70 replies to this topic

[Egghead]

Remember:
5. The players do not cheat.

To put the problem another way, all players of the team are housed in separate rooms. Everyone is informed about the colour of all others' caps by an arbiter. There is absolutely no possibility of learning more information than those colours.

Mirror in the room, not cheating; just not handled by the rules.
Arbiter gives away the cap's color, not a team member; so again, not handled by the rules.

The statement "There is absolutely no possibility of learning more information than those colours," makes the game impossible to win without wild guessing; so, either the statement is false and one can deduce the color of his/her own cap or it is true and there is no point in trying to figure out the game as it is always a 1/3 chance of winning (possibilities are correct, incorrect, abstain).

[amateur]

How about this strategy

Spoiler for

Let's make it easy on ourselves and call the state of the hats 0 and 1.

Every participant look and sum up all the hat state. The sum should be between 0 and 14.

If a participant see a sum that is odd, abstain from guessing.

If a participant see a sum that is 0, 2, 4, or 6, guess that his hat state is 1.

If a participant see a sum that is 8, 10, 12, or 14, guess that his hat state is 0.

This way, the winning rate is about 3/5.

Spoiler for nice try, but can you explain

how you got the probability of 3/5?

[Kornrade]

How about this strategy

Spoiler for

Let's make it easy on ourselves and call the state of the hats 0 and 1.

Every participant look and sum up all the hat state. The sum should be between 0 and 14.

If a participant see a sum that is odd, abstain from guessing.

If a participant see a sum that is 0, 2, 4, or 6, guess that his hat state is 1.

If a participant see a sum that is 8, 10, 12, or 14, guess that his hat state is 0.

This way, the winning rate is about 3/5.

This strategy may give a "little more than 50%" winning chance, but it is still far from the optimal solution.

Edited by Kornrade, 07 December 2010 - 09:33 PM.

[Kornrade]

Mirror in the room, not cheating; just not handled by the rules.
Arbiter gives away the cap's color, not a team member; so again, not handled by the rules.

The statement "There is absolutely no possibility of learning more information than those colours," makes the game impossible to win without wild guessing; so, either the statement is false and one can deduce the color of his/her own cap or it is true and there is no point in trying to figure out the game as it is always a 1/3 chance of winning (possibilities are correct, incorrect, abstain).

No mirror in the room, no other sources of indirect information
The arbiter communicates which team member was which colour, just as if they stood in the circle and everyone could see everyone else.

"makes the game impossible to win without wild guessing"
Whoever chooses to write RED/BLUE has a 50% chance of being wrong because of the statistical independence of coin toss events (that was already established). I never said that the team wins 100%!

[bushindo]

This strategy may give a "little more than 50%" winning chance, but it is still far from the optimal solution.

Well, I see what you're saying.

Spoiler for

Modify the decision table as below. Only the bottom half is listed, the top part is similar

Sum       Decision
...
7         Abstain
8         Guess 0
9         Guess 1
10        Abstain
11        Guess 0
12        Guess 1
13        Abstain
14        Guess 0

The winning change is now 2/3. Is that still far from optimal?

Edited by bushindo, 07 December 2010 - 10:02 PM.

[Kornrade]

Yes...

Spoiler for Hint

The winning percentage for the optimal strategy is well over 70%

[jagdmc]

Spoiler for My guess

If the distribution of the caps is entirely random, then the probability of having a red/blue cap is ALWAYS 0.5 regardless of the other teammembers' color. Hence, the best strategy is for all but one to write down "---" while the 15th player will either write "blue" or "red", whatever he likes most. That team will have a probability of 0.5 of winning the game.
It doesn't matter if a player sees in his teammates that there are 14 blue caps, if it was determined by random, there is a possibility of having the 15th blue cap on my head.
The distribution of reds and blues will be even only after tossing an infinite number of times a fair coin.

[puzzled_again]

Well, I see what you're saying.

Spoiler for

Modify the decision table as below. Only the bottom half is listed, the top part is similar

Sum       Decision
...
7         Abstain
8         Guess 0
9         Guess 1
10        Abstain
11        Guess 0
12        Guess 1
13        Abstain
14        Guess 0

The winning change is now 2/3. Is that still far from optimal?

Spoiler for

I agree that there is only a 50% chance that a cap will be blue/red regardless of what colour the other caps are, but Bushindo's algorithms are correct in that, the above decision table will bring the probability of winning to 10/16...
It works out
all red hats - lose
1 blue hat - win
2 blue hats - win
3 blue hats - lose
4 blue hats - win
5 blue hats - win
6 blue hats - lose etc...........

[phillip1882]

over 70%?

Spoiler for i don't see how that's possible

with bushido's algorithm you have the following states.
let's make red 0 and blue 1.
all red:
everyone guesses blue, lose
1 blue:
the person who is blue will guess blue, all other abstain. win.
2 blues:
both blues will abstain, all reds will guess blue. lose.
3 blues:
all blues will guess blue, all reds abstain, win
4 blues:
all reds guess blue all blues abstain, lose.
5 blues:
all reds abstain, all blues guess blue, win.
etc.
seems pretty 50/50 to me.

[puzzled_again]

over 70%?

Spoiler for i don't see how that's possible

with bushido's algorithm you have the following states.
let's make red 0 and blue 1.
all red:
everyone guesses blue, lose
1 blue:
the person who is blue will guess blue, all other abstain. win.
2 blues:
both blues will abstain, all reds will guess blue. lose.
3 blues:
all blues will guess blue, all reds abstain, win
4 blues:
all reds guess blue all blues abstain, lose.
5 blues:
all reds abstain, all blues guess blue, win.
etc.
seems pretty 50/50 to me.

Spoiler for

10/16 is only 62.5%, so if the answer is "over 70%" then it can't be the answer.. but this is how I read the algorithm..
decision 0 = red, decision 1 = blue

so extrapolating from the table (as only the last half is shown)..
Sum Decision
0 1 (blue)
1 - (abstain)
2 0 (red)
3 1 (blue)
4 -
5 0 (red)
etc....

So, if there were two blue hats, each of the blue hats would abstain BUT the red hat users would see 2 blue bats and thus vote red not blue so it would be a win, continuing this through the 16 possible scenarios gives 10 scenarios where the team would win