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Sir Arthur Eddington's Cricket Problem

Sir Arthur Eddington's Cricket Problem

Extract from the score of a cricket match between Eastershire and Westershire:

EASTERSHIRE - Second Innings:

A. A. Atkins6
Bodkins8
D. D. Dawkins6
Hawkins6
Jenkins (J)5
Larkins4
Meakins7
Hon. P. P. Perkins11
Capt. S. S. Simkins6
Tomkins0
Wilkins1
Extras0
Total60

Bowling Analysis:

  OversMaidensRunsWickets
Pitchwell12.12148
Speedwell60151
Tosswell75311

The score was entirely in singles and fours, with no catches, no-balls, or short runs. Speedwell and Tosswell each had only one spell of bowling. Pitchwell bowled the first over, with Mr Atkins taking the first ball. Speedwell was the other opening bowler.

Q:

  1. Whose wickets were taken by Speedwell and Tosswell?
  2. Who was not out?
  3. What was the score at the fall of each wicket?

[Editor's note: This is a brilliant and complex puzzle. The spoiler has two complete solutions, but each is well over 1000 words. I do urge you to have a serious crack at it before you look at the spoiler. Here are some hints to get you started if you want, hidden by rot13 so as not to spoil the fun:

  1. Jub objyrq rnpu bire?
  2. Jung fpbevat fgebxrf jrer gurer va Gbffjryy'f aba-znvqra biref? Jub pbhyq unir fpberq gurz?
  3. Jung qbrf guvf gryy lbh nobhg Obqxvaf?
  4. Jung pna lbh fnl nobhg jung unccrarq va gur svefg srj biref?]

A:

  1. Speedwell took Jenkins' wicket. Tosswell took Bodkins' wicket.
  2. Wilkins.
  3. 6 - 12 - 18 - 23 - 31 - 41 - 44 - 59 - 59 - 60.

Here are two complete solutions, one by me and the other by David Louisson from New Zealand. We were working independently and took rather different approaches, and each solution has its pros and cons, although happily we did get the same answers to the questions asked.

Thomas Bending's solution

Speedwell was an opening bowler but he did not bowl over 1, and he had one spell, so he bowled overs 2, 4, 6, 8, 10 and 12. Pitchwell bowled a partial over, which must have been the last over, over 26. Tosswell is the only bowler who could have bowled over 25, and he had one spell, so he bowled overs 13, 15, 17, 19, 21, 23 and 25.

Call overs 1-12 Part A of the innings, and call overs 13-26 Part B.

Tosswell conceded at least seven 4s (otherwise it would have taken him at least 13 deliveries to concede 31 runs, in which case he couldn't have got five maidens). Tosswell conceded at most seven 4s (otherwise he'd have conceded at least 32 runs). So Tosswell conceded precisely seven 4s, and hence three singles. Tosswell had only two non-maiden overs and got only one wicket, so at most five batsmen scored off Tosswell. These five batsmen scored seven 4s off Tosswell between them, so they must have included Bodkins and Perkins scoring two 4s each (since every other batsman scored at most one 4).

Bodkins scored only off Tosswell, so he did not score and was not out during Part A. Atkins faced Pitchwell initially, so whenever he faced Speedwell he was on an odd score (otherwise the batsmen would be at the wrong ends). Atkins' final score was even, so Speedwell didn't dismiss Atkins. By the same argument Speedwell didn't dismiss Dawkins or Hawkins.

Atkins, Dawkins and Hawkins were out before the end of Part A, otherwise there's no batsman that Speedwell could have dismissed. The first wicket (Atkins) fell at 6. The second wicket (Dawkins) fell at 12. The third wicket (Hawkins) fell at 18.

Jenkins came in when Pitchwell dismissed Hawkins, so Jenkins faced Pitchwell initially, so whenever Jenkins faced Pitchwell in Part A he was on an even score. Jenkins' final score was odd, so Pitchwell didn't dismiss Jenkins. Speedwell dismissed someone, so he must have dismissed Jenkins. The fourth wicket (Jenkins) fell at 23.

Larkins came in when Speedwell dismissed Jenkins, so Larkins faced Speedwell initially, so whenever Larkins faced Pitchwell in Part A he was on an odd score. Larkins' final score was even, so Pitchwell didn't dismiss Larkins. Speedwell took only one wicket so he didn't dismiss Larkins. So Larkins was not out at the end of Part A.

