Since Justin kept on nagging me I’ll post the solution now. The puzzle is reproduced below:
There are three magicians performing a fantastic trick. The trick is as follows: They stand in a circle with their eyes closed. Then 3 audience will put the number 1, 2, or 3 on each of their forehead. Each magician then open their eyes, and see the number on the forehead of the other two magicians, but not their own number. Without any communication, they each write simultaneously on separate sheets of papers what they think their numbers are, and then reveal the 3 papers at the same time to the audience.
Then BAM!! AT LEAST ONE of them will be right about his number. How can this be??? How can it possibly be that you look at 2 random numbers and make some conclusion about another random number???
Short, concise, rigorous, mathematically sound answer: Consider the number formed by the digits. For example, if the audience put the numbers 1, 2, and 2, consider the number 122. The sum of the digits of this number modulo 3 can only have 3 possible outcome, so each magician simply writes down one possibility of the digit sum.
Ok, chances are you don’t understand the solution above, so here’s a more elaborate answer:
So let’s say the audience puts down the digits 1, 2, and 2 on magicians A, B, and C, respectively. so in order for C to answer correctly, he’ll open his eye, sees 1 and 2, and then writes down 2 as his answer.
SO let’s say before the performance, A, B, and C get together, and they agree that whenever C sees the number 1 and 2, he’ll always write down a 2. That way, if the audience happens to write down 1, 2, 2, the performance will be a success.
Note that because of the symmetry, if the audience write down 2, 1, and 2, the performance will still be a success!! Because C sees the numbers 2 and 1 (although on different persons now), so he’ll still write down the 2 and guess correctly.
But you say, if we do it like this, it’ll only work if the audience write down 2,1,2 or 1,2,2!! How about other numbers?
Well, how many other numbers are there? There are 111, 222, 333, 112, 113, 122, 223, 133, 233, and 123 (ignoring permutations). That’s not too many, certainly we can find a rule that makes A, B, and C always guesses correctly.
For example, let’s say A follows the rule: Whenever I see 11, 22, 33, I write 1, 2, and 3. Otherwise, whenever I see 12, 23, 13, I write down 3, 1, 2, respectively. Then A will always guess correctly when the audience write down any one of the numbers 111, 222, 333, 123, or their permutations. So we know for sure that A will get the answer right for 4 of the 10 cases. Doing the same thing for B and C we conclude the following as the golden rules, where 11-1 means if you see 2 1s, write down a 1:
A: 11-1, 22-2, 33-3, 12-3, 13-2, 23-1 (covers 111, 222, 333, 123)
B: 11-2, 22-3, 33-1, 12-1, 13-3, 23-2 (covers 112, 133, 223)
C: 11-3, 22-1, 33-2, 12-2, 13-1, 23-3 (covers 122, 113, 233)
Following these rules will guarantee that at least one of A, B, or C guess his own digit right. Try it out!!
Now if you want to understand the short solution, note that A always answer the number that makes the digits sum to a multiple of 3, B always answer the number that makes the digits sum to 1 + a multiple of 3, and C always answer the number that makes the digits sum to 2 + a multiple of 3.
