3 ms·

If 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 tos

by hmexx 14y ago

If 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!?

dennisgorelik 14y ago
It's surprising and also means that A and B can choose a strategy when they lose 100% of the time: - A always uses A's result as a guess for B's result. - B always uses opposite to B's result as a guess for A's result. They would always lose then.
datdatruth 14y ago
It doesn't, you're simply wrong. See my other comments.
- Derander 14y ago
  Think of it like this: What is the probability that two coins come up with the same face value? 0.5 If we use the guessing scheme where each player guesses his own guess as the result of the other player's coin, then players A and B win if their coins are the same. Their coins are the same with probability 0.5. Thus, Expectation[game] = Expectation[Game|A & B Win] * P(A & B win) + Expectation[Game|A & B Lose] * P(A & B Lose) = 1 * .5 + -2 * .5 = -1, so C should not play. If the players guessed randomly then the expectation of the game would be as you say it is.
- theonewolf 14y ago
  If you want to say someone is wrong, it would be better to provide an explanation of exactly why they're wrong rather than point everyone elsewhere and make then figure out the hole in the logic. Point out the hole directly.
gabemart 14y ago
>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 guess randomly, they attempt to guess the total set without using the information they have available to them. When A and B both guess the results of their own coin, they use the information they have regarding the set of results (i.e. the results of their own coin) to reduce the problem space and increase their chances of success. Clearly, if you have heads, you know the chances of the set of results being heads-heads are much higher than choosing a random set of results. In other words, the proposition for A is not "What are the odds of B having heads given that you have heads?" but rather "What are the odds of the set of results being heads-heads given that you have heads?"
- hmexx 14y ago
  Nicely broken down!
- _dps 14y ago
  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.
sold 14y ago
Conversely, they can use a negation of their result. In this case they win given TH and HT and lose given TT and HH.
josephlord 14y ago
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.

by hmexx 14y ago

dennisgorelik 14y ago

It's surprising and also means that A and B can choose a strategy when they lose 100% of the time: - A always uses A's result as a guess for B's result. - B always uses opposite to B's result as a guess for A's result. They would always lose then.

datdatruth 14y ago

It doesn't, you're simply wrong. See my other comments.

Derander 14y ago
Think of it like this: What is the probability that two coins come up with the same face value? 0.5 If we use the guessing scheme where each player guesses his own guess as the result of the other player's coin, then players A and B win if their coins are the same. Their coins are the same with probability 0.5. Thus, Expectation[game] = Expectation[Game|A & B Win] * P(A & B win) + Expectation[Game|A & B Lose] * P(A & B Lose) = 1 * .5 + -2 * .5 = -1, so C should not play. If the players guessed randomly then the expectation of the game would be as you say it is.
theonewolf 14y ago
If you want to say someone is wrong, it would be better to provide an explanation of exactly why they're wrong rather than point everyone elsewhere and make then figure out the hole in the logic. Point out the hole directly.

gabemart 14y ago

>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 guess randomly, they attempt to guess the total set without using the information they have available to them. When A and B both guess the results of their own coin, they use the information they have regarding the set of results (i.e. the results of their own coin) to reduce the problem space and increase their chances of success. Clearly, if you have heads, you know the chances of the set of results being heads-heads are much higher than choosing a random set of results. In other words, the proposition for A is not "What are the odds of B having heads given that you have heads?" but rather "What are the odds of the set of results being heads-heads given that you have heads?"

hmexx 14y ago
Nicely broken down!
_dps 14y ago
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.

sold 14y ago

Conversely, they can use a negation of their result. In this case they win given TH and HT and lose given TT and HH.

josephlord 14y ago

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.