Sir Arthur Eddington's Famous Cricket Puzzle: A Solution

Here's my solution to yesterday's puzzle:

-First, set up the bowling order: the only possible order is this - Pitchwell and Speedwell alternate for 6 overs each, then Tosswell and Pitchwell alternate overs, with Tosswell bowling 7 overs and Pitchwell bowling 6.1 overs.

-Tosswell bowls 5 maidens off his 7 overs, therefore his other 12 balls yield 31 runs. This means that there must be at least 7 fours off Tosswell’s overs.

-Since Tosswell only takes one wicket, the maximum amount of batsmen he can bowl to during those 12 balls is 5. Therefore, 2 batsmen must hit 2 fours each off Tosswell. The only two batsmen who could possibly do this are Bodkins and Perkins (no-one else scores more than 7 runs).

-So Bodkins must survive to face Tosswell, and he can not score in the first 12 overs.

-As Speedwell does not bowl a maiden this means that Bodkins can not face him at any point. Therefore, Speedwell must bowl 3 overs which yield 4 runs each, and 3 overs that yield 1 run each (any other possible combination means that Bodkins would have to score at least 1 run off him).

-We know that Pitchwell only bowls two maidens. Therefore, of the 3 overs that Speedwell bowls which yield 4 runs each, Bodkins must face Pitchwell after only 2 of those overs (and score nothing), and the other 4 runs must come off Speedwell’s last over.

-By the time that Speedwell finishes his last over, there must have been at least 19 runs scored - 15 off Speedwell, and at least 4 off Pitchwell (as he would have bowled 6 overs, with only 2 of them being maidens). Therefore, Atkins, Dawkins and Hawkins must all be out by this point (since Bodkins hasn’t scored), and Jenkins must have scored at least 1 run. But Jenkins can’t have scored only 1 run by this point, as we know that 4 runs are taken off Speedwell’s last over. Therefore, 23 runs have been scored from the first 12 overs.

-This means that the 5 batsmen who scored the 7 fours off Tosswell are Bodkins (2), Perkins (2), Larkins, Meakins and Simkins.

-It also means that a four was taken somewhere off Pitchwell’s first 6 overs, and no fours were taken off his last 6.1 overs (no other combination of scoring shots is possible). And this means Pitchwell’s last 6 full overs yield 1 run each (since 8 runs were taken off his first 6 overs, and there were no maidens in his last 6 full overs).

-Bodkins must be on strike for Tosswell’s first over as 4 runs were scored off Speedwell’s last over. Can it be a maiden? No, because a single has to be scored off Pitchwell’s next over, and none of Bodkins, Jenkins or Larkins (if Jenkins has gone out) can do that. Therefore, Bodkins must score 2 fours off Tosswell’s first over, and then go out, and then someone else must score a four, which can not be Jenkins. So Bodkins is Tosswell’s only wicket, and Jenkins is Speedwell’s only wicket. Fall of wickets so far: 1/6 (Atkins), 2/12 (Dawkins), 3/18 (Hawkins), 4/23 (Jenkins), 5/31 (Bodkins).

-Meakins replaces Bodkins, and he must score a four off Tosswell’s first over (as there are at least 3 fours in that over). Perkins and Simkins must score their fours in Tosswell’s second non-maiden over. What about Larkins? Since Tosswell has already taken his wicket, only two batsmen (Perkins and Simkins) can face him in his second non-maiden over. So Larkins must score his four in Tosswell’s first over, and Meakins must score a 1 to get him on strike. Therefore, 17 runs are scored off Tosswell’s first non-maiden over and 14 off his second one.

-The next few overs go as thus. Meakins scores a single off Pitchwell’s next over, Tosswell’s next over is a maiden (since Meakins doesn’t face Tosswell in his second non-maiden over), and then another single is scored off Pitchwell’s next over. It can’t be from Larkins, so he must be the next man out, at 6/41. Hence, the single comes from Perkins. Tosswell’s next over is a maiden (as Meakins is still in), and then Meakins scores his last single off Pitchwell’s next over, followed by another Tosswell maiden, and a Perkins single off Pitchwell. Then Tosswell bowls yet another maiden, and Meakins is dismissed by Pitchwell (since he can’t score another run), making it 7/44.

-Simkins comes in and scores a run off Pitchwell. Perkins and Simkins then plunder 14 off Tosswell, with Simkins hitting 5 of the runs, and Perkins 9. That takes Simkins up to 6 runs and Perkins to 11, meaning they can’t score anymore. In Pitchwell’s next over, he takes Perkins at 8/59, and then Tomkins at 9/59. Wilkins then comes in and scores the final single. Tosswell then bowls his final maiden over to Wilkins, and then Pitchwell comes back and takes Simkins with his first ball of the over (and last ball overall).

Final answers:

a) Tosswell dismissed Bodkins, Speedwell dismissed Jenkins, Pitchwell dismissed the rest.
b) Wilkins was left not out.
c) The fall of wickets: 6, 12, 18, 23, 31, 41, 44, 59, 59, 60.