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This is my favourite brain teaser, and one of the most difficult that I know. I apologize if it has been posted before and I missed it.

A team of 15 takes part in a contest. They know each other and all the rules of the contest, and they have time to make their strategy before the actual contest starts.

The rules of the contest are:

1. The players sit in a circle, so that everybody sees everybody. Then the players are blindfolded.

2. On each player's head, either a red or a blue cap is placed. The colour of the cap is determined completely arbitrarily, e.g. by coin toss.

3. The blindfold is removed, and everybody has to write on a piece of paper the answer to the question "What is the colour of your cap?". The answer must be "RED", "BLUE" or "---". Everybody has to write down their answer at the same time.

4. Everybody shows the paper at the same time.

5. The players do not cheat. (They do not exchange information through words, utterances, signs, facial expressions, delays in writing their answer, etc.)

The procedure of the contest is done only once. The team wins the contest if at least one answer is correct and no answer is incorrect. "---" is considered a neutral answer, neither right nor wrong.

Examples:

if team member #12 correctly answers RED, and everybody else answers "---", they win.

if all members answer correctly (RED/BLUE), except for member #12 (who said RED and had blue), they lose.

Which is the strategy with the biggest chance to win, and how high is that chance?

Have fun!

[P.S.] For those who know Romanian, this and many other brainteasers are to be found at www.chichitza.com and www.chichitza.com/forum

As for chess, a section in English is www.chichitza.com/chess.html . It contains puzzles and studies. Some studies may be more difficult for computers than for humans :D

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Posted · Report post

It seems all the respondents here are treating the coin toss as random event.

A fair coin toss would follow the binomial distribution for probability of a predefined pattern in successive tosses.

So, let's say what are the odds of seeing 7 reds and 7 blues? Coz that's the only 50-50 possibility for the chosen leader to guess wrong.

Whenever this balance is tilted in favor of one color, the leader should chose the other color, to account for balancing nature of binomial distribution.

"The colour of the cap is determined completely arbitrarily, e.g. by coin toss."

Coin toss is only an example of random event that may be used to determine the colour on each cap independently of the other colours. Thus, all configurations from RRR...R to BBB...B are equally probable.

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Posted (edited) · Report post

See next post...

Edited by Kornrade
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Posted · Report post

Key to a lot of these types of problems: Make it simple, then expand.

So consider all but 3 people abstain from answering.

These 3 can follow the simple strategy:

1. If the other two colors differ then abstain.

2. If the other two colors are the same, choose the opposite.

With 3 people you have 2^3=8 possibilities:

BBB GBB

BBG GBG

BGB GGB

BGG GGG

For 2/8 (BBB, and GGG) all three will guess incorrectly.

For the other 6, two will abstain and one will guess correctly.

There's your 75%.

Now, I started simple and picked 3. Maybe someone can build on this by take into people into account to raise the percentage?

Well done so far! :)

going from the trivial 50% to the "simple" 75%. Now start the hard part: how to extrapolate?

Key question: what is new in this strategy compared to all other strategies suggested?

Btw, the winning percentage is well over 75%, but I shall say no more :D

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Posted (edited) · Report post

Key to a lot of these types of problems: Make it simple, then expand.

So consider all but 3 people abstain from answering.

These 3 can follow the simple strategy:

1. If the other two colors differ then abstain.

2. If the other two colors are the same, choose the opposite.

With 3 people you have 2^3=8 possibilities:

BBB GBB

BBG GBG

BGB GGB

BGG GGG

For 2/8 (BBB, and GGG) all three will guess incorrectly.

For the other 6, two will abstain and one will guess correctly.

There's your 75%.

Now, I started simple and picked 3. Maybe someone can build on this by take into people into account to raise the percentage?

We have to be careful with how we calculate probabilities on this. If each individual cap colour was decided by a coin toss as per the OP, then these probabilities would work. However, Kornrade has since clarified that "all configurations from RRR...R to BBB...B are equally probable." If we start from this position, then the chance of any 3 caps having the same colour is actually 50%. If the chance of each distribution was based on binomial probabilities (i.e. in the coin toss scenario), then your result would be correct, as the chance of 3 caps of the same colour would only be 25%. Attaching a spreadsheet just to illustrate the difference:

Binomial vs uniform distribution.xls

Edited by rajat_magic
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Posted · Report post

Provided the user knows information of other 14 hats, wouldn't it be much safer to calculate probability using this information. In other words, what is the probability that both the children are girls when you already know one of them is a girl? Most people say it is 1/3 while few debate it would remain 1/2. For this puzzle I will go with the first option, as otherwise the best strategy will be - all except one predesignated person taking a guess with 1/2 probability. Not much of a strategy if you ask me.

