30 replies to this topic

[ujjagrawal]

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, 25 June 2012 - 05:19 AM.

[voider]

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.

[ujjagrawal]

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, 25 June 2012 - 11:18 AM.

[ujjagrawal]

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

[MikeD]

Spoiler for

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.

[TripleOption]

Spoiler for random guess

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

[MikeD]

Spoiler for

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.

Spoiler for

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

[andaryfaysal]

Spoiler for Using probabilities

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, 25 June 2012 - 08:21 PM.

[bonanova]

Spoiler for Looks like

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.

[voider]

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.