they will need to score another 20 runs. Lose another two wickets and they will need to score another 40 runs off these last five overs of their reduced innings.

Conversely, if their total sat on 6 for 160 off 25 overs, then off the final five overs they would require a further three runs without losing another wicket, or if they were to lose one more wicket they would need to add another 23 runs to the team’s total before the end of the 30 overs their innings has been reduced to.

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Obviously, the best tactic in such a scenario would be to aim for another 23 runs without undue risk, just in case they do lose another wicket in freakish circumstances – this would be smarter than just trying to block out the last five overs for three measly runs.

Rachael Haynes bats for Australia

(Photo by Paul Kane/Getty Images)

Tables 1 and 2 showed earlier that the par diagonal of equilibrium, which is the fundament that underpins McWarehouse, requires no wickets to fall in the first five overs – of a 50-over innings and conversely the first ten per cent of any maximum overs amount – and then allows one wicket every five overs, or ten per cent of maximum overs. There are several reasons why that first ten per cent of maximum overs does not permit a wicket in order to keep the run chase in equilibrium while maintaining, in mathematical theory, the required run rate per over across the innings.

One wicket down after five overs in a 50-over innings represents a VBO scenario. Hence, in order to be on par at that stage, a team needs to have lost one less wicket than VBO. Additionally, five overs is well before the minimum required overs that need to be bowled in order to constitute a match – 20 with the DLS method. In the 50-over competition of the association I umpire, we set it at 25. I believe it should be set at half the number of maximum overs for a match, and I will discuss this in Part 6.

Furthermore, if the minimum overs for a 50-over match is 25, then we can eliminate the 15- and 20-over rows from Table 2, and then the only thing of relevance is that the par diagonal of equilibrium allows for four wickets to fall in the first 25 overs -or 50 per cent of the maximum overs number – to remain on par, keeping up the required run rate per over.

A misconception is that McWarehouse weights all ten partnerships of an innings equally, i.e. assumes the same proportion for both batsmen and tailenders. Table 4 below dispels this misconception. The most important columns to pay attention to are the middle column as well as the one on the far right. This clarifies that the permissible WAF (wicket affordability factor) as well as the subsequent runs per partnership requirement decreases marginally for the first five wickets of an innings and then more considerably for the last five.

Overs Bowled Wickets Standing Progressive WAF Runs Still to Get: Target 250 Runs Per P’ship Needed: Target 250
0 All 30 250 25
5 10 27 225 22.5
10 9 26.7 200 22.2
15 8 26.3 175 21.9
20 7 25.7 150 21.4
25 6 25 125 20.8
30 5 24 100 20
35 4 22.5 75 18.8
40 3 20 50 16.7
45 2 15 25 12.5
50 1 n/a 0 Target Achieved on Par = 9 for 250 (50 overs)

Part 3b will complete this third section in the series with some more hypothetical chase situations. I will also revisit some controversial matches from the 1992 World Cup where that ridiculous highest-scoring overs method was used. In so doing, I will demonstrate the applicability of the McWarehouse solution to the infamous semi-final between England and South Africa.

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