To answer the actual question: whether C should play this game depends on C's estimated probability of A and B coming up with a winning strategy. In the real world, there would be many people with whom I would play this game.
Coin Puzzle: Predict the Other's Coin
31–40 of 50 posts
Re: Coin Puzzle: Predict the Other's Coin
#32Dysfunctional teams never can agree. By that I mean, if A decides to use the strategy "guess what I flip"; B decides on the strategy "guess opposite what I flip". And vice versa. These teams can never win! So for dysfunctional teams, C can play the game no matter the payout.
[deleted]
Re: Coin Puzzle: Predict the Other's Coin
#33Dysfunctional teams never can agree. By that I mean, if A decides to use the strategy "guess what I flip"; B decides on the strategy "guess opposite what I flip". And vice versa. These teams can never win! So for dysfunctional teams, C can play the game no matter the payout.
Actually I've done the hash on this approach and this still results in a losing game for C.
Instead of HH and TT being winners with this approach, HT and TH become winners with this approach. Any combination of opposite or agreement will result in the same net odds.
The way for C to win is for A or B to treat it randomly.
A very interesting experiment would be to test how long it would take a random human subject A with no contact with a consistent B (but a memory for past results) to figure out how to win (in cases where they were told winning was possible, impossible, or neither).
Re: Coin Puzzle: Predict the Other's Coin
#34The question should really be rephrased to "Will C make money if he/she plays this game?" Maybe C doesn't like money, has too much of it but really only likes giving it out after coordinated acts of random chance. Then C would love this game regardless of A and B's strategy. Maybe C spends all of his/her free time at RPS tournaments, throwing singles at the winners like a rap video.
Re: Coin Puzzle: Predict the Other's Coin
#35Here is a bit harder version of the puzzle, with 100 players instead of 2.
100 prisoners are given either a blue or a red hat, at random. Each prisoner is told colors of every hat except their own, and has to make a guess about his hat with absolutely no communication. They win only if they all guess correctly. They can agree on a strategy beforehand. What is the optimal probability of success?
Re: Coin Puzzle: Predict the Other's Coin
#36If they both agree to use their own coins result as the guess of the OTHER person's coin, they should get it right 50% of the time instead of 25%. Possible tosses: TT win HT lose TH lose HH win Counter-intuitive though isn't it. Why does using your own result improve the odds of winning the game!?
Conversely, they can use a negation of their result. In this case they win given TH and HT and lose given TT and HH.
Re: Coin Puzzle: Predict the Other's Coin
#37If they both agree to use their own coins result as the guess of the OTHER person's coin, they should get it right 50% of the time instead of 25%. Possible tosses: TT win HT lose TH lose HH win Counter-intuitive though isn't it. Why does using your own result improve the odds of winning the game!?
>Why does using your own result improve the odds of winning the game!? I find the result very confusing as well. I suspect the source of this counterintuitiveness is that I over simplified the puzzle when I first read it. I simplified the puzzle to "A attempts to guess B's coin, and B attempts to guess A's coin", whereas in fact the true puzzle is "A and B together try to guess the total set of results". When A and B…
I'm not sure the conditional probability argument explains it. Specifically, given A=H, the probability of (H,H) is the same as the probability of (H,T) so that alone shouldn't inform your guess.
I believe the core mechanism at play here is that coordination eliminates the possibility of one player being right and the other being wrong, because one can partition the coin outcomes into "We have the same result" and "We have different results"; by agreeing in advance to only guess "we have the same result" every time you eliminate the "I'm right, you're wrong" and "You're right I'm wrong" failure modes (50% of the probability space under random guessing).
One could accomplish the same by always guessing the opposite of what you obtain (i.e. always bet on "we got different results" which again has a 50% probability).
This is related to the "guess your own hat's color" riddle:
"You and a friend are on a game show. The host sets you facing each other at a table, blindfolded. A hat, known to be either white or black, is placed on each of your heads. The host removes the blindfolds, and asks each of you to write down the color of your own hat without communicating with each other. If either of you guesses correctly, you both win a prize. How do you guarantee success by pre-communicating a strategy?"
The answer here uses the same partitioning into "either we're the same, or we're different". One friend agrees to always guess his own hat to be the same as the friend's, while the other always guesses that his hat is different.
Re: Coin Puzzle: Predict the Other's Coin
#38Dysfunctional teams never can agree. By that I mean, if A decides to use the strategy "guess what I flip"; B decides on the strategy "guess opposite what I flip". And vice versa. These teams can never win! So for dysfunctional teams, C can play the game no matter the payout.
Actually I've done the hash on this approach and this still results in a losing game for C. Instead of HH and TT being winners with this approach, HT and TH become winners with this approach. Any combination of opposite or agreement will result in the same net odds. The way for C to win is for A or B to treat it randomly. A very interesting experiment would be to test how long it would take a random human subject A w…
Yes, but my point was that one team member uses HH & TT, while the other team member uses HT & TH. This is always a loss for the team regardless of how the flips turn out.
Re: Coin Puzzle: Predict the Other's Coin
#39If they both agree to use their own coins result as the guess of the OTHER person's coin, they should get it right 50% of the time instead of 25%. Possible tosses: TT win HT lose TH lose HH win Counter-intuitive though isn't it. Why does using your own result improve the odds of winning the game!?
I was struggling with why this helps but the answer is that this strategy correlates their correct guesses. They are either both wrong or both right.
Re: Coin Puzzle: Predict the Other's Coin
#40Here is a bit harder version of the puzzle, with 100 players instead of 2. 100 prisoners are given either a blue or a red hat, at random. Each prisoner is told colors of every hat except their own, and has to make a guess about his hat with absolutely no communication. They win only if they all guess correctly. They can agree on a strategy beforehand. What is the optimal probability of success?
Each should guess whatever would make an even number of red hats. That makes it a 50% chance they're all wrong, and a 50% chance they're all correct -- thus consolidating all their "rightness" into a single block of probability.