There are three people standing in a queue so that the last person can see both people in front, the middle person can see one person in front, and the first person can't see anyone. Three hats, chosen out of three red and two blue hats, are placed one each on their head. The last person is asked the color of hat on his head, and he says "I don't know". The middle person is asked the same and he also says "I don't know". The first person is asked the same question, and he says "I know". How?Answer: Since the last person says he doesn't know, the front two have at least one red hat (else the blue hats would be exhausted and the last person would know he has a red hat). Given that the front two have at least one red hat, the fact that the second person says he doesn't know means that the first person has a red hat (otherwise the second person would know he has a red hat).
Now assume that each of the three people can see the other two, and each has a blue or a red hat (there are no limitations on total number of blue or red hats). When the bell rings, each of them must simultaneously state their hat color, or say "pass". If none of them is wrong but at least one is right, all of them win a big prize. "Pass" doesnt not count as either right or wrong. What is the strategy to maximize their chances of winning?
Answer: 75% chance. All people act as follows: if they see two hats of the same color on other two people, they state the color opposite in color. If they see different color hats on other two people, they say "pass".
Instructor Notes: http://www.relisoft.com/science/hats.html Hint - lead them to understand that random guess has 50% chance so the idea is to choose a better strategy than that. Given that saying "pass" is always not incorrect, it should be possible. Next, draw all possibilities, and make kids realize that those can be partitioned in two sets (all three colors the same, and 2/1 split). By focusing on 2/1 split, can they solve the problem?
Homework problem: We are back to three black and two white hats, and every person can see the other two people. At every bell, whoever knows his hat color has to announce it. What will happen, and how quickly will everyone have announced their hat color?
Answer:
If there are two people wearing white hats, on the first bell, the third person will announce that he know color of his hat (black), and on the next bell, the other two, having seen what the third person said, will announce that they have white hats.
If there is one person with white hat, and two with black, on the first bell, no one will announce anything. On the next bell, people with black hat (who can see one white hat) will each announce that they have a black hat (otherwise someone would have seen two white hats and announced at first bell). On the third bell, the remaining person will announce the white hat.
If all three have white hats, no one will announce on first two bells, but all three will realize that everyone must have a white hat at the third bell (otherwise one of the top two scenarios would have happened), and everyone will simultaneously announce on third bell.
Note: The symbol "." is used as a multiplication sign below
Principal 1: If the thing we are counting is an outcome of a multistage process, then the number of outcomes is the product of the number of choices for each stage
(MC Chap 2, Prob 3) Three towns A, B, C in Wonderland. Six roads go from A to B and four roads go from B to C. In how many ways can one drive from A to C?
Answer: 6.4 = 24Prob 4 - Add town D, with 3 paths from A to D, and 2 from C to D - now how many paths? (Answer: 24+3.2=30)
(MC Chap 2, Prob 6) A natural number is odd-looking if all its digits are odd. How many odd looking 4 digit numbers are there?
Answer: 5.5.5.5 = 625
(MartinShCol - 1.1) A date is ambiguous if it can be validly written both as DD/MM/YYYY and MM/DD/YYYYY (for example, 08/09/2001 is ambiguous because it can mean 8th Sep or 9th Aug). How many ambiguous dates are there in a year?
Answer: 12*11 = 132 (Note that 03/03/2001 is not ambiguous so every month has 11 ambiguous dates)
(MartinShCol - 1.2) How many ten digit numbers can you construct using each of the digits 0-9 exactly once.
Answer: 9.9.8.7.6.5.4.3.2.1
Principal 2: If the thing we are counting can happen in different exclusive ways, then the number of outcomes is the sum of the number of outcomes through each way
Instructor Notes: Key is for students to understand when to add and when to multiply.
(MC Chap 2, Prob 10) Hermetian alphabet consists of three letters A, B, C. Words can have 1-4 letters. How many words are possible in Hermetian alphabet?
Answer: 3+3.3+3.3.3+3.3.3.3 = 120
(Shakuntala - 229) How many rectangles does a chessboard have (1296)
Answer: 1296One way is to try and count rectangles of each size and add them upAn easier way is to see that each horizontal line segment along a side of the chessboard (of different sizes 1, 2, 3...8) and each vertical line segment along the perpendicular side produce a rectangle. Hence number of rectangles is square of the number of line segments. Number of line segments along a side are 8 + 7+ 6+... +1 = 36. The number of rectangles is 36.36 = 1296
Instructor Note: First step is for kids to realize that there are rectangles of different sizes. Then they will start counting all, which is fine. Introduce the notion of how a rectangle is formed by two line segments perpendicular to each other
(Manas contributed this puzzle) Put mathematical signs between four 9's to make 100
Answer: 99+9/9 , 99/.99 (Here "." is a decimal!)
(Shakuntala - 12) At a party, every person shakes every other person's hand exactly once, and there are 28 handshakes. How many people are there
Answer: 8
Instructor Notes: Get kids to understand how to count these and not to double-count
(MC Chap2, Prob 13) In how many ways can you place a white and a black rook on a chessboard so that they can't capture each other?
Answer: 64.49What if both rooks are identical? Answer: 64.49/2
Homework Problem (MC Chap2, Prob 14) In how many ways can you place a white and a black king on a chessboard so that they can't capture each other?
Answer: 3612 (4.60+24.58+36.55)
Instructor Notes: First insight is that the number of squares attacked depends on the position of the king. Second, kids should correctly "add" the different scenarios.
References:
Mathematical Circles (Russian Experience), by Dmitri Fomin, Sergey Genkin, Ilia Itenberg
The Colossal Book of Short Puzzles and Problems, by Martin Gardner