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# Which Chair you will choose ?

## Question

Prof. Sushi Rama gives out only one A+ each year in her course on Logic. This year, you and two other students got 100 on all the exams, and you REALLY want that A+. The professor says that she will give the 3 of you a test to determine the winner.

There are 3 chairs in her office. After lunch each of you will sit in one of the chairs, and then she will paste two stamps on your forehead. She shows you the 4 white and 4 black stamps that she has. You will be able to see the stamps on the other two students' foreheads, but not your own, or the two leftover stamps. Then she will ask the student in the first chair to tell what color stamps are on his or her forehead. If the student cannot logically deduce the colors, she will move on to the second chair, then the third chair. If that does not decide the issue, she will continue around the circle of chairs until one of you gives the correct response, with correct reasoning, based on the stamps that are visible and the other students' answers.

After lunch you are the first to arrive at the professor's office. Which chair do you choose and why?

Edited by ujjagrawal
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## 30 answers to this question

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I assume she will randomly choose the stamps because e.g. if you chose the first chair + the four stamps you see are the same color. So you want to maximise your chances.

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I assume she will randomly choose the stamps because e.g. if you chose the first chair + the four stamps you see are the same color. So you want to maximise your chances.

what are the chances of seeing 4 stamps of same colors on first chair... Is this the maximum ? what if you take 2nd or 3rd chair? Edited by ujjagrawal

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100+ views and just 1 attempt... this is not so tough guys... just try it out... definitely you will find out the correct answer...

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MikeD    175

I would say sitting on the second chair would be the best choice.

The people sitting on the chairs have the chance of working out their colour of the stamps of their foreheads in this ratio: 6:8:5.

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I'm not sure if i have all the combos. but if i do this is what i'm thinking

bb/wb/wb

ww/wb/wb

ww/wb/bb

ww/ww/bb

ww/bb/bb

wb/wb/wb

that gives us six. two of which we can eliminate because they just wouldn't be fair to all students. (ww/ww/bb, bb/bb/ww)

so if you sit in the first chair this gives you a 2 out of 4 chance to guess right. so first chair

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MikeD    175

I would say sitting on the second chair would be the best choice.

The people sitting on the chairs have the chance of working out their colour of the stamps of their foreheads in this ratio: 6:8:5.

I recheck my working, i think that the chair is right but the ratio should be 14:22:14

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The first can only guess it right if the two othes have the same color so:

P_1 = P(ww.bb.bb)+P(bb.ww.ww) = (3/14)^2 * 2 = 9/98 = 9.2%

similarily:

P_2 = P(bb.ww.bb) + P(ww.bb.ww) = P_1 = 9.2%

the third has the following extra advantage:

P_3 = P(bb.bb.ww) + P(ww.ww.bb) + P(bb.ww.bw) + P(ww.bb.bw) = P_1 * 2 = 18.4%

the forth turn (first chair)

P_4 = P(bw.ww.bw) + P(bw.bb.bw) = (3/14) * 2 = 42.9 %

the fifth turn (second chair) or more

P(>5) = 1 - (P_1 + P_2 + P_3 + P_4) = 20.3%

which is lower than the previous seat. therefore, I would choose the first seat since the student on the forth turn has the highest probability.

Please let me know if I missed something.

Edited by andaryfaysal

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bonanova    76

Second seat wins most frequently, but I get different odds from previous solvers.

I'll enumerate the cases so that if I'm wrong at some point, it will be clear where.

Let b [black], m [mixed], w [white] represent BB, (BW or WB), and WW respectively.

There are nineteen permissible b,m,w triplets, some, like bbb, bbm, etc. are not permissible.

If six of the eight stamps are distributed at random to the students, we can determine the relative frequency of their occurrence.

We can then group them in terms of how many students receive stamps of mixed color.

Case 1: No mixed students: 6 triplets, each occurring 1 time

Case 2: One mixed student: 6 triplets, each occurring 4 times

Case 3: Two mixed students: 6 triplets, each occurring 4 times

Case 4: Three mixed students: 1 triplet, occurring 16 times.

Case 1. The student wins who differs in color from the other two: If bww, Seat 1 wins.

Case 2. The student wins who has the mixed stamps: e.g. If bwm, Seat 3 wins.

Case 3. The student wins who follows the non-mixed student: If bmm, Seat 2 wins.

