Probability BrainTeaser (toughie)

[chaosuk]

Now no one-up-man-ship here guys, don't spoil it for other folk if you simply know the answer. Good luck.

Every year, after the Christmas social at the Woodenhead Bridge Club, a party game is played. Four playing cards are randomly selected from a big box of used cards and stuck on the forehead of each of the four players sitting round a bridge table. Each person can only see the cards on the foreheads of the other three players. They each then attempt to guess correctly the colour, red or black, of their own cards. They do this by writing down their guesses on a piece of paper, so the others can't see what they've written. Up to three of them may also elect to write 'pass'. If, when the papers are revealed, one or more players have guessed the colour of their cards correctly the table gets a bottle of champagne. Any wrong guess, however, means they lose the prize. They may agree a strategy in advance but they mustn't speak or signal after the cards have been placed. They have a 50% chance of winning when one of the players guesses at random and the other three pass. How can they improve this to 75%?

[RocketManJames]

When you say that they cannot speak or signal to each other, does this mean that they must all write down their guesses simultaneously?

-RMJ

[LetYouDown]

My first thought is that you need to have someone automatically right pass no matter what, effectively making this a three person game with "Let's Make a Deal" undertones. I'll look into it on my last break. Interesting problem.

[fnord_too]

I am assuming the following:

1. The box is sufficiently full of cards, and the distribution of colors is unknown such that seeing the cards of your team mates does give you enough information to improve to any significant level the chance of guessing your own colour.

2. No information may be transmitted in any way between the players. For example, there can be no timing of writing an answer by any player to reveal information.

My first smart ass idea is to play twice. But I think I have an idea:
There is one way the cards can be all red and one way they can be all black.
There are 8 ways to have 3 of one color and 1 of the other.
There are 6 ways to have two of each color.

So...
2 ways to have 4 of one color
8 ways to have a 3-1 split
6 ways to have a 2-2 split

So, everyone except south passes. If south sees a 2-1 split, he writes the 1. If he sees 3 of the same color, he guesses at random.
So, for the 2 4-0 cases, he wins once.
For the 6 2-2 splits he wins all 6.
For the 8 3-1, he loses the 6 he does not have the 1 and wins one of the two he does hace the one.

That is 50/50 again. I have typed every strategy I can think of and erased it because I keep coming back to 50/50 if there is no information passed in any way. After wracking my brain, I do not think this is possible without some way to pass information. I think, given my assumptions, I could probably prove that.

Note: If south say looked at the other three cards, and if he saw a 2-1 split, which he will 12 times out of 16, then wrote the 1, which would be correct 6 times in 12 within 10 seconds, that gets 6 right answers and 6 wrong. If he writes something in that time, everone else writes 'pass' if he doesn't east knows he has the same color as west and north, writes that down and everyone else writes pass. That gets you 62.5% correct BUT is cheating by my assumptions since information is passed. (Here the signal is writing or not writing an answer in a given time frame). You can get 100% using that or similar singalling methods.

Can you PM me the answer if you are not ready to post it, I really don't see how this can be done given my assumptions.

[tallstack]

[ QUOTE ]
does this mean that they must all write down their guesses simultaneously?


[/ QUOTE ]

I think that this is an important question. If they do not have to answer simultaneously then I am pretty sure they can be correct 100% of the time.

So, I am guessing that they do have to be simultaneous.

Dave

[Koss]

Assuming you will always have 3 pass and one person guess I don't see any way it can be done. It's like seeing the results of 3 coinflips and using those to predict the 4th. No way. I just don't get how seeing the other 3 cards helps here.

Each guess is 50/50, and having multiple people guess just cuts the odds of them both being right in half. So to get a 75% probability that everyone who guesses will guess right seems impossible!

[chaosuk]

Guys it aint impossible, there is a very logical/not lateral explanation. I'm out to dinner and I'll post a clue tonight if folk want it. It got the better of me, I got stuck on the premise of the inevitable 50:50 too.
It isn't a trick.

[chaosuk]

yes both your assumptions are correct.
I said not lateral, but thats a little unfair. You probably need to go a little lateral then logical.
The proper clue tonight will help.

[LetYouDown]

Spoiler below in white (I think, heh)

<font color="white">Take one person out of the equation entirely...just determine who it is beforehand.

Of the other three, there's 8 possible combinations:

Red Red Red
Red Red Blue
Red Blue Red
Red Blue Blue
Blue Blue Blue
Blue Blue Red
Blue Red Blue
Blue Red Red

The 75% jumps out because 75% of these have two of one color and one of another. So for each person...if they see Red, Red or Blue, Blue...they write down the other color. If they see a mix of both, they pass.

What they'd write (respectively)...

Blue Blue Blue (lose)
Pass Pass Blue (win)
Pass Blue Pass (win)
Red Pass Pass (win)
Red Red Red (lose)
Pass Pass Red (win)
Pass Red Pass (win)
Blue Pass Pass (win)

I tried to figure this out where all 4 people have to be involved, but I'm hitting a stumbling block and I'm not sure it can be done. Either way, the answer still doesn't break the rules.</font>

[chaosuk]

Well I'm still here...

You got it, very well done. Your 'first thought' was the leap needed.