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The team of A and B can increase the chance from ¼ to ½ by consistently taking their output from step 2 and using it as the guess for step 3. Same thing if they both agree to guess the inverse of step 2.
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Humm let me see if I get this right - Assuming everybody is honest The chance of A getting the guess right is 0.5, same for B, so the joint probability should be 0.25 (or 1/4) You're out of 2 dollars (3 - the 1 you get) for something with a chance of 1/4 (in the other 3/4 you're being payed $1) SO Looks like it's worth playing. Expected losses: 2 * 1/4 Edit: see comment below Edit 2, see the other comment below where…
> You're out of 3 dollars for something with a chance of 1/4 (in the other 3/4 you're being payed $1) You're out $2 (since you get the $1 from the other players at the beginning, no matter what)
-------------------------- The fact that there is no communication ensures that whether A is right and whether B is right are completely indpendent. Conceptually, predicting a coin flip is like making a statement "1" or "0" which will then be XOR'd with a 1 or 0 from a securely, randomly, uniformly generated OTP of which no copies exist. In other words, the plaintext is immediately lost forever and you just have the ciphertext.
As a result of this, we must truly consider that A being right is a 50/50 proposition. It is also indpendent of B being right.
Thus we have the following four cases:
A right, B right - Result value * Percent chance = EV
0, 0 = +1 * 0.25 = +0.25
0, 1 = +1 * 0.25 = +0.25
1, 0 = +1 * 0.25 = +0.25
1, 1 = -2 * 0.25 = -0.5
------------------------------- sum of above: +0.25. Therefore, as long as each team member is independently predicting their own coin (which means that their prediciton will be xor'd by a random bit) C should play this game long-term.
(The values of +1 is because each round starts with the team giving C +1. If at the end of the roudn C must return 3 then this is +1 -3 = -2 for the round.)
Now here is another interesting question. What if under the same conditions A and B both try to predict C's coin toss, of which there is only one? Should C now play? Here is the answer is: "No", because A can predict heads, B can predict heads, and then it looks like this: A right, B right - Result value * Percent chance = EV
0, 0 = +1 * 50% = 0.5
0, 1 = +1 * 0% = 0 } not possible
1, 0 = +1 * 0% = 0 }
1, 1 = -2 * 50% = -1
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-0.5
In this case, C should not play. This is because in this case the events are not truly independent, there is a way to break the 25% 25% 25% 25% into 50% and 50% - namely by picking the same prediction together.
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Anyone find a solution? I have an idea but want to check.
The team of A and B can increase the chance from ¼ to ½ by consistently taking their output from step 2 and using it as the guess for step 3. Same thing if they both agree to guess the inverse of step 2.
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!?
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!?
It doesn't, you're simply wrong. See my other comments.
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.
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!?
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?"
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!?
>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…