The first three cases have no chair preference, thus comprise 18 wins for each student.

Case 4. Seat 2 wins. There are 16 such occurrencess.

This case tips the odds to favor Seat 2.

Enumerating the nineteen cases:

Seat 1 wins wbb[1], bww[1], mbw[4], mwb[4], mmb[4], mmw[4] = 18 times

Seat 2 wins bwb[1], wbw[1], bmw[4], wmb[4], bmm[4], wmm[4], mmm[16] = 34 times

Seat 3 wins bbw[1], wwb[1], bwm[4], wbm[4], mbm[4], mwm[4] = 18 times

So the relative chances of winning seem to be 9, 17, 9.

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The probability of four same color (thus first chair winning):

8/8 * 3/7 * 2/6 * 1/5 = 1/35. Same answer results from using combinations.

There are 70 (ordered) permutations of the 8 stamps, each with equal chance.

There are 16 (ordered) permutations of 4 of the 8 stamps, not with equal chance.

If this doesn't occur, then first chair cannot know the answer.

Second chair now knows that first chair does not see: YYXXXX

Specifically it leaves 68 permutations with equal chance.

Then, if he sees XXYYXX he wins. As in the first case, there are two permutations that satisfy this.

So he wins this way with 1/34 chance.

I believe he learns nothing else.

Third chair now knows that second chair does not see: XXYYXX

This eliminates 2 possibilities, leaving 66 with equal chance.

If he sees XXXXYY he wins, 2/66 chance.

Also if he sees XXYY?? then he wins knowing he has XY (since the others haven't won yet).

This could happen as BBWW**** or WWBB**** where **** is permutations of BBWW. 2 * 6 = 12 ways

However XXYY?? intersects with XX??XX at XXYYXX so there are 8 ways left.

So if he didn't win by XXXXYY he wins this with 8/64.

Equivalent overall: 2/66 + 64/66 (8/64) === 10/66 = 5/33

I believe he learns nothing else.

First chair:

He's survived this far, so there are 56 possibilities left. YYXX?? is not possible.

He applies same thing as third chair, he wins if he sees ??XXYY where he must have XY. Still 12 possibilities, minus 4 intersections.

Wins with chance 8/54 = 4/27.

I can't see him learning anything else.

Second chair:

48 possibilities left, YYXX?? and ??XXYY are excluded.

If he sees XX??YY he must have XY and he wins. 8 ways => 8/48 = 1/6 chance of winning like this.

If he doesn't win like that, he knows the XX, YY, XY pairs aren't there (no one sees them). XXYYXY and XXYYXX and XXYYYY aren't there, it must mean XX YY isn't there. What about XX XY? (XX XY YY, XX XY XX can't be). One mixed pair doesn't exist, this means there are two or three mixed pairs.

Thus if he sees XX??MM or MM??XX where MM = mixed pair, then he knows he has XY.

XXMMMM has 8 forms: 2x for swapping X with Y, 4x for two MM pairs as XY or YX. No more multipliers because the rest must be YY.

MMMMXX has 8 ways also.

He wins with 16/40 chance now = 2/5

Overall: 8/48 + 40/48 (16/40) = 24/48 = 1/2 chance.

Third chair: 24 left.

So I think there are 2 or 3 mixed pairs. Second chair would have won if one of 1st or 3rd pair was not mixed, therefore they must both be mixed.

I believe possibilities left are MMMMMM or MMXXMM. This would mewan third must have XY no matter what.

Can check this by seeing how many possibilities MMMMMM and MMXXMM form:

MMXXMM: 2x for XY or YX on third pair, 2x for XY/YX on 1st pair, 2x for identify of X, 1x for the rest must be YY. 8 ways.

MMMMMM: 8x for three pairs XY/YX. Remaining is also a pair, 2x for that. 16 ways.

Totals 24 so correct.

Patterns:

1 XXXX??, 2 ways

2 XX??XX, 2 ways

3 ??XXXX, 2 ways

3 XXYY??, 8 new ways

1 ??XXYY, 8 new ways

2 XX??YY, 8 new ways

2 XXMMMM or MMMMXX, 16 ways

3 MMXXMM or MMMMMM, 24 ways

By intuition or otherwise, the order of events is independent of the ways of winning (must be better way to explain). Anyhow, you can just add up the number of ways.

