Chess Exhibition

Since each set involves two players, the total number of sets played will be the sum of the sets played by each player divided by $2$ , which comes out to $38/2 = 19$ .

Now no player can sit out more than one set in a row, so with $19$ sets played, the fewest sets a player can be involved in is $9$ , which can only occur if such a player is involved in every other set starting from the 2nd set. As Magnus fits this criteria, we know that he is involved in only the even-numbered sets, and so $\boxed{\text{Garry and Fabiano}}$ must have played the 13th set.

As mentioned by Brian Charlesworth, the total numbers of sets played is 19.

Since Fabiano played 15 sets, 19 - 15 = 4 sets did not involve him, meaning Magnus and Garry played 4 sets with each other. Therefore:

M~G = 4, which implies that M~F = 9 - 4 = 5, and also that G~F = 15 - 5 = 10.

Based on the rule of the game, no same pairing can occur consecutively. Hence, if we consider the 10 sets of G~F matches, there must be at least one M~G or M~F set occurring between each of the G~F sets. The total of 9 M~G and M~F sets is the minimum possible, which means that the 10 G~F sets must have been the odd-numbered sets (1, 3, 5, ..., 19), while the even-numbered sets are either M~G or M~F.

Since set 13 is odd-numbered, it must have been G~F.

(Considering the names: If Carlsen is really this bad at badminton against both Caruana and the aging Kasparov, then it's a good thing he chose chess:)

Final Scores: Fabiano vs Magnus: 5 to 0; Garry vs Magnus: 4 to 0; Fabiano vs Garry: 5 to 4 with unknown results in the 19th game.

$\frac{9 + 14 + 15}{2} = 19$ so there were $19$ sets played, since each set involves two people.

If we want to minimise the number of games played, then they must lose each game. This is because if they win, they play another game straight after, so losing will minimise their games. This strategy then ensures that they play a game, miss a game, play a game, miss a game, and so on. Also, it is obvious that this is further minimised by sitting out the first set. Therefore, the minimum possible number of sets played can be achieved by playing sets $2,4,6...18$ . This amounts to $9$ sets (and it's a unique way to achieve the minimum when the total number of sets is odd), and since Magnus only played $9$ sets, he must have played in only the even numbered sets. Therefore he didn't play in set $13$ meaning the answer is Fabiano & Garry.

I just wanted to share a more visual approach - that lets us know who played who at any step.

As mentioned respectively by Brian Charlesworth, Leonard Ng and by definition:

• total number of sets is 19,

• sets that were played: M~G = 4, F~M = 5, F~G = 10

• no same pair can play twice in a row

Let's construct a diagram of the games.

Let M~G be "-", F~M be "|", F~G be "^".

Let's place M~G (4 sets): - - - -

To place F~M (5 sets), since no same pair can play consecutively, there is only one way to arrange them: | - | - | - | - |

Same goes for F~G (10 sets, compared to the 9 sets already drawn). So here's our final diagram: ^ | ^ - ^ | ^ - ^ | ^ - ^ | ^ - ^ | ^

Now all you need to do is count and you get who plays who at any set. You also get cat emoticons, but that's just bonus.

If Marcus played 9 games, Fabiano played 15 games, and Gary played 14 games, then Gary and Fabiano must've been the pair who played the 13th game.

Consider the situation as a venn diagramm. One of the pairs; either MG, GF, or MF, must've been the first the play. Now, pick any pair of players; let's say Gary and Fabiano. If they were not the first pair, then the number of games played by this pair must be equal to the number of losses of Fabiano. This is because one of the players changes every match, as they lose. If the current playing pair were Gary and Fabiano, then Marcus must've lost the previous set, as he's now sat down.

Assuming this is true, Gary and Fabiano would've had to have played less then 9 matches, because the number of matches they play is also the number of matches Marcus has lost, and so the no. of their matches cannot be higher than the total number of matches played by Marcus. However, there is no configuration of matches which satisfies that venn diagramm. Therefore, it must be the case that Gary and Fabiano were the first pair. This means that they would have played one more match than Marcus' losses. The only configuration that satisfies this venn diagramm is as follows: MG 4 matches, GF 10 matches, MF 5 matches.

Next, consider this as the no. of wins and losses. Marcus lost 9, won 0, and so on. Next, arrange this into a continous series of 19 pairs, one for each match. As Marcus lost all his sets, but was not first, he must've been in every even numbered set. Therefore Gary and Fabiano were in every odd numbered set. Hence set 13 was played by Garry and Fabiano.

Credit to others who developed a more erudite and concise solution to the problem.

I took a much longer approach where I looked at how many games each pair of players played. So here it goes:

Let $n_{MG}$ , $n_{GF}$ and $n_{FM}$ be the number of games between (Magnus and Garry), (Garry and Fabiano) and (Fabiano and Magnus) respectively. We can solve these using the total number of games by each player: $n_{MG} + n_{FM} = 9\\ n_{MG} + n_{GF} = 14\\ n_{GF} + n_{FM} = 15$

to get $n_{GF} = 10$ , $n_{FM} = 5$ and $n_{MG} = 4$ . The total number is 19, but Garry and Fabiano have played 10 games. Hence, this pair has played every alternate game and is the first pair to be matched up, i.e. every odd pair. The 13th set is odd, and therefore it was played by Garry and Fabiano.

I beleave magnus was disqualified after 9 sets , they all had to play 8 sets and the 9 disqualifed magnus 9 - 9 = 0 is out garry plays 14 sets - 8 = 6 sets left fabiano plays 15 sets - 8 = 7 sets left Garry play fabiano 6 + 7 = 13 sets this is how i get it

The total number of sets played equals 38 and if you divide that by 2 then the outcome is 19. Magnus only played 9, Garry played 14 and Fabiano played 15, therefore Magnus wouldn't have endured past set 10 (as the we have been informed), so Garry and Fabiano played the 13th streak.

There is a total of $\frac{9+14+15}{2} = 19$ sets. Since F played 15 sets, M and G played 4 sets with each other. Since M played a total of 9 sets, it must be that M and F played 5 sets. Moreover, F and G played 10 sets. Now note the following:

1. F, G, and M start either on Set #1 or Set #2.
2. If M wins a game or starts on Set #1, then there will be consecutive sets that have to be played by F and G. Since this violates the rules of the problem, M must start on Set #2 and lose every set of his 9 sets.

Therefore, M was involved in Sets # 2, 4, 6, 8, 10, 12, 14, 16, and 18, leaving Set #13 for F and G.

I stll don't get why is this problem too hard? Magnus DID NOT play set 13, because he stop playing after set 9. The other two persons played more than 13 games, therefore they MUST play game 13th