Given the batsmen dismissed in Part A, the seven 4s conceded by Tosswell must have been scored by Bodkins (two), Larkins, Meakins, Perkins (two) and Simkins.

Bodkins and Larkins scored only off Tosswell, so they were both on 0 at the start of Part B. Bodkins was facing Tosswell at the start of Part B (because neither Bodkins nor Larkins scored in Part A after Larkins came in to face Speedwell). Neither Larkins and Bodkins scored any singles or scored any runs off Pitchwell. So the first runs in Part B were a 4 by Bodkins off Tosswell in over X.

There were ten scoring strokes off Tosswell, and these all occurred in Tosswell's two non-maiden overs. So there were at least four scoring strokes in over X. So Bodkins must have scored both his 4s in over X and then been dismissed to allow someone else to score in that over. So Tosswell dismissed Bodkins, and all the remaining batsmen were dismissed by Pitchwell. The fifth wicket (Bodkins) fell at 31. Meakins came in to replace Bodkins.

Bodkins was the only wicket in over X, so Perkins and Simkins must have scored their 4s in Tosswell's other non-maiden over, over Y. There were no wickets in over Y, so no other batsman scored in over Y. So Larkins and Meakins must have scored their 4s in over X. So Meakins must have scored a single in over X to bring Larkins to bat. Larkins' 4 was the last scoring stroke in over X (because he can't have scored a single to bring anyone else to bat, and he can't have been out). So Meakins must have scored his 4 before his single in over X. So the scoring strokes in over X were Bodkins 4, Bodkins 4, Meakins 4, Meakins 1, Larkins 4.

At the start of over X+1 Meakins was facing Pitchwell on 5, and until over Y any runs were scored off Pitchwell. Meakins was dismissed before over Y, so he must have scored his remaining two singles before over Y. After he scored the first of these Larkins was facing Pitchwell, and he must have been dismissed to allow someone else to score to bring Meakins back to bat. The sixth wicket (Larkins) fell at 41.

Perkins came in to replace Larkins, then scored a single to allow Meakins to face Pitchwell (Perkins can't have scored a 4 first because he scored all his 4s in over Y). Meakins then scored his last single to bring Perkins back to face Pitchwell. Perkins must then have scored another single to bring Meakins back to allow Meakins to be dismissed. The seventh wicket (Meakins) fell at 44.

Simkins came in to face Pitchwell, and the other batsman was Perkins who was on 2. There were no other wickets before over Y (although there may have been scoring strokes before over Y).

The scoring strokes in over Y were three 4s and two singles, scored by Perkins and Simkins. Perkins must have scored two of the 4s and Simkins the other 4. The batsmen must have scored one single each. So the batsmen were at the same ends at the end of over Y as they had been at the start of over Y. Apart from his runs in over Y, Simkins scored one single. If this was before over Y then over Y+1 started with Perkins facing Pitchwell on 11. If it was after over Y then over Y+1 started with Simkins facing Pitchwell on 5. In the latter case Simkins must have scored his last single before anyone could be dismissed, bringing Perkins to face Pitchwell on 11. In either case, the eighth wicket didn't fall before the point when Perkins first faced Pitchwell after over Y. At this point Perkins was on 11 and Simkins was on 6. So there was no score before the next wicket, so the batsmen didn't move before the next wicket, so the next wicket was Perkins. The eighth wicket (Perkins) fell at 59.

Tomkins then came in to face Pitchwell. Tomkins didn't score and Simkins was on 6, so there was no score before the next wicket, so the batsmen didn't move before the next wicket, so the next wicket was Tomkins. The ninth wicket (Tomkins) fell at 59.

Wilkins then came in to face Pitchwell. He can't have been out before he scored a single which brought Simkins to face Pitchwell. After this there was no score before the next wicket, so the batsmen didn't move before the next wicket, so the next wicket was Simkins. The last wicket (Simkins) fell at 60.

Wilkins was left not out.

