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.

[Thythe]

For some reason I don't think this is correct. Maybe I'm an idiot. More thoughts later....

[fnord_too]

Ok, here is my last line of attack:
S and E are the "players".
They each can see N and W.
Half the time N and W will have the same color card and half they wont.
The half they don't, E knows that S sees a 2-1 distribution, and S knows the same about E.
There are a total of 12 ways S can see a 2-1 distribution: 6 2-2 splits and 6 3-1 splits.
Given that N and W have different cards, we can calculate the conditional...
After a lot of typing I got back to this does not help.

Bah!

Ok, one more attack I can think of involves answering or not answering based on the two colors to your immediate left. Say if they are the same, you pass, otherwise you do something. First, this loses two ways for the monotone boards.

With the 3-1 split, only 1 person will see RB (or BR).
With a 2-2 split we have RRBB, RBRB as combinations. In the later, everyone sees two varying ones. In the former, 2 see varying ones.

So
With 3-1, the one person goes with the same as the 2-1 and we win 8 times.
If in the 2-2 the person goes with the same as the 2-1 and we lose 6 times and we are back to 50/50.

BAH!

Which of the following is not true?

1. At least one person has to write a non pass answer
2. The only information that person has is what color the other three cards are.
3. Here is how someone may end up writing a color:
a. Fixed before cards are drawn, and color is chose deteministically based on what writer sees.
b. Writer writes or does not write based on what they see.
4. The arranging of colors makes no difference. That is RRRB = RRBR = RBRR = BRRR for all intents and purposes. (I have a hard time not seeing all the 3-1, 2-2 and 4-0 cases being isomorphic.)

I keep looking at conditional probability and keep coming back to nothing helps. If the above are all true, there is no way I think to get about 50/50 (that is, I am fairly certain I can prove it. Of course your clever answer will probably make me slap my head, but I just don't see how this is possible.)

[fnord_too]

Nice!

Thank you for saving my afternoon.

[JoshuaD]

[ QUOTE ]
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>

[/ QUOTE ]

Doesn't that not work? Something with not being able to back pedal probability like that?

[fnord_too]

[ QUOTE ]


Doesn't that not work?

[/ QUOTE ]

That question hurts my brain. Are you asking "Does that work?"

I think it does. The guy iterated through all possibilities, so I am pretty sure he is correct.

[LetYouDown]

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

[/ QUOTE ]
Thanks. Good problem...wish we had more like this. I need something to get my brain working throughout the day.

[PrayingMantis]

[ QUOTE ]
For some reason I don't think this is correct. Maybe I'm an idiot. More thoughts later....


[/ QUOTE ]

I'm with you. It looks like an absurd anwer, with some backward probability in it. But maybe I'm an idiot too, and I don't have enough time now to get into it.

[Gabe]

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.

[Koss]

I threw together a quick computer simulation of the supposedly correct answer and I keep getting 50%. I'm looking for any possible errors in my code but I can't find any. If anyone would like I can send them my code, I wrote it in VBA because that's all I've got here at work.

edit: Scratch that. I found the error. It works, and is now returning a 75% win rate. Looks like the solution is correct.

[LetYouDown]

LOL, didn't see the edit until after I wrote my own in C#. Oh well... ::delete::

[LetYouDown]

[ QUOTE ]
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.

[/ QUOTE ]
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 &amp; 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]

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

[/ QUOTE ]

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]

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

[/ QUOTE ]
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]

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

[/ QUOTE ]

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.

[/ QUOTE ]

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.

[bighomage]

[ QUOTE ]
Cool.

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

PairTheBoard

[/ QUOTE ]

Ya, actually, the first time I was shown this problem, the professor asked if with more than 4 people, could the table be even more sure of winning? I answered no, but he claimed there was, and launched into an explanation, but I was zoned out and missed most of it. I'm very useful, as you can see. But anyway, according to him, the table's chance of winning should go up as you add more people. Anyone have the math?