First chair: 10 ways

Second chair: 26 ways

Third chair: 34 ways

So the logical choice is third chair, with 34/70 chance of winning, assuming the other two students are not (color)blind and are equally logical.

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Got to say, it's __________ that five of us have completely different solutions, and no working in common.

Someone find the right word for me

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bonanova    76

As noted, there are 70 equally probable distributions of the stamps.

They fall into Analysis for each case follows.

1. No student has mixed stamps. [6] For example bwb, [where chair 2 would win.]
The student who differs in color from the other two wins: he sees four stamps of one color.
There is no chair preference for this case. Two for each chair.

2. Exactly one student has mixed stamps. [24] For example bwm, [where chair 3 would win.]
The mixed-stamp student thinks it might be case 1. But after the other two pass, he knows otherwise, and wins.
There is no chair preference for this case. Eight for each chair

3. Exactly two students have mixed stamps. [24] For example, bmm [wherechair 2 would win.]
Here the chair that would win case 2 can't know he is mixed [he sees another mixed student] so he passes.
That second mixed student thus knows it's not case 2. So she know she is mixed, and wins.
There is no chair preference for this case. Eight for each chair.

4. All three students are mixed. [16] Only example is mmm, [where chair 2 will always win.]
Everyone thinks this might be case 3, but after chair 1 passes the third time, it's not.
They all know it's case 4, and the chair that has first chance to say so is chair 2, who wins.
This is the only case with a chair preference: chair 2 always wins. Sixteen for chair 2.

The relative frequencies for the three chairs are thus 18/34/18, or 9/17/9.

Winning probabilities are thus .257 / .486 / .257

Interestingly, only in case 1 does the winning chair have same-color stamps.

91.4% of the time the winner is mixed-color.

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Got to say, it's __________ that five of us have completely different solutions, and no working in common.

Someone find the right word for me

"interesting". This means the puzzle is so good that it got each one of us to think differently. Thank you for the brain teaser, ujjagrawal.

Although after going through the posts, I am now convinced that bonanova has nailed the answer.

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"interesting". This means the puzzle is so good that it got each one of us to think differently. Thank you for the brain teaser, ujjagrawal.

Although after going through the posts, I am now convinced that bonanova has nailed the answer.

Personally I find it bizarre. I've noted that the "signal to noise ratio" (if signal means correct/true/good reasoning/answer/process, noise means bad/incorrect/false) on brainden is worse than most ad hoc forums or messages, even for the trivially easy puzzles. One of these forms is where people give answers that don't mean a thing to anyone else; the words and logic is practically gibberish in English. How does this happen? Even for most people to acknowledge a correct logical answer seems to be impossible here. If this place lacks common sense, you have to wonder what "communal" value there is given the objectives of contributing here.

In this case, the fact that we have completely different solutions means most of us are completely wrong, and yet believe we are all on the right track.

I have no problem accepting I could be wrong, but I'm the only one who has given a proof, and I have implicitly disproved all other answers. But in my experience, proofs and disproofs mean nothing to most people.

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The probability of four same color (thus first chair winning):

8/8 * 3/7 * 2/6 * 1/5 = 1/35. Same answer results from using combinations.

There are 70 (ordered) permutations of the 8 stamps, each with equal chance.

There are 16 (ordered) permutations of 4 of the 8 stamps, not with equal chance.

If this doesn't occur, then first chair cannot know the answer.

Second chair now knows that first chair does not see: YYXXXX

Specifically it leaves 68 permutations with equal chance.

Then, if he sees XXYYXX he wins. As in the first case, there are two permutations that satisfy this.

So he wins this way with 1/34 chance.

I believe he learns nothing else.

Third chair now knows that second chair does not see: XXYYXX

This eliminates 2 possibilities, leaving 66 with equal chance.

If he sees XXXXYY he wins, 2/66 chance.

Also if he sees XXYY?? then he wins knowing he has XY (since the others haven't won yet).

This could happen as BBWW**** or WWBB**** where **** is permutations of BBWW. 2 * 6 = 12 ways

However XXYY?? intersects with XX??XX at XXYYXX so there are 8 ways left.

So if he didn't win by XXXXYY he wins this with 8/64.

Equivalent overall: 2/66 + 64/66 (8/64) === 10/66 = 5/33

I believe he learns nothing else.

