CTY: My Envelope Has More Than Your Envelope

This is my last week at CTY. In class, we looked at this problem:

Someone hands you an envelope and asks you to look at the contents (let's say it is some dollar amount). They then say that they are holding a second envelope that has either double of yours or half of yours. You must decide whether to switch envelopes or keep the one you currently have.

To solve this mathematically, you would use the expected value formula we used last week. Say your envelope has X dollars. You can either:

EV (Stay) = X
EV (Switch) = 1/2(1/2X) + 1/2(2X) = 5/4X

In other words, you will make 25% more money on average by switching. However, that is not realistic in these situations.

You must examine the problem from a logical standpoint. If you have a very generous person offering this deal and showing an envelope with $10, it might be worth the gamble to switch. If a more conservative person offered the same deal, you would be better off staying.  I found this problem cool because once again, math failed to give a reliable answer.

Bonus: Here is another problem to solve. I will give the answer in a month.

There are a group of monks who all vowed not to communicate to one another in any way (speaking, codes, sings, etc.). Every morning, the monks all gather in a circle and the head monk speaks to them.  One morning, the head monk said that there were sinners among them. He waved his hand and a mark appeared on all of the foreheads of the sinners (everyone knew who had marks, but could not see if they had one). He asked for anyone who knew they were a sinner to leave.

The second morning, the head monk announced that there were still sinners left and put the marks back. Once again, he asked for anyone who knew they