Prof. Tai Rant 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?
First, it's worth noting that my choice may well be moot. If Prof Rant wants one of us to win, she can always assign stamps to engineer that outcome. (The problem doesn't say she'll paste stamps randomly.)
But it's only interesting if the stamp selection *is* random, so let's assume that.
Now, how may configurations are possible?
If the 2 unused stamps are both black, then there are two black stamps among the six used stamps. they can either be on the same student or different students. If they're on the same student, there are 3 equally likely possibilities, and if there's on different students, there are 3 equally likely possibilities for the student without one, for a total of 6.
Ditto if the 2 unused stamps are both white.
If the unused stamps are black and white then there are three black stamps among the students. Either one student has two and another has 1, of which there are 6 permutations since each student has a different number of black stamps, or all three students have 1 each which contributes one additional for a total of 7. 6+6+7 = 19
If I or one of my fellow students see four of the same color stamps, we clearly know our own in the first round. There's no other way in the first round for a student to know their stamps, even with the knowledge that previous students could not identify theirs. As a result, there's no advantage for one seat or the other. If all the stamp pairs are homogenous, then exactly one student will solve in round 1.
In round two, all of the students now know there are not two same-colored pairs in play. NOW, if any student sees two pairs of DIFFERENT colors (one BB and one WW) then they know their own stamps -- they can't be a pair of the same color or the odd student out would have won in the previous round.
Student 1 wins in round 2, then if they see BB and WW in some order on students 2 and 3.
Suppose I'm the second student in round 2. If I see BB or WW on student 3 then I know I have BW. If I had that same pair then student 1 would have won in round 1, and if I had the other pair then student 1 would have won just prior.
So the only way I can't win in round 2 is if student 3 has BW (or somebody won before my 2nd turn.)
If I'm student 3 then and it gets to me in round 2, then I know I have BW because if I had anything else somebody else would already have won.
So now how many of the 19 configurations are won by each position?
When there are two BB or two WW present (6 cases: BB, BB, WW with WW in each position and BB, WW, WW with BB in each position) each position racks up 2 wins.
In round 2, position 1 wins if positions 2 and 3 are BB and WW in some order, and nobody else has already won. There are two configurations like this depending on that order.
In round 2, position 2 wins if position 3 is BB or WW and nobody else has already won. There are 12 of 19 positions like this, but ALL of the previous wins are among them, so there are only 4 additional win opportunities that open here.
In round 2, position 3 wins if nobody else has won. At this point, there are 7 such cases.
Overall, then, position 1 wins in 4 cases, position 2 wins in 6 cases, and position 3 wins in 9 cases.
As long as the cases are equally likely, I'd rather be in position 3. My odds are still worse than even money, but they're much better then the two alternatives!
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Posted by Paul
on 2024-07-10 17:35:05 |
A starting point?