First chair:

He's survived this far, so there are 56 possibilities left. YYXX?? is not possible.

He applies same thing as third chair, he wins if he sees ??XXYY where he must have XY. Still 12 possibilities, minus 4 intersections.

Wins with chance 8/54 = 4/27.

I can't see him learning anything else.

Second chair:

48 possibilities left, YYXX?? and ??XXYY are excluded.

If he sees XX??YY he must have XY and he wins. 8 ways => 8/48 = 1/6 chance of winning like this.

If he doesn't win like that, he knows the XX, YY, XY pairs aren't there (no one sees them). XXYYXY and XXYYXX and XXYYYY aren't there, it must mean XX YY isn't there. What about XX XY? (XX XY YY, XX XY XX can't be). One mixed pair doesn't exist, this means there are two or three mixed pairs.

Thus if he sees XX??MM or MM??XX where MM = mixed pair, then he knows he has XY.

XXMMMM has 8 forms: 2x for swapping X with Y, 4x for two MM pairs as XY or YX. No more multipliers because the rest must be YY.

MMMMXX has 8 ways also.

He wins with 16/40 chance now = 2/5

Overall: 8/48 + 40/48 (16/40) = 24/48 = 1/2 chance.

Third chair: 24 left.

So I think there are 2 or 3 mixed pairs. Second chair would have won if one of 1st or 3rd pair was not mixed, therefore they must both be mixed.

I believe possibilities left are MMMMMM or MMXXMM. This would mewan third must have XY no matter what.

Can check this by seeing how many possibilities MMMMMM and MMXXMM form:

MMXXMM: 2x for XY or YX on third pair, 2x for XY/YX on 1st pair, 2x for identify of X, 1x for the rest must be YY. 8 ways.

MMMMMM: 8x for three pairs XY/YX. Remaining is also a pair, 2x for that. 16 ways.

Totals 24 so correct.

Patterns:

1 XXXX??, 2 ways

2 XX??XX, 2 ways

3 ??XXXX, 2 ways

3 XXYY??, 8 new ways

1 ??XXYY, 8 new ways

2 XX??YY, 8 new ways

2 XXMMMM or MMMMXX, 16 ways

3 MMXXMM or MMMMMM, 24 ways

By intuition or otherwise, the order of events is independent of the ways of winning (must be better way to explain). Anyhow, you can just add up the number of ways.

First chair: 10 ways

Second chair: 26 ways

Third chair: 34 ways

So the logical choice is third chair, with 34/70 chance of winning, assuming the other two students are not (color)blind and are equally logical.

nice work.... your approach seems most correct to me... third chair is my choice too...

This is not a pure probability problem... it is a combination of probability and logic... the person sitting on a chair will not make any guesses based upon probability but he will try to deduce the answer by logic... incase he is not able to... he passes on... and in worst case senario... correct answer can be logically deducable until 6th turn i.e by person sitting on 3rd chair... that also hold maximum probability(chances) for a person to logically deduce the answer. Anyone with a different views/answer with explanation is most welcome...

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Yoruichi-san    116

Personally I find it bizarre. I've noted that the "signal to noise ratio" (if signal means correct/true/good reasoning/answer/process, noise means bad/incorrect/false) on brainden is worse than most ad hoc forums or messages, even for the trivially easy puzzles. One of these forms is where people give answers that don't mean a thing to anyone else; the words and logic is practically gibberish in English. How does this happen? Even for most people to acknowledge a correct logical answer seems to be impossible here. If this place lacks common sense, you have to wonder what "communal" value there is given the objectives of contributing here.

In this case, the fact that we have completely different solutions means most of us are completely wrong, and yet believe we are all on the right track.

I have no problem accepting I could be wrong, but I'm the only one who has given a proof, and I have implicitly disproved all other answers. But in my experience, proofs and disproofs mean nothing to most people.

I've been following this thread kind of haphazardly, but I have to come in and say that what you're saying, whether you're trying to or not, is discouraging people from posting with their ideas/thoughts on a puzzle, which I think is a key aspect of BD being a 'community' rather than just a website.

Even if people are wrong, they may say something that inspires thoughts in others, and that in itself is valuable. I am a scientist, and highly value collaboration. Pretty much all great discoveries, although often credited to one person or the other, were made by a collaboration or at least one person building the past works of others.

