30 replies to this topic

[MikeD]

Spoiler for now that I had time

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.

[MikeD]

@bushindo I like you answer.

[bonanova]

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.

Spoiler for now that I had time

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.
The (see above) part of your analysis is what I asked about above.

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.

[MikeD]

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

Spoiler for reasoning

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, 28 June 2012 - 07:40 AM.

[bushindo]

Spoiler for now that I had time

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, 28 June 2012 - 07:49 AM.

[bonanova]

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

Spoiler for reasoning

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.

Spoiler for I agree

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!

[Godsryche]

With this being a Logic class, perhaps we don't need to do any math at all...

Spoiler for No Math

Assuming that the teacher wants to be fair about the test, then the only fair way she could go about assigning the stamps is to put one of each color on each students head, that way each student will have the same information as the others. Therefore, the first chair is the choice to make so you can answer first, logical fairness dictates that your head must have a stamp of each color.

[bonanova]

With this being a Logic class, perhaps we don't need to do any math at all...

Spoiler for No Math

Assuming that the teacher wants to be fair about the test, then the only fair way she could go about assigning the stamps is to put one of each color on each students head, that way each student will have the same information as the others. Therefore, the first chair is the choice to make so you can answer first, logical fairness dictates that your head must have a stamp of each color.

The difficulty with this approach is that if the students do as the OP requires, the student who made your additional assumption of fairness would pick chair #2.

[MikeD]

I think though that for fairness the teacher would randomly assign the stamps. If she chose the order of stamps, she would be favouring a chair and therefore a student. For example, if she chose 1 black one and 1 white one for each student, then she would be favouring the student in the second chair.

[Godsryche]

The difficulty with this approach is that if the students do as the OP requires, the student who made your additional assumption of fairness would pick chair #2.

But the student in the first chair gets to guess first, no? So, if I walk in with the fairness assumption, then I don't really even need to see what the other two students have on their heads in order to determine what mine are.