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PrintBelorusija 2012
Belarus 2012 counting and probability
Problem
Ten chess players take part in a chess tournament. Each player plays exactly one game with any other player. For marking results a new system is used at this tournament: a player receives 3 points for a win, 1 point for a draw, and 0 point for a loss (in accordance with the old system a player receives 1 point for a win, 0.5 point for a draw, and 0 point for a loss). Grand master takes first place, he wins more than a half of his games and a number of his points is greater than a number of any other player points. Grand master says that in accordance with the old marking system grand master does not find himself among the first seven players (i.e. the number of his points is less than the number of points of the seventh player), while grand master says that in accordance with the old system grand master does not find himself even among the first eight players (i.e. the number of his point is less than the number of points of the eighth player).
a) Can grand master be right ? b) Can grand master be right ?
a) Can grand master be right ? b) Can grand master be right ?
Solution
a) The following table shows the example of the tournament such that grand master takes the eighth place only. So grand master can be right.
b) By condition, grand master wins more than a half of his games, i.e. he wins at least 5 of his games, so he takes at least 5 points in accordance with the old system of marking. So if is not among the first eight players, i.e. the eighth player has more points than , then at least eight players have more points than grand master , i.e. any of them has at least 5.5 points. Thus the sum of the points of the first eight players is greater than or equal to . But the total number of points in the tournament is in accordance with the old marking system, a contradiction. Thus grand master cannot be right.
| N | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | Points (Σ) | Points (Σ) | |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| New system | Old system | |||||||||||
| N | ● | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 15 | 5 |
| 1 | 1 | ● | 0.5 | 0.5 | 0.5 | 0 | 0.5 | 0.5 | 1 | 1 | 14 | 5.5 |
| 2 | 1 | 0.5 | ● | 0.5 | 0.5 | 0.5 | 0 | 0.5 | 1 | 1 | 14 | 5.5 |
| 3 | 1 | 0.5 | 0.5 | ● | 0 | 0.5 | 0.5 | 0.5 | 1 | 1 | 14 | 5.5 |
| 4 | 0 | 0.5 | 0.5 | 1 | ● | 0.5 | 0.5 | 0.5 | 1 | 1 | 14 | 5.5 |
| 5 | 0 | 1 | 0.5 | 0.5 | 0.5 | ● | 0.5 | 0.5 | 1 | 1 | 14 | 5.5 |
| 6 | 0 | 0.5 | 1 | 0.5 | 0.5 | 0.5 | ● | 0.5 | 1 | 1 | 14 | 5.5 |
| 7 | 1 | 0.5 | 0.5 | 0.5 | 0.5 | 0.5 | 0.5 | ● | 0.5 | 1 | 13 | 5.5 |
| 8 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0.5 | ● | 1 | 4 | 1.5 |
| 9 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | ● | 0 | 0 |
Final answer
a) Yes. b) No.
Techniques
Counting two waysInvariants / monovariants