Also as a scientist, I've noticed that in science, economics, game theory, and even math, there are still problems 'experts' disagree on the correct approach to.

And personally, as a puzzle maker who likes to make somewhat complex puzzles, I highly encourage people to work together and share their working thoughts. It usually 'gets the thread going' and builds a kind of happy, fun atmosphere as well.

To be honest, I have not given this particular puzzle enough analysis to form an opinion on who is 'correct', my beginning line was with probabilities such as your first line, but I never took the time to finish. I recognize that you've put a lot of work and thought into your answer, and you gave a really good and thorough explanation...but I have to point out, it does not constitute a proof...your own 'comments' in red show the parts that need to be completed for it to be a proof.

Welcome to the den ;P.

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bushindo    14

Personally I find it bizarre. I've noted that the "signal to noise ratio" (if signal means correct/true/good reasoning/answer/process, noise means bad/incorrect/false) on brainden is worse than most ad hoc forums or messages, even for the trivially easy puzzles. One of these forms is where people give answers that don't mean a thing to anyone else; the words and logic is practically gibberish in English. How does this happen? Even for most people to acknowledge a correct logical answer seems to be impossible here. If this place lacks common sense, you have to wonder what "communal" value there is given the objectives of contributing here.

In this case, the fact that we have completely different solutions means most of us are completely wrong, and yet believe we are all on the right track.

I have no problem accepting I could be wrong, but I'm the only one who has given a proof, and I have implicitly disproved all other answers. But in my experience, proofs and disproofs mean nothing to most people.

I understand your frustration of receiving no feedback after a well-reasoned and meticulous post of the solution. It is obvious from your post that you are quite smart, and I think you would fit right in this community. Personally, I generally find puzzles here to be excellent (this topic is an perfect example), and the community here includes some very fastidious, clever, and brilliant logicians. I have found myself more than once awed and humbled by the creativity and the sheer elegance of some puzzles and solutions that the Denizens come up with. I hope you will give Brainden some more time to change your opinion of its worth.

I'll have to admit that sometimes it is hard to get some feedback on this board. In this case, however, you could probably initiate some discussions about the correctness of the solution by examining other solutions (bonanova's in for instance) to see whether it is incorrect or whether you missed anything. I have examined both of your solutions, and I believe both have some errors. I may be wrong, and I often am, but I'll include those possible errors here

bonanova's solution

1. Exactly two students have mixed stamps. [24] For example, bmm [wherechair 2 would win.]
Here the chair that would win case 2 can't know he is mixed [he sees another mixed student] so he passes.
That second mixed student thus knows it's not case 2. So she know she is mixed, and wins.
There is no chair preference for this case. Eight for each chair.

The possible error is highlighted in red. I find that there is indeed a chair preference here in that the third 3rd chair can not win if there are two students with mixed stamps. Consider the two states bmm and mmb, my calculations show that they both give will the win to the second chair.

Also, I believe that if there are 3 students with mixed stamps, then the third chair would win.

voider's solution (the spoiler tags are added by me)

Third chair: 24 left.

So I think there are 2 or 3 mixed pairs. Second chair would have won if one of 1st or 3rd pair was not mixed, therefore they must both be mixed.

I believe possibilities left are MMMMMM or MMXXMM. This would mewan third must have XY no matter what.

Can check this by seeing how many possibilities MMMMMM and MMXXMM form:

MMXXMM: 2x for XY or YX on third pair, 2x for XY/YX on 1st pair, 2x for identify of X, 1x for the rest must be YY. 8 ways.

MMMMMM: 8x for three pairs XY/YX. Remaining is also a pair, 2x for that. 16 ways.

Totals 24 so correct.

The possible error is highlighted in green. If the states are MMXXMM, then the 3rd chair would say 'No' on turn 3, allowing the 1st chair to rule out the state XX and YY for his own state. The 1st chair would then win on turn 4.

Having said that, I'll contribute a solution of my own. This solution will give a winning chance of 42/70 or 60%.

Choose the 1st chair. During the first turn, count the number of black stamps you see on the other two students. Then raise your hand and choose your state among the following table

0 black: BB

1 black: MM

2 black: MM

3 black: MM

4 black: WW

The chance of getting (0, 1, 2, 3, 4) blacks are (1/70, 16/70, 36/70, 16/70, 1/70), respectively. The chance of winning using the table above for (0, 1, 2, 3, 4) blacks are (1, 1/2, 2/3, 1/2, 1). So that gives us an expected value of 42/70.

