Chess games

Logic Level 3

In a round-robin tournament (all contestants play against all others), a win is scored with 1 point, a draw with 0.5 points and a loss with 0 points. In a game of 5 contestants, it is known that:

A, B, C, D, E received distinct scores
The ranking order (from top to bottom) was A, B, C, D, E
B is the only contestant that didn't lose a game
E is the only contestant that didn't win a game

Can you deduce the final scores of these contestants?

Input your answer by concatenating the scores of each contestant A, B, C, D, E and removing any decimal points. E.g. if their score was 3.5, 3, 2.5, 2, 1, then enter your answer as 3532521.

The answer is 3252151.

1 solution

A A
Jun 24, 2016

The order in which the players won starting from the winner to the last place is as stated A , B , C , D , E.

Considering this it would be useful to start in an organized manner by observing the possible scores A could have gained such that he anyway is in the 1st place.

Therefore , if A won the maximum number of points he could have obtained at most 4 points which would have meant that A won in all his matches with the other players. Yet , it is known that B is the only one who never lost a game meaning that A couldn't have won in all the matches and therefore A couldn't have obtained 4 points.

If A would have obtained 3,5 points then it would have meant that A won in all his matches but had a draw with B but then again A wouldn't have lost any game.

Now , considering the match between A and B there are 2 possible cases. Because B never lost a game the match between A and B can end either as a lose for A (1) or in a draw(2) which have to be analyzed.

Case 1. The only possible result of the match between A and B is 0 - 1 (A losses and B wins) and A has obtained at most 3 points. Now (make the following reasoning) , because B didn't lost any match and won the match with A he must have obtained at least (by making a draw with all the other players and therefore obtaining the minimum number of points from 4 games in 3 games) 1 + 3 * 0,5 = 2,5 points meaning that A couldn't have obtained less than 3 points because then A would have either lesser or an equal score with B.

This means therefore that A has 3 points losing to B and winning to C , D , E anyway.

Also , as observed from the above for B to not have the same or a greater score than A B must have obtained 2,5 points drawing in his matches with C , D , E and obtaining 0,5 points in each.

C and D have by now the same score of 1,5 points since they have a draw with B and have a win known from the problem's statement as E is the only one who never lost a game.

The game between C and D can't be a draw because then they should both win against E and would obtain the same number of points.

Between C and D then either C won or D won. If D would have won then he would have obtained at least 2 points since he can't lose against E (as E doesn't have any win) and C at most 1,5 points which would be contradictory.

Therefore C won against D and D won against E as D must have won at least a game and he didn't win from the matches with A , B or C any. If C would have won against E then C would have obtained from having wins (with D and E) and 1 draw (with B) 2,5 points which is the same score as B. Because of course C has lesser points than B C can't win against E and because he can't defeat E C must make a draw anyway.

By this the final order is A 3 points B 2,5 points C 2 points D 1,5 points E 1 point anyway.

Case 2. A and B have a draw. Then A can obtain at most 2,5 points and B's minimum score must be as B didn't lose any match obtained only from draws in all his 4 matches obtaining 2 points. To have the order considered in the table the only possible score would be C 1,5 points D 1 point E 0,5 points. Yet anyway considering this D has at least 1,5 points a draw with B and at least a win because just E never won. Therefore this leads to a contradiction this case being inconsistent and therefore impossible anyway.

By the above analysis the only possible consistent situation is the one presented in case 1 anyway.

That meaning A 3 points , B 2,5 points C 2 points D 1,5 points E 0,5 points the answer being 32521505 anyway.

As an observation it can be interesting to understand such problems more generally that is understanding for what kind of restrictions and given information and some number of contestants is a unique solution.

You should explain what "A couldn't have obtained less than 3 points because then the order of the table wouldn't be respected." means, and state why it's true.

Of course, once we know A has 3 points, we know the answer is 3252151 by the above. A better answer format might be to ask what the result of the game between C and D was, or asking for the number of ties, which would require working out more of the game.

"This means therefore that A has 3 points losing to b and winning to C , D , E anyway." need not be true as yet. A could have 3 point by (say) losing to C.

Calvin Lin Staff - 4 years, 11 months ago

1 and 3. A can't obtain 4 or 3.5 points as then it would mean that he doesn't lose any match and would contradict what is known from the problem (that only B never lost a game) I hope you agree. Therefore A must obtain at most 3 points and lose some match (thanks for the correction) and it is known that B can obtain since he never lose a match at least 4 * 0.5 =2 points.

Consider better the result of the game between A and B which is or not a draw. If A and B end in a draw A can obtain at most 2,5 points (1 lost match and 1 draw). Because B has 0.5 points already from the draw with A and to "respect the order from the table" B can obtain at most 2 points. Then indeed out of necessity the other players should have the scores C 1,5 D 1 E 0,5. Nonetheless D has at least a win and a draw with B (since B has 4 draws) therefor D has at least 1,5 points. This results in a contradiction showing that the match played between A and B can't be at draw. Because A and B can't draw and A can't win against B it can be concluded that B wins against A and then follows the reasoning presented upper.

Point 2. E can obtain either 1 or 0.5 or 0 points and as such it is not necessary concluded (if this is the reasoning you had in mind when you said it is necessary) that that is the solution.

It can be showed better that A and B have either a draw or one of them losing and by this the format is nice enough for the reasoning to run smoothly think but thanks for the correction anyway.

A A - 4 years, 11 months ago

1) I meant for you to add the explanation into the solution.

If A would have obtained 3,5 points then it would have meant that A won in all his matches but had a draw with B

No, even if A obtained 3.5 points, that does not guarantee that he must have drawn with B. (Yes, the subsequent logic of "he did not lose any games" is correct. However, you threw in a false conclusion, which makes your argument wrong).

Because B never lsot a game either the match between A and B is a draw (1) or a lose case (2) and this cases will be analyzed.
Case 1. The only possible result of the match between A and B is 0 - 1

My reading of your first sentence is that case 1 is a draw, and not a loss. You don't want to mix them up for the reader to have to read your mind.

Please work on a better presentation of the ideas and thought process, which would make your solution easier to read.

Calvin Lin Staff - 4 years, 11 months ago

@Calvin Lin – Point 2. So , I can't conclude that if A won 3,5 points he has a draw with B because the table of scores can look different , namely for A (3,5 points) there are other available tables of scores anyway. But this is wrong in my reasoning. For A to obtain 3,5 points , A can obtain them just by having a configuration 3 wins and a draw. It is the known that from those matches , A can't win against B because B never lost a match and therefore A must have a draw with B anyway.

The table of scores will be 0.5 - 0.5 with B and 1 -0 with C , D , E.

1 (and 3). I just made some changes in the form of the argument. I explained that A can't have a draw with B since if that would be the case it leads to contradiction anyway.

By this the format is there are 2 cases , 1 is wrong because is inconsistent therefore only one is consistent therefore just the consistent case is sound and can be accepted and as such by necessity it anyway can be concluded that it is right therefore this covers the sufficient material for making a deduction by he uniqueness of the possible solution. Changing the format like this and therefore considering the conclusion sound it can be explained why A can't have a draw with B which therefore appears in the solution but ok , maybe I will work again at chiseling the format of the answer until it is completely clear and "crystallized" in it's confuse structure anyway.

Oh , and yes I will correct that error in inverting the cases anyway.

Sorry and of course thanks for point it out anyway.

A A - 4 years, 11 months ago

Chess games

The answer is 3252151.

1 solution

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