Note that this solution answers the questions that were asked assuming that there is a legal game that meets all the conditions, but it does not prove that such a game exists. We haven't used or checked the fact that Pitchwell had two maidens, for example. See David Louisson's solution below for an over-by-over analysis showing one possible legal game (there are other possible legal games as well).

David Louisson's solution

COMPLETE SOLUTION USING A PROCESS OF LOGICAL DEDUCTION

1. Establish who bowled which overs -

2. Establish possible scoring shots off each bowler -

3. Establish batsmen who can hit fours -

[Hence maximum possible fours hit in the innings = 11]

4. It follows that, to obtain 7 fours off Tosswell, at least 5 different batsmen (and only 5, provided two of these are Bodkins and Perkins) must have faced his two non-maidens. This is also only possible if he takes his (solitary) wicket in one of these overs. (To have 6 batsmen face his two non-maidens would require two wickets in these overs, and so forth). Hence we know that Bodkins and Perkins each hit two fours, and that these were all hit off Tosswell.

5. Hence Bodkins (8 runs) only scoring shots are his 2 fours. Since Tosswell does not begin bowling until the 13th over, we must keep opening bat Bodkins AT THE WICKET AND SCORELESS throughout overs 1 to 12. This means that a single in any of these overs may not be followed by either another scoring shot, or a wicket.

6. Hence Speedwell (6 overs) can not have conceded more than 6 singles; therefore he must have conceded 3 fours and 3 singles (see point 2). As none of his overs were maidens, there must be at least 6 scoring shots, i.e. 3 of his overs conceded 1 four, and the other 3 overs conceded 1 single. [Also, 3 fours off Speedwell + 7 off Tosswell = a minimum of 10 fours hit in total].

7. From point 6, each of Speedwell's overs conceding boundaries must be followed by a Pitchwell maiden (else we have Bodkins scoring runs). Hence Pitchwell's two maidens occur in his first 6 overs. Thus runs conceded off Pitchwell's first 6 overs must be between 4 and 8 inclusive. As 15 were hit off Speedwell, the team total at the end of over #12 must lie between 19 and 23 inclusive. The third four hit off Speedwell must be in his 6th over (the 12th in the innings), to prevent Pitchwell from having to bowl a third maiden.

8. Hence runs scored in the first 12 overs must be - 6 by each of Atkins, Dawkins and Hawkins, and then either 1 or 5 by Jenkins (enabling Bodkins to remain scoreless at the other end). That makes at least 7 singles in these overs - as 3 of these are off Speedwell, then Pitchwell must have conceded at least 4 singles in his opening 6 overs. Any more runs conceded than these must be a four, as no more than 1 single may be conceded in any one of overs 1-12 (see point 5). Thus we have Pitchwell conceding either 4 singles (4 runs), or 4 singles and a four (8 runs), in his first 6 overs. [If the latter, then the four must have been followed by a single in the same over (since he bowled only 4 non-maidens in his first 6-over spell)].

9. The only scoring shot in the 12th over is a four (point 7), and given that Bodkins is scoreless (point 5), this must have been hit by his partner at this time, whom we know (from point 8) must be Jenkins. Hence Jenkins scores all 5 of his runs in the first 12 overs, the team total at the end of 12 overs is 23, and Bodkins is the striker at the start of Tosswell's first over. This means that Pitchwell did concede a four in his first spell (point 8), and given that his first spell also included both maidens (point 7), he must concede exactly one single off every over during his second spell. Given that Atkins, Dawkins, Hawkins and Jenkins have all hit their fours in the first 12 overs, the 7 fours hit off Tosswell must be from Bodkins and Perkins (2 each), and Larkins, Meakins and Simkins (1 each) - see point 3.

10. If Tosswell's first over is a maiden (played out by Bodkins - point 9), then we need a batsman in strike against Pitchwell in over #14 who can take a single (point 9). Jenkins has already used his 5 runs, and the next batsman, Larkins, scores his only runs (a four) off Tosswell (point 9). Hence Tosswell's first over can not be a maiden. Given that Bodkins hits both his fours off Tosswell, he must do so, and then be dismissed in the same over, to allow the required number of runs (7 fours + 3 singles = 10 scoring shots in Tosswell's two non-maidens) to be hit off Tosswell. Hence the first 3 'action balls' in over #13 must be 44X, and TOSSWELL'S ONLY WICKET IS BODKINS.

