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The table has two obvious problems: the total drawn by the teams is odd, even though each draw should count in two of the teams' totals, and the number of wins exceeds the number of losses, even though there should be a loss for every win.
The rows show the correct total number of games played for each team, 4, so any corrections in the Won, Drawn and Lost columns must maintain this balance.
The noted problem with wins, losses and draws can be corrected either by reducing the number of wins by 3 and increasing the number of draws by 3, or by reducing the number of wins by two and adding one each to the number of draws and losses, since the total of all games is correct. As only three numerical corrections are to take place, and any subtraction in one category must be offset by an addition in another for any individual team, the win/loss/draw stats can be adjusted for only one team. Only Arnsley and Boldham have enough wins shown to be the adjusted team in the W/D/L columns.
If the adjustment is to decrease wins by 2 and increase draws and losses each by 1, then all three adjustments are to W/D/L, so the goals columns remain the same. If the adjustment is to Arnsley, then it would be left with no wins, 2 draws and 2 losses, but its goals for and against are the same, so that's impossible. If the adjustment is to Boldham, that gives it a record of 1 win, 2 draws and 1 loss. This is at least a possibility, with a 2-0 win, two 0-0 draws and a 0-1 loss.
If the adjustment is to decrase the wins by 3 and increase the draws by the same amount, then it must be Boldham to be adjusted, to 0 won and 4 drawn. It follows that it must be the team to have its goals adjusted, as it has only drawn games, so the goals for and goals against must be the same. However, this leaves the third correction to be in one of the goals columns. But, as is, the two columns balance. A change to one of these two columns would upset that balance, so this can't be a solution.
So the corrected table of stats is:
Won Drawn Lost Goals Goals
for against
Arnsley 2 1 1 2 2
Boldham 1 2 1 2 1
Cleeds 1 3 0 1 0
Drochdale 1 2 1 1 1
Erby 1 0 3 1 3
Arnsley's wins must have been 1-0 each, and its draw 0-0, with its loss being 0-2.
Boldham must have had a 0-1 loss, a 2-0 victory and two 0-0 ties.
Cleeds must have had a 1-0 win and three 0-0 ties.
Drochdale must have had a 1-0 win, a 0-1 loss and both its draws were 0-0.
Erby had a 1-0 victory and three 0-1 losses.
So Arnsley's 0-2 loss must have been against Boldham. Its two 1-0 wins must have been against Drochdale and Erby, making its 0-0 draw be against Cleeds.
Thus Boldham's 2-0 victory was against Arnsley; its two ties against Cleeds and Drochdale and its 0-1 loss to Erby.
Erby, thereby, suffered its three 0-1 losses against Arnsley, Cleeds and Drochdale.
So Cleeds 1-0 win was that game against Erby and its three 0-0 ties were against Arnsley, Boldham and Drochdale.
Drochdale's 1-0 win was against Erby; its 0-1 loss against Arnsley, and its two 0-0 ties against Boldham and Cleeds.
From Enigma No. 1438 by Susan Denham, New Scientist, 14 April 2007. |
