Coin Puzzle: Predict the Other's Coin

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%.<p>Possible tosses:<p>TT win<p>HT lose<p>TH lose<p>HH win<p>Counter-intuitive though isn't it.<p><i>Why does using your own result improve the odds of winning the game!?</i>

Humm let me see if I get this right<p>- Assuming everybody is honest<p>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)<p>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<p>Looks like it's worth playing.<p>Expected losses: 2 * 1/4 &#60; Expected wins: 1 * 3/4<p>Edit: see comment below<p>Edit 2, see the other comment below where the probability of prediction can be of 1/2, in this case: Expected losses: 2* 1/2, wins: 1/2 so you should not play

Here is a bit harder version of the puzzle, with 100 players instead of 2.<p>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 <i>all</i> guess correctly. They can agree on a strategy beforehand. What is the optimal probability of success?

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.<p>Same thing if they both agree to guess the inverse of step 2.

Spoiler. The below contains an exact response<p>-------------------------- 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.<p>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.<p>Thus we have the following four cases:<p>A right, B right - Result value * Percent chance = EV<p>0, 0 = +1 * 0.25 = +0.25<p>0, 1 = +1 * 0.25 = +0.25<p>1, 0 = +1 * 0.25 = +0.25<p>1, 1 = -2 * 0.25 = -0.5<p>------------------------------- 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.<p>(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.)<p>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<p>0, 0 = +1 * 50% = 0.5<p>0, 1 = +1 * 0% = 0 } not possible<p>1, 0 = +1 * 0% = 0 }<p>1, 1 = -2 * 50% = -1<p>-------------------<p>-0.5<p>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.<p>--------------------------

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.

Both A and B guess the same as the results of their own toss. Their answers are correct if the tosses are HH and TT, 50% chance.<p>So no, C should not play this game.

Dysfunctional 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.

Maybe you could do even better than the obvious "guess the same as your coinflip" strategy, by using quantum tricks.

The 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.

Anyone find a solution? I have an idea but want to check.

Hmm. If A and B are unwitting investors, and C is a quant with inside information, OF COURSE C should take this "bet".<p>It's not like C would ever everrrr....Lie....<p>And when G(overnment) comes around to fix this evil fraud, they fine for 10% of the profits. Everybody but A and B make out like a bandit. Sound like '09, doesn't it?<p>(tongue in cheek, only because there were multiple vald analysees of this question elsewhere in the comments.)