11. We must now fit 5 fours and 3 singles into the remaining 9 balls of Tosswell's 2 non-maidens [remembering that Perkins is now the only remaining batsman capable of hitting more than one four]. If Jenkins is still batting when Bodkins is dismissed, then Bodkins' replacement will be Larkins, which means that the only scoring shot possible in the remaining 3 balls of over #13 is a solitary four, leaving 7 scoring shots for Tosswell's second non-maiden, which is impossible. So Jenkins must have been dismissed before the start of over #13. Given that we know he struck his four during over #12 (point 9), he must also have been dismissed in this over. Hence JENKINS IS SPEEDWELL'S ONLY WICKET, and we can now confirm the FALL OF THE FIRST FIVE WICKETS: 1/6 (ATKINS), 2/12 (DAWKINS), 3/18 (HAWKINS), 4/23 (JENKINS), 5/31 (BODKINS).

12. The remaining 3 balls in over #13 must be negotiated by Meakins (replacing Bodkins), and Larkins. Given that Larkins, Meakins, Perkins (twice) and Simkins all hit fours off Tosswell (end of point 9), the remainder of over #13 must be 41 (Meakins) and 4 (Larkins), while Tosswell's second non-maiden must result in two fours and a single to Perkins and a four and a single to Simkins. See points 4 and 9 regarding the fours.

13. Since we know that Pitchwell took the other 8 wickets, the key to solving the rest of the puzzle is to get each batsman in strike to Pitchwell after he has made his required score. We also know that he conceded exactly one single off each over during his second spell (point 9), and have to arrange all of this around Tosswell's 5 maidens, and also include his second non-maiden, which must be faced by only Perkins and Simkins.

14. Meakins has only 2 singles remaining to him, after over #13. Perkins must take 9 of his runs from Tosswell, and Simkins 5 (point 12). That means that the 6 singles from Pitchwell's last 6 overs must be hit by Meakins (twice), Perkins (twice), Simkins (once), and Wilkins (once), to complete the scoring for each batsman. Meakins is in strike to Pitchwell in over #14, so he takes the required single. Over #15 (Tosswell) must be a maiden, because Meakins can not score from Tosswell. Over #16 (Pitchwell) requires a single, which Larkins can not provide. Hence he must have been dismissed (by Pitchwell) in either over #14 or #16 - whichever, it makes no difference, given the questions posed, i.e. the FALL OF WICKET IS 6/41 (LARKINS). So his replacement, Perkins, takes the single from over #16.

15. Perkins is in strike to over #17 (Tosswell). Given that Meakins (now 6 not out) is his partner, this over must also be one of Tosswell's maidens (see point 13). So Meakins is in strike for over #18, and he takes the required single from Pitchwell, and bring him in strike for over #19 (Tosswell). Given that Meakins has scored all of his 7 runs, this must be Tosswell's third maiden. Hence, in over #20, Perkins must take the required single from Pitchwell, bringing him to 2 not out. It doesn't matter in which Pitchwell over (#20 or #22), Meakins is now dismissed, the FALL OF WICKET IS 7/44 (MEAKINS), and Perkins will have the strike at the start of over #21.

16. It makes no difference to the required answers, as to whether over #21 or #23 is Tosswell's second non-maiden (but if it is #21, then Meakins must have been dismissed in #20). Given that Perkins must hit 2 fours and a single from this over, and Simkins a four and a single (point 12), the only thing left to determine is which of Simkins or Wilkins took the singles off overs #22 and #24, by Pitchwell. For last man Wilkins to take the single as early as over #22, Meakins must be dismissed at the end of over #20, Perkins must hit 441 and the Simkins 41 off over #21, leaving Simkins in strike, and given that he has only scored 5 of his 6 runs, can not be dismissed (along with Tomkins) to bring Wilkins to the crease. So Simkins must take the single from over #22, and Wilkins from over #24.

17. Given that either Perkins or Simkins must be dismissed to allow Wilkins to the crease, Tosswell's second non-maiden must occur before Wilkins' single in over #24 (point 16). So Wilkins will have strike for over #25, and play out what must be Tosswell's final maiden. Hence he can not be dismissed from the only ball bowled by Pitchwell in over #26. SO WILKINS IS LEFT NOT OUT.

