20150702, 13:23  #1 
Aug 2002
3·2,837 Posts 
July 2015

20150703, 23:42  #2 
May 2013
East. Always East.
6BF_{16} Posts 
What terrible wording. "Please provide your answer as a list of 216 digits, 16 for a guess and 0 for illegal cases"
What on earth does that even mean? In any kind of graded school work a person would get wrecked for wording it like that. Sure they explain it later but there must be a smarter way to explain it the first time. The game isn't explained much better. It requires a lot of extra thinking and assumptions and I call bullshit on anyone saying that it's deliberately clouded. Clearly a player should not be able guess one of the numbers they see because that would make it impossible for one of the others to win (because they aren't allowed to guess the same number twice). I'm going to assume this is one of the hidden rules... The game probably does not loop back around for the same reason. Although it could go until nobody can guess. Does a player benefit from others losing? Does a player benefit from others winning? Do the players know if a previous guess was right or wrong? This looks like a shitty version of classic examples of game theory. 
20150703, 23:56  #3 
May 2013
East. Always East.
11×157 Posts 
As an example let's have:
2 4 5 6 The guy with 6 has to guess first. Obviously he has no guaranteed success since his guesses are 1 3 or 6. At this point it's just 33.3%. There's no way to deduce his own number from the others if it's all random. He guesses 1. Mr 5 (probably, if the rules are nonstupid) knows that his own hat cannot be 1. That leaves 3 5 and 6. He guesses 3. Mr 4 knows that his own hat cannot be 1 or 3. That leaves 4 5 and 6. He might guess 4. If he knows that guesses 1 and 3 are incorrect, then he knows the other two bozos have 5 and 6, so he would be certain that his hat says 4. (If the game ends when someone guesses correctly then that's enough: the game continuing is proof that a guess was incorrect) From the start of the game, the first player to guess can only eliminate 3 of 6 guesses. The second player can only eliminate 2 of 6 guesses, but can eliminate a third based on the first player's guess. The third eliminates 1 of 6, but eliminates two based on the previous players' guesses. Without some systematic strategy among the players I see no way for the second player to have guessed successfully. Again, does the first player WANT the second player to guess correctly? The problem statement SAYS that all other players CAN guarantee a correct guess but then asks for a strategy for the first player. Is this a strategy to HELP or HINDER the second guesser?? And without communicating among themselves, how does that even help? There's another version of this where there are three players with coloured hats. There are two red and two blue; one for each, and then a spare which nobody can see. The players win if ANYONE guesses correctly which colour hat is on their head. The idea of this game is that it is trivial for the guy at the back if he sees two similarcoloured hats: "Hmm there's two blue in front of me, therefore mine is clearly red". It's not so easy if he sees one of each, but if the second guy sees that the first isn't immediately jumping to a conclusion, then he should know: "The guy in front of me is red and the guy behind me would easily have guessed that his was blue if mine was also red. Because he hasn't, that means he doesn't know, so mine must be blue!" That one is clever and takes some thinking to understand. I just can't see how a similar strategy applies here... Last fiddled with by TheMawn on 20150704 at 00:10 
20150704, 01:25  #4  
"Forget I exist"
Jul 2009
Dumbassville
2^{4}·3·5^{2}·7 Posts 
Quote:
z+x+y=14a assume player one guesses 5 9(x+y)=z so player two knowing the two in front of him can safely say his number is such and such based on the knowledge about 14 and player 1's guess. though I guess this is still a probabilistic way at best. so since it's asking for a function I'm guessing f(a,x,y) means 14(a,x,y)=z is my best guess but I'm stupid. and I still don't even understand the answer format other than involving w^3 digits for 1w. 

20150704, 07:13  #5  
"Robert Gerbicz"
Oct 2005
Hungary
3^{2}×5^{2}×7 Posts 
Quote:
Quote:
Assume the following: they are clever enough to make sure that the first three person can make a correct guess, for the fourth person you can't make sure this (in some cases he will make a correct guess, in others a wrong). To achieve this, before they stand in a line they (all 4 person) are agreeing in a strategy what they will follow in the game, that is the meaning of the "Our challenge this month is to show that all the rest can." sentence for me. When standing in line they can't make any further communications, just the guessing. And they hear each others guess. Interesting question, for my strategy I don't need it. (so for example we don't know in the game that the fourth person made a correct guess or not.) 

20150704, 07:23  #6  
"Robert Gerbicz"
Oct 2005
Hungary
3^{2}·5^{2}·7 Posts 
Quote:


20150706, 23:09  #7 
May 2013
East. Always East.
11·157 Posts 
But that's just a bunch more assumptions WE have to make that they could just make for us. A problem is difficult because it involves complex lines of thought, not because you can't parse the words!
"The four players may discuss a strategy beforehand, and their goal is to ensure that the second player to guess is always guaranteed to be correct. When the game begins, the players are allowed no communication except for saying their guess aloud." And how easy was that? I wrote five words, erased them, and then wrote that, all in one go with zero edits. Now I need to think about this problem. It's suddenly more interesting. 
20150707, 04:45  #8 
"Robert Gerbicz"
Oct 2005
Hungary
627_{16} Posts 
The problem needs slightly more, the guess of the first three in the line should be correct (so the second, the third and the fourth person in the guessing order). In other parts it is a correct interpretation of the problem.
Last fiddled with by R. Gerbicz on 20150707 at 04:46 
20150708, 04:44  #9 
May 2013
East. Always East.
11×157 Posts 
Right. It's not too difficult to make the second guesser be correct but the next two must also somehow be able. That's trickier.

20150708, 13:49  #10 
"Robert Gerbicz"
Oct 2005
Hungary
627_{16} Posts 
It is also important to not forget this condition:
"they need to guess the number on their own head, but they are not allowed to use any number that was previously guessed. " So they need to guess different number. Note that without this the problem would be easy (and I think that it is well known in mathematics). 
20150709, 15:53  #11 
"Robert Gerbicz"
Oct 2005
Hungary
627_{16} Posts 
There is an update on Ibm (we also discussed on this issue):
"Update (9/7, thanks James): The last person (first to guess) is helping the other players to win, and that they have agreed on a strategy in advance." And I was again the first solver of the puzzle. 
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