It’s moderately amusing – nowhere near as hard to solve as I had supposed from “it’s really really cool and kind of shocking” – but worth doing.
It took me about a minute to solve this puzzle but I didn’t realise it at the time (perhaps I was looking for the “shock”) – I had read the question too fast and supposed it asked for the winner of game two.
So I’d wasted a few more minutes drawing tables and thinking about combinatorics and parity and blah blah before coming to the conclusion that one couldn’t know, so I re-read the question, and saw that I was right even if I was wrong.
HERE’S THE CHALLENGE :
Can we add some information – possibly disguised and as oblique as possible* – that allows “my” misread puzzle to be solved?
I did it longhand, like the 3 Children puzzle you linked to.
I started the amount of times A played as AB + AC = 10.
The amount of times B played is BA + BC = 15 and the amount of times C played is CA + CB = 17.
Then I wrote down all possible outcomes for each set of equations and found they all shared one outcome: AB = 4, BC = 11, and CA = 6, so altogether they played 21 games. Since B won over half the games, B is the control group.
I then wrote up outcome blocks for the first few games, from 3 to 5. They all had to start with A vs B and end up as B winning, and have B win as many as possible. The first 3 outcomes had B losing the 2nd match. The 4th outcome had B winning the 2nd match. If we stick with just the first 3 outcomes, Outcome 1 occurs once, Outcome 2 once, and Outcome 3 occurs 3 times. Inputting the charts I made won’t work too good on this message board UI, so maybe I can attach a spreadsheet later if you really want to see the breakdown.
Anyhoo, B has to lose the second match in each outcome for it all to add up. There are other combinations, but the question forces us to use only the outcomes that match for the second game. So, I don’t think any additional information is needed.
The total number of games played by the three = 10 + 15 + 17 = 42; since each game has two players, there were actually 21 games played.
Since whoever sits out game N plays game N+1, no one can sit out two consecutive games. A sat out 11 of the 21 games; the only way this is possible without sitting out two in a row is if A sat out games 1, 3, …, 21, which means A lost all of games 2, 4, …, 20.
The number of times a player sits out = the number of times a player loses, if you don’t count game 21. B sat out 6, and C sat out 4, so B lost 6 games in the first 20, all to C (since A lost every game played), and C lost 4 games in the first 20, all to B. There is no way of knowing who won game 21.
I think I have a way of writing it so you can determine who won game 2:
Add the condition that the winner of the first game between any two players always had more wins in the games against each other.
This will result in
Since B has more wins against C, B won the first game between the two (otherwise C would start 1-0 against B but end up 4-6), which was the first game played (as has already been shown), so game 2 was A-B, which B won, since A lost every game played.
As shown above, A lost all even numbered games including game 2.
As to B and C, of the odd numbered games 1 through 19 inclusive, B won 4 times and C won six times, then whoever won the preceding odd game wins the even game against A. (We also don’t know who won the 21st game.) Any order of those victories works within the information given.
My suggestion for an additional constraint would be:
C mentions that if he’s playing, he always wins prime numbered games… except for the unlucky number 13.
Or you could change the games played totals to 10 14 18 and remove the unlucky 13 business.
If we’re getting mathsie, I think I’d prefer, B says “I always win games that are multiples of three”.
In my “ideal” I see the extra clue(s) not mentioning numbers directly (e.g. “C had always wanted red hair like B used to have before he started dating A” type-of-gumf).
If I read right what you are saying, you are saying that, in the original puzzle, B must lose the second game. If this is what you are saying, it’s an incorrect answer.
Besides, what The Great Unwashed wants to know is if it is possible to determine definitively who WON game two, and if not, what added info would make that determination possible.
As the puzzle is initially posited, A loses every game he plays, those games being all the even numbered games. The game is played 21 times. Thus, game 1 is between B and C. Whoever loses sits out a game, then comes back in and plays. The winner plays A, as well as the subsequent game after beating A. The last game, game #21, is also between B and C.
As Some Call Me… Tim has pointed out, to determine who wins the first game requires extra knowledge. I will point out that in essence, it becomes a whole different puzzle, solvable only if you have figured out what happened to poor old A.
Sorry, didn’t see this first time around: A cannot play in the first game, or else they will have played (at least) 11 times (contrary to what we are told).
I kept trying all possible outcomes and finding a way to add up those combinations of wins & losses to figure out the exact playing order. As the solutionpoints out, this is unnecessary if you consider the minimum number of games possible for any player to play and lose them all. That would mean if player X started the 1st game and lost, he’d be playing in all the odd-numbered games, which amounts to 11.
A played 10 times, so consider that he sat out the first game, and played all the even-numbered games. That amounts to 10. Since he lost them all, that includes the 2nd game, so the answer is A.
I wanted to make the puzzle better – it falls a bit flat the moment you notice that 10 is less than half of 21 – I’d like that to be the beginning of the solution, not all of it.
[spoiler]It took me a minute or so to figure out A and B played four games, A and C played six games, and B and C played eleven games. This is the only combination where A played ten games, B played fifteen, and C played seventeen.
You could never have the same pair of players playing twice in a row. So the eleven games between B and C had to be all the odd numbered games. And if B and C were playing in the third game, then A must have lost the second game.[/spoiler]