Edited by bushindo
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I've been following this thread kind of haphazardly, but I have to come in and say that what you're saying, whether you're trying to or not, is discouraging people from posting with their ideas/thoughts on a puzzle, which I think is a key aspect of BD being a 'community' rather than just a website.

Even if people are wrong, they may say something that inspires thoughts in others, and that in itself is valuable. I am a scientist, and highly value collaboration. Pretty much all great discoveries, although often credited to one person or the other, were made by a collaboration or at least one person building the past works of others.

Also as a scientist, I've noticed that in science, economics, game theory, and even math, there are still problems 'experts' disagree on the correct approach to.

And personally, as a puzzle maker who likes to make somewhat complex puzzles, I highly encourage people to work together and share their working thoughts. It usually 'gets the thread going' and builds a kind of happy, fun atmosphere as well.

To be honest, I have not given this particular puzzle enough analysis to form an opinion on who is 'correct', my beginning line was with probabilities such as your first line, but I never took the time to finish. I recognize that you've put a lot of work and thought into your answer, and you gave a really good and thorough explanation...but I have to point out, it does not constitute a proof...your own 'comments' in red show the parts that need to be completed for it to be a proof.

Welcome to the den ;P.

Referring to the website, not this particular thread: (Maybe in the wrong section of the forum, but it is a reply.)

Firstly I agree on the value of collaboration and openness to creativity. But I believe in the context of logic puzzles there is usually a distinction between "something I haven't thought about" and "something that makes no sense, and is based on assumptions that are difficult to intelligently comprehend". Obviously I must have some pride in what I believe to be logical or at least reasonable thinking (even if it is inaccurate) because it affects me when others can't see this distinction, or get it the wrong way round. In fact, this is the first thing I discovered on brainden, other than that it had puzzles and problems on it. Of 200+ comments on the Epimenides Paradox, I dare say 20% or less showed accurate comprehension of its meaning and solution. I discussed this with rookie and proposed a way to improve people's thinking, not just define an answer, a rare opportunity. It didn't come together. I know that generally you can't change people nor the way they think, and that's why my prior post is just an expression of my observations. If it is assumed that nothing can be done about it, and no one has tried, then what does it say about an appropriate environment on brainden? Is it appropriate that most people don't "get" one of the most famous paradoxes, doesn't it virtually encourage the notion that "it doesn't matter what's true, as long as others agree with you", or even "it doesn't matter what's true, as long as you agree with yourself [, and have fun in the process]" (deliberate hint of Hedonism, it's hard to initiate a discussion on that level) ? I would imagine a problem-solving context to do a bit better than that.

Can you express whatever you think about a logic puzzle on brainden? Yes

Should you be able to express it without second thought, and expect it to be accepted for what it is? I don't know.

Do all expressions have value to the "system"? I don't believe so. Spam, redundancy, etc.

Although not directly relevant, I will point out that the ideal logic problem solver would probably read problems, solve alone (never give up) and never come to a forum. But if I'm communicating any sense to you, you'll see that I think the worth of the community with respect to its social and objective values (if any) is a conundrum. You can't have everything, and I'm observing what brainden doesn't have.

Edited by voider

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bushindo    14

Also, I believe that if there are 3 students with mixed stamps, then the third chair would win.

Oops, upon second thoughts, I'll revise the claim I made here

It seems that bonanova was right. If there are 3 students with mixed stamps, then the second chair would win.

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bonanova    76

I have examined both of your solutions, and I believe both have some errors.

I may be wrong, and I often am, but I'll include those possible errors here

bonanova's solution

The spoiler did not get into the quote, but you suggest that mmb gives the win to chair 2.

It may help to take three cases in sequence, to see how adding m's adds complexity.

w b b Simplest case.

1. bbb mbb wbb WHITE All four black stamps are visible elsewhere.

w m b Add an m.

1. bmb mmb wmb Pass. I am not black, that would take 5 black stamps.

2. wbb wmb wwb Pass. I am not black, else 1 would have said WHITE [see above case]

3. wmb wmm wmw Pass. I am not white, else 2 would have said BLACK

1. bmb mmb wmb Pass. In both cases, 2 and 3 would pass.

