Probability BrainTeaser (toughie)

[LetYouDown]

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I don’t have time to work it out exactly, but it must be something like this: The first two players pass every time. The next to the last player guesses unless the last player’s card is red.

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How would the last player know whether to guess or not?

[chaosuk]

The key to the sense of this solution is the redundant wrong-guesses.

Each of the three active players guesses half the time and and of those guesses only half of those are right. Since 3 players are guessing half the time there is clearly overlap - some are guessing at the same time. The cleverness of the solution lies in the fact that when a player guesses correctly, s/he is the only one guessing. But when he's guessing incorrectly they are all guessing incorrectly. Each player is guessing 1 in 2 & so correctly 1 in 4. Since they are correct guesses are indepndent: their strategy is correct 3 x 1/4.

[OrangeKing]

Very clever. The overall correct guessing percentage for the team is still only 50%; each individual only guesses correctly 50% of the time; and yet the team actually wins 75% of the time, because their wrong guesses are clustered in only two of the eight possibilities.

[PrayingMantis]

Yes, I see that now. Very nice. At first glance this answer looked like a confusion of conditional and non-conditional probabilities, but from a closer look it is very elegant and does get 75% success in this particular game.

[Gabe]

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How would the last player know whether to guess or not?

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Because he's last. I'm assuming that it goes around the table and each player decides in turn to fold or guess. With 4 players, they should have a strategy where they are better than 15-1 favorites to beat the house, if the deck isn't stacked and the deck is smaller than infinity. Each player passes unless one or more of the remaining players has a red card, if none of them do, they "guess" red. If they all pass the last player "guesses" red.

[probman]

I didn't see the solution posted earlier until after I posted. Here it is anyway.


Here's a simple solution. I can achieve 75% success without using all four players, but simply three. Here is the strategy. I assume that all four players must make their decision simultaneously.

To begin with, one player is chosen to always pass. This player will be the inactive one and the other three are designated active players.

For the other three, the rule is simple. If an active player sees the same color card on the two other active players he says his card is the opposite color. For example, if he sees that the other two players are wearing red cards he says his card color is black. Otherwise he passes.

Analysis:
Since there are three active players there will always be two of them with the same color card. Therefore at least one player will aways guess his card color. The only way for the strategy to fail is if all three have the same color card. That will only happen 2/8 times. Hence a win rate of 75%.

[PairTheBoard]

Cool.

Seems like there should be some way to improve on the solution by including all 4 players. Maybe not. Anybody know?

PairTheBoard

[PairTheBoard]

Ok how about this. Suppose the rules are that all 4 players Must guess their card. If done randomly they only get the champaigne 1 time in 16. How can they make their guesses to improve that to 50%?

16 possibilities:

4 same color : 2 ways
3 same color : 8 ways
2 same color : 6 ways

If you see 3 the same color guess opposite. If you see 2-1 guess the same as the 2.

Or doing just the opposite works as well.

PairTheBoard

[LetYouDown]

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I'm assuming that it goes around the table and each player decides in turn to fold or guess.

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That's an assumption I don't think you can make. I think that would fall under the category of "signaling" as well. I assumed no direct or indirect communication.

[HesseJam]

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How would the last player know whether to guess or not?

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Because he's last. I'm assuming that it goes around the table and each player decides in turn to fold or guess. With 4 players, they should have a strategy where they are better than 15-1 favorites to beat the house, if the deck isn't stacked and the deck is smaller than infinity. Each player passes unless one or more of the remaining players has a red card, if none of them do, they "guess" red. If they all pass the last player "guesses" red.

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If you let only play three guys, there is no combination possible, where they all three see a 1-1 distribution. Thus, at least one will guess.