"With the players having each appeared 10+15+17 =42 times there must have been 21 matches. Poor Attika is awful at this game because even the worst rock-paper-scissors player in this system gets to play every second game. That means if she was in the very first match she would have racked up a minimum of 11 games (even if they were losses). So she must have lost 10 times and spectated 11 times. That means Burt and Celia played the first and last match, with Attika getting beaten in every even round match. So Attika lost game two!"
Stephen, Reddit
Wednesday, November 5, 2025
"In total, the kids were involved in a match 42 times. So there were 21 matches, giving rise to 21 wins and 21 losses. The most significant constraint comes from Attika being in only 10 of those matches. This means that at least 11 matches must have been between Burt and Celia. Whenever Burt plays Celia, the winner then must play against Attika. So after each of the 11 B-C matches, someone will play Attika. But Attika was in only 10 matches, which means that one of the B-C matches has to have happened at the very end. The only way this lines up is with Attika playing in each of the even matches, while Burt and Celia face off in each of the odd matches. If Attika were to win any of her matches, she would be in the next match as well, but we figured out that all odd numbered matches are B-C. So, Attika must lose all her matches, and in particular she must lose the second match of the day. Note: in the picture, we see that the girl on the left is winning the match, so this must be Celia. Attika is cheering in the back, her spirit unbroken by her string of losses. (Or perhaps they're only just starting.)."
Refreshing Username, Reddit
Thursday, November 6, 2025
"Alternate, slightly simpler route to the solution:
The sum of matches played between the players is 42, so with 2 players in each match there were 21 matches. You play, at a minimum, in every other match because the spectator replaces the loser. The sole way to play in only 10 matches is to lose every even numbered match (2, 4, 6, ... 18, 20). Therefore, Attika played in and lost the 2nd match."
Rick, United States
Tuesday, November 11, 2025
"The sum of all the matches played by the three participants is 10 + 15+ 17 = 42. Two participants are required per match, so there were 21 matches. Since Attika played in 10 matches, that means that they did not play in 11 matches. The only way this is possible is if Attika sat out each odd numbered match and played in each even number match. However, Attika could not be the winner of the second match, since that would mean they would not sit out match three. Therefore, Attika lost the second match."
Chris, Scotland
Saturday, November 1, 2025
"With the players having each appeared 10+15+17 =42 times there must have been 21 matches. Poor Attika is awful at this game because even the worst rock-paper-scissors player in this system gets to play every second game. That means if she was in the very first match she would have racked up a minimum of 11 games (even if they were losses). So she must have lost 10 times and spectated 11 times. That means Burt and Celia played the first and last match, with Attika getting beaten in every even round match. So Attika lost game two!"
Stephen, Reddit
Wednesday, November 5, 2025
"In total, the kids were involved in a match 42 times. So there were 21 matches, giving rise to 21 wins and 21 losses. The most significant constraint comes from Attika being in only 10 of those matches. This means that at least 11 matches must have been between Burt and Celia.
Whenever Burt plays Celia, the winner then must play against Attika. So after each of the 11 B-C matches, someone will play Attika. But Attika was in only 10 matches, which means that one of the B-C matches has to have happened at the very end. The only way this lines up is with Attika playing in each of the even matches, while Burt and Celia face off in each of the odd matches.
If Attika were to win any of her matches, she would be in the next match as well, but we figured out that all odd numbered matches are B-C. So, Attika must lose all her matches, and in particular she must lose the second match of the day.
Note: in the picture, we see that the girl on the left is winning the match, so this must be Celia. Attika is cheering in the back, her spirit unbroken by her string of losses. (Or perhaps they're only just starting.)."
Refreshing Username, Reddit
Thursday, November 6, 2025
"Alternate, slightly simpler route to the solution:
The sum of matches played between the players is 42, so with 2 players in each match there were 21 matches. You play, at a minimum, in every other match because the spectator replaces the loser. The sole way to play in only 10 matches is to lose every even numbered match (2, 4, 6, ... 18, 20). Therefore, Attika played in and lost the 2nd match."
Rick, United States
Tuesday, November 11, 2025
"The sum of all the matches played by the three participants is 10 + 15+ 17 = 42. Two participants are required per match, so there were 21 matches. Since Attika played in 10 matches, that means that they did not play in 11 matches. The only way this is possible is if Attika sat out each odd numbered match and played in each even number match. However, Attika could not be the winner of the second match, since that would mean they would not sit out match three. Therefore, Attika lost the second match."