Is there a legitimate answer to this logic puzzle or is it impossible?

Question

Two contestants are in separate rooms. The host goes in each room and flips a coin. Both people guess if coin was heads or tails in the opposite room. If both are wrong, game is over. If at least one is right, we do another round. Contestants both win a billion dollars if game lasts 100 rounds. What strategy will make them winners?
They can speak to each other before the 1st round but can't communicate after game begins.

Puzzling · Accepted Answer

The first thing to clarify is that there are *two* coins being flipped and each person is guessing the state of the coin in the *other* room, not in their own room.

It sounds impossible since they can't see the other person's coin and we assume they can't even hear what the other person is guessing. There apparently seems to be no way to communicate the state of a coin from one room to the other room.

But the beauty is they don't have to both be right. They only need *one* of them to be right.

There are really two cases to consider.
- The coins in each room could be the same (TT or HH)
- The coins in each room could be different (TH or HT)

So a simple strategy is that person 1 could guess the SAME as the coin in their room, and person 2 could guess the opposite of the coin in their room. No matter what, one (and exactly one) of them will be right. But they could do this indefinitely.

Now, if they are worried that their strategy could be found out. They could instead agree to some other strategy. For example they could alternate who guesses the same and who guesses the opposite.

Or if they were worried that was too obvious (and assuming these are smart people that might know π to 100 decimal places), they could walk down the digits of π. An odd digit (e.g. 3 or 1) would mean person one guesses the same and person two guesses the opposite. And if the digit was even (e.g. 4) then they would switch the pattern.

If the obfuscation isn't necessary, then I would go with the simplest strategy so they don't accidentally lose track of the pattern and mess things up.

The following video shows an example of this in action. Here they were guessing whether a card was red or black, but it's essentially the same problem. Can you figure out what pattern they used?

Dixon · Answer

Regardless of head and tails, all outcomes can be split into two sorts; both the same, or both different.
  
In which case, if one contestant always guesses they are both the same (based on his coin), and the other always guesses both are different (based on her coin), then one of them will be correct on any given turn.

llaffer · Answer

If they know the rules in advance and can plan before the game starts, all that they need to do is have one person call heads and the other person call tails on every coin flip.  One of them will be right every time.

Anonymous · Answer

One person always calls what they flip while the other person calls the opposite of what they flip I cannot remember the proof for this but I'm 95% sure I'm remembering the same riddle from years ago. So at the beginning when they can speak once to each other they must agree who will call what they flip and who will call the opposite of what they flip. I think that the proof is similar to the two guarded doors riddle and the guard who always tell the truth and the guard who always lies.

Is there a legitimate answer to this logic puzzle or is it impossible?

4 Answers