2. wbb wmb wwb MIXED I am not white, else 3 would have said BLACK

m m b Add another m.

1. bmb mmb wmb Pass. I am not black.

2. mbb mmb mwb Pass. I am not black.

3. mmb mmm mmw Pass. I could be anything.

1. bmb mmb wmb Pass. I only know I am not black.

2. bmb mmb mwb Pass. I could be white. Since 1 is m, 3 can't say BLACK [see above case]

3. mmb mmm mmw Pass. I could be anything. All would pass in each case.

1. bmb mmb wmb MIXED I am not white, else 2 would have said MIXED.[see above case]

So I have chair 1 winning this distribution of stamps.

If chair 2 wins, it would be on the second round.

How does she deduce by then that she is not white?

Having said that, I'll contribute a solution of my own. This solution will give a winning chance of 42/70 or 60%.

Choose the 1st chair. During the first turn, count the number of black stamps you see on the other two students. Then raise your hand and choose your state among the following table

0 black: BB

1 black: MM

2 black: MM

3 black: MM

4 black: WW

The chance of getting (0, 1, 2, 3, 4) blacks are (1/70, 16/70, 36/70, 16/70, 1/70), respectively. The chance of winning using the table above for (0, 1, 2, 3, 4) blacks are (1, 1/2, 2/3, 1/2, 1). So that gives us an expected value of 42/70.

This is a nice analysis to maximize probability that your answer is correct. I had understood the puzzle to require that when you answer you must be certain that your answer is correct. Then, each chair gives you a particular probability of being the first to be able to answer with certainty of your color.

If the student cannot logically deduce the colors, she will move on to the second chair, then the third chair. If that does not decide the issue, she will continue around the circle of chairs until one of you gives the correct response, with correct reasoning, based on the stamps that are visible and the other students' answers.

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MikeD    175

Case 1: No Mixed

WBB (2) - Seat 1

BWB (2) - Seat 2

BBW (2) - Seat 3

(Person in the seat can see four black stamps, so both of his must be white)

Case 2: 1 Mixed

BWM (8) - Seat 3

MBW (8) - Seat 1 (on second turn)

BMW (8) - Seat 2 (on second turn)

(If his were White or Black one of the other would of know, since no other person spoke it must be mixed.)

Case 3: 2 Mixed

MBM (8) - Seat 1 (on second turn)

(If seat 1 were black or white, player three would have know that his was mixed, therefore seat 1 has to be mixed.)

BMM (8) - Seat 2 (on second turn)

MMB (8) - Seat 2 (on second turn)

(Seat 2 can see a person with black stamps, since the other person did not know, then the other person can't see two black or one black and one white, therefore seat 2 must be mixed.)

Case 4: All Mixed

MMM (16) - Seat 2 (on second turn)

(For seat 2 the only other possiblilty would be MBM (or MWM), if that was the case then seat 1 would have known (see above), therefore seat 2 has to be mixed.)

Seat 1 - 18

Seat 2 - 42

Seat 3 - 10

So seat 2 would be the best to choice he would have 42/70 chance that the stamp you laid out in a combination that he could work out.

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MikeD    175

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bonanova    76

Hi MikeD. We seem to be converging on a solution.

I think we differ only in one case: mmb.

See previous post for my analysis of that case.

See comment in red inside your spoiler for a question about yours.

Thanks.

Case 1: No Mixed

WBB (2) - Seat 1

BWB (2) - Seat 2

BBW (2) - Seat 3

(Person in the seat can see four black stamps, so both of his must be white)

Case 2: 1 Mixed

BWM (8) - Seat 3

MBW (8) - Seat 1 (on second turn)

BMW (8) - Seat 2 (on second turn)

(If his were White or Black one of the other would of know, since no other person spoke it must be mixed.)

Case 3: 2 Mixed

MBM (8) - Seat 1 (on second turn)

(If seat 1 were black or white, player three would have know that his was mixed, therefore seat 1 has to be mixed.)

BMM (8) - Seat 2 (on second turn)

MMB (8) - Seat 2 (on second turn)

(Seat 2 can see a person with black stamps, since the other person did not know, then the other person can't see two black or one black and one white, therefore seat 2 must be mixed.)