:) that wasn't what I proposed. I understand conditional probability, but this puzzle doesn't seem so simple. For e.g, given that I (one of the 15 members) see a specific pattern of colors (say all Bs.."BBB...B"), what is the probability that MY hat is blue? It is still half, irrespective of the pattern. If I'm appointed the leader, and every one else will automatically abstain, the success rate can't get better than 0.5.

Now, if there is no appointed leader.. situation is different.

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Posted (edited) · Report post

We have to be careful with how we calculate probabilities on this. If each individual cap colour was decided by a coin toss as per the OP, then these probabilities would work. However, Kornrade has since clarified that "all configurations from RRR...R to BBB...B are equally probable." If we start from this position, then the chance of any 3 caps having the same colour is actually 50%. If the chance of each distribution was based on binomial probabilities (i.e. in the coin toss scenario), then your result would be correct, as the chance of 3 caps of the same colour would only be 25%. Attaching a spreadsheet just to illustrate the difference:

I think you misunderstood the name "configuration". A configuration is for me a succession of colours starting from a predesignated person and going anti-clockwise. Thus, a shift in the colours results in a different configuration. Configurations BRR..RRR and RRR..RRB are different and the probability of any particular configuration to appear is 1/2^15.

In your words, the binomial distribution is necessary for this problem. vinays84's suggestion yields 75% winning chances.

Edited by Kornrade
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Posted · Report post

Decide one Leader & Mark one Person for Leaders assistance who Sit in Front of Leader (Better) any suitable place (as everybody can see everybody).

After the Blinds will be removed from eyes, if leader's assistant will look up then leaders hat is Blue & if Leader's Assistant Looks Down to Earth then its Red.

Others will write "- -" and only one person (Leader) will write correct Exact Color. 100% Chances of Success.

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Posted (edited) · Report post

:duh:

Signaling of ANY kind is cheating!!!!

It is neither allowed, nor attempted by the players.

Edited by Kornrade
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I think you misunderstood the name "configuration". A configuration is for me a succession of colours starting from a predesignated person and going anti-clockwise. Thus, a shift in the colours results in a different configuration. Configurations BRR..RRR and RRR..RRB are different and the probability of any particular configuration to appear is 1/2^15.

In your words, the binomial distribution is necessary for this problem. vinays84's suggestion yields 75% winning chances.

Ah. Sorry. I understood configuration to mean "combination" in the permutation-combination sense.

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You've had me confused there. On one hand you are saying that the events are "arbitrary" while on the other you are talking about probability. :unsure:

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Posted (edited) · Report post

Three of them will suggest the following:

if any one of them see two identical colors..RR or BB

so he should write the opposite color

but if he see different colors he should abstain.

In this method we will obtain 75%

with three persons we have:

RRR

RRB

BRR

BBB

BBR

RBB

so in four ordes out of six we will have the right answer

where the other 12 will abstain.

Edited by wolfgang
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Three of them will suggest the following:

if any one of them see two identical colors..RR or BB

so he should write the opposite color

but if he see different colors he should abstain.

In this method we will obtain 75%

with three persons we have:

RRR

RRB

BRR

BBB

BBR

RBB

so in four ordes out of six we will have the right answer

where the other 12 will abstain.

It has been already suggested by vinays84 (btw, you forgot to include 2 cases RBR and BRB).

Now, what conclusions do you draw from that?

If so few people can do so much, imagine what you can do with all 15.

Key to a lot of these types of problems: Make it simple, then expand.

So consider all but 3 people abstain from answering.

These 3 can follow the simple strategy:

1. If the other two colors differ then abstain.

2. If the other two colors are the same, choose the opposite.

With 3 people you have 2^3=8 possibilities:

BBB GBB

BBG GBG

BGB GGB

BGG GGG

For 2/8 (BBB, and GGG) all three will guess incorrectly.

For the other 6, two will abstain and one will guess correctly.

There's your 75%.

Now, I started simple and picked 3. Maybe someone can build on this by take into people into account to raise the percentage?

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93.75% - all but 5 abstain, if one of the five sees four of the same color then choose the other color. or, if any one of the five sees three of one color and one of the other color then choose the same color as the single colored cap seen, otherwise abstain.