18. Simkins and Perkins have scored all of their runs (point 17) before Wilkins gets to bat, so the FALL OF THE FINAL 3 WICKETS IS: 8/59, 9/59 (TOMKINS), 10/60.

A POSSIBLE OVER-BY-OVER ANALYSIS

A possible analysis follows. Figures in parentheses give batsmen's scores, and bowlers' analyses, at the END of each over.

Over #1 - Atkins (1*) takes a single off Pitchwell (1-0-1-0). Total: 1.

Over #2 - Atkins (2*) takes a single off Speedwell (1-0-1-0). Total: 2.

Over #3 - Atkins (6) hits a four, then is dismissed. FOW: 1/6. Incoming Dawkins (1*) takes a single off Pitchwell (2-0-6-1). Total: 7.

Over #4 - Dawkins (5*) hits a four off Speedwell (2-0-5-0). Total: 11.

Over #5 - Bodkins (0*) plays out a maiden from Pitchwell (3-1-6-1). Total: 11.

Over #6 - Dawkins (6*) takes a single off Speedwell (3-0-6-0). Total: 12.

Over #7 - Dawkins (6) is dismissed. FOW: 2/12. Incoming Hawkins (1*) takes a single off Pitchwell (4-1-7-2). Total: 13.

Over #8 - Hawkins (5*) hits a four off Speedwell (4-0-10-0). Total: 17.

Over #9 - Bodkins (0*) plays out a maiden from Pitchwell (5-2-7-2). Total: 17.

Over #10 - Hawkins (6*) takes a single off Speedwell (5-0-11-0). Total: 18.

Over #11 - Hawkins (6) is dismissed. FOW: 3/18. Incoming Jenkins (1*) takes a single off Pitchwell (6-2-8-3). Total: 19.

Over #12 - Jenkins (4) hits a four, then is dismissed by Speedwell (6-0-15-1). FOW: 4/23.

Over #13 - Bodkins (8) hits two fours, then is dismissed by Tosswell. FOW: 5/31. Incoming Meakins (5*) hits a four, then a single, then Larkins (4*) hits a four, all off Tosswell (1-0-17-1). Total: 40.

Over #14 - Meakins (6*) takes a single, then Larkins (4) is dismissed by Pitchwell (7-2-9-4). FOW: 6/41.

Over #15 - Meakins (6*) plays out a maiden from Tosswell (2-1-17-1). Total: 41.

Over #16 - Perkins (1*) takes a single off Pitchwell (8-2-10-4). Total: 42.

Over #17 - Perkins (1*) plays out a maiden from Tosswell (3-2-17-1). Total: 42.

Over #18 - Meakins (7*) takes a single off Pitchwell (9-2-11-4). Total: 43.

Over #19 - Meakins (7*) plays out a maiden from Tosswell (4-3-17-1). Total: 43.

Over #20 - Perkins (2*) takes a single off Pitchwell (10-2-12-4). Total: 44.

Over #21 - Perkins (2*) plays out a maiden from Tosswell (5-4-17-1). Total: 44.

Over #22 - Meakins (7) is dismissed. FOW: 7/44. Incoming Simkins (1*) takes a single off Pitchwell (11-2-13-5). Total: 45.

Over #23 - Simkins (6*) hits a four, then a single. Perkins (11*) hits two fours, then a single, all off Tosswell (6-4-31-1). Total: 59.

Over #24 - Perkins (11), then the incoming Tomkins (0) are dismissed. FOW: 8/59, 9/59. Incoming Wilkins (1*) takes a single, off Pitchwell (12-2-14-7). Total: 60.

Over #25 - Wilkins (1*) plays out a maiden from Tosswell (7-5-31-1). Total: 60.

Over #26 - Simkins (6) is dismissed off the first ball from Pitchwell (12.1-2-14-8). FOW: 10/60. Wilkins remains not out.

This is one of several possible analyses, although all yield the same answers to the questions asked. Other possible solutions involve interchanging the scoring shots of Atkins, Dawkins and Hawkins (who scored 6 runs each). However, because Bodkins must remain scoreless during this period, this does not alter the score at the fall of each of the first 3 wickets.

These are the only viable solutions that I am aware of.

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