I guess that is the point I'm not getting.

How would seeing one black and one white disclose your color?

I think at this point Seat 2 knows only that she is not black.

What tells her that she also is not white?

Case 4: All Mixed

MMM (16) - Seat 2 (on second turn)

(For seat 2 the only other possiblilty would be MBM (or MWM), if that was the case then seat 1 would have known (see above), therefore seat 2 has to be mixed.)

I think Seat 2 can't know she is mixed in this case until her third turn.

Seat 1 - 18

Seat 2 - 42

Seat 3 - 10

So seat 2 would be the best to choice he would have 42/70 chance that the stamp you laid out in a combination that he could work out.

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MikeD    175

Hope this helps, I sometimes have a hard time explaining things.

MMB

Seat 1

Can see M and B, can be anything.

Seat 2

Can see M and B, can't be B. Can be M or W.

Seat 3

Can see M and M, can be anything.

Seat 1

Since seat 3 passed,can't be black. Can be M or W.

Seat 2

Have to be W or M.

If I was white, seat 1 would have seen one Black and one White, and reasoned: since there is one black and one white,

and each person passed, meaning that no one saw two of one colour, I can't have Black or White I must have mixed.

Therefore since seat 1 didn't act, he could not have seen white and black. (and Black, black is out already)

I must have mixed.

Edited by MikeD
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bushindo    14

Case 1: No Mixed

WBB (2) - Seat 1

BWB (2) - Seat 2

BBW (2) - Seat 3

(Person in the seat can see four black stamps, so both of his must be white)

Case 2: 1 Mixed

BWM (8) - Seat 3

MBW (8) - Seat 1 (on second turn)

BMW (8) - Seat 2 (on second turn)

(If his were White or Black one of the other would of know, since no other person spoke it must be mixed.)

Case 3: 2 Mixed

MBM (8) - Seat 1 (on second turn)

(If seat 1 were black or white, player three would have know that his was mixed, therefore seat 1 has to be mixed.)

BMM (8) - Seat 2 (on second turn)

MMB (8) - Seat 2 (on second turn)

(Seat 2 can see a person with black stamps, since the other person did not know, then the other person can't see two black or one black and one white, therefore seat 2 must be mixed.)

Case 4: All Mixed

MMM (16) - Seat 2 (on second turn)

(For seat 2 the only other possiblilty would be MBM (or MWM), if that was the case then seat 1 would have known (see above), therefore seat 2 has to be mixed.)

Seat 1 - 18

Seat 2 - 42

Seat 3 - 10

So seat 2 would be the best to choice he would have 42/70 chance that the stamp you laid out in a combination that he could work out.

Nice work! I think you have the correct answer.

This is a nice analysis to maximize probability that your answer is correct. I had understood the puzzle to require that when you answer you must be certain that your answer is correct. Then, each chair gives you a particular probability of being the first to be able to answer with certainty of your color.

If the student cannot logically deduce the colors, she will move on to the second chair, then the third chair. If that does not decide the issue, she will continue around the circle of chairs until one of you gives the correct response, with correct reasoning, based on the stamps that are visible and the other students' answers.

That was an attempt to think outside the box. It definitely doesn't fit the constraint of the OP unless we considerably relax the definition of 'logically deduce'

Edited by bushindo

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bonanova    76

Hope this helps, I sometimes have a hard time explaining things.

MMB

Seat 1

Can see M and B, can be anything.

Seat 2

Can see M and B, can't be B. Can be M or W.

Seat 3

Can see M and M, can be anything.

Seat 1

Since seat 3 passed,can't be black. Can be M or W.

Seat 2

Have to be W or M.

If I was white, seat 1 would have seen one Black and one White, and reasoned: since there is one black and one white,

and each person passed, meaning that no one saw two of one colour, I can't have Black or White I must have mixed.

Therefore since seat 1 didn't act, he could not have seen white and black. (and Black, black is out already)

I must have mixed.

You explained it well; I just was slow to follow it.

I should have seen that.

Since we established that in Case 1 the [only] mixed chair wins, therefore,

Chair two knows that if she were white, then Chair 1 would be the only mixed, and he would have won.

Since he didn't, then she must be mixed, also.

And Chair 2 now becomes the overwhelming favorite: 60% win probability.

Nicely done!

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