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Posted · Report post

Three of them will suggest the following:

if any one of them see two identical colors..RR or BB

so he should write the opposite color

but if he see different colors he should abstain.

In this method we will obtain 75%

with three persons we have:

RRR

RRB

BRR

BBB

BBR

RBB

so in four ordes out of six we will have the right answer

where the other 12 will abstain.

I meant 6 out of 8 will be right...which makes 75%

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93.75% - all but 5 abstain, if one of the five sees four of the same color then choose the other color. or, if any one of the five sees three of one color and one of the other color then choose the same color as the single colored cap seen, otherwise abstain.

there are 10 orders with 4 to 1 out of 31 odds which are about 32.258%

and 20 orders with 3 to 2 out of 31 odds making about 64.516%

and by joining both of them you will have only 10 odds to be right!

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93.75% - all but 5 abstain, if one of the five sees four of the same color then choose the other color. or, if any one of the five sees three of one color and one of the other color then choose the same color as the single colored cap seen, otherwise abstain.

this answer is wrong. the fool who hastily wrote this should go back to bed or have another cup of coffee (and maybe a tasty pastry).

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If 3 participants have 2^3=8 possibilities and only only 2 are wrong leaving you with 75% probability, then 15 participants should have 2^15 = 32768 possible combinations.

if only 2 still lead to wrong answers then you have 32766/32768=0.999938965 or 99.99% probability.

This can't be right but seems like only logical progression that leads to something above 75%

As for who answers and when, I'm stuck.

It has been already suggested by vinays84 (btw, you forgot to include 2 cases RBR and BRB).

Now, what conclusions do you draw from that?

If so few people can do so much, imagine what you can do with all 15.

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1/2.

No matter what you do, you can't discern information about the color of your own hat.

*It was chosen randomly

*There can be no other form of communication amongst your peers (facial expressions, etc.)

*everyone has to make their guess simultaneously

*the fact that the majority of other hats is a different color is irrelevant.

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1/2.

No matter what you do, you can't discern information about the color of your own hat.

*It was chosen randomly

*There can be no other form of communication amongst your peers (facial expressions, etc.)

*everyone has to make their guess simultaneously

*the fact that the majority of other hats is a different color is irrelevant.

The facts you mention are right, but the implication [1/2] is not. Browse through the thread to find strategies up to 75% winning chance (so far).

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1/2.

No matter what you do, you can't discern information about the color of your own hat.

*It was chosen randomly

*There can be no other form of communication amongst your peers (facial expressions, etc.)

*everyone has to make their guess simultaneously

*the fact that the majority of other hats is a different color is irrelevant.

And I meant to add that: "The team wins the contest if at least one answer is correct and no answer is incorrect."

I see a lot of answers where that isn't taken into account. My having multiple people guess you're almost guaranteeing that someone will be wrong, thus losing the game for everybody.

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It has been already suggested by vinays84 (btw, you forgot to include 2 cases RBR and BRB).

Now, what conclusions do you draw from that?

If so few people can do so much, imagine what you can do with all 15.

with more cases,we get less %, I reached to the 7th order,and I`ve got only 55%

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this answer is wrong. the fool who hastily wrote this should go back to bed or have another cup of coffee (and maybe a tasty pastry).

lol

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Posted · Report post

approx 99.994%?

and..

with an even number of people?

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assume they are sitting in a circle.

with 5 we have the following combos.

BBBBB BBRRB BBRRR RRBRB

BBBBR BRBBR BRBRR RRRBB

BBBRB BRBRB RBBRR BRRRR

BBRBB BRRBB BRRBR RBRRR

BRBBB RBBBR RBRBR RRBRR

RBBBB RBBRB RRBBR RRRBR

BBBRR RBRBB BRRRB RRRRB

BBRBR RRBBB RBRRB RRRRR

if you see 4 of the same color, gueess that color

if you see 2 red, 2 blue, abstain.

if you see 3 of the same color, guess the oppisite color.

22/32 wins, 68.75%, is there a way to get it higher?

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Each person looks to see if there are 3 persons with consecutive hats of the same color. If so, they abstain from answering. If not, and the hats on each side of themselves is the same color, answer the same color (assuming everyone else will abstain from guessing because you are the guy). Haven't calculated the odds of this happening and only once in a group of 15, but I guess it is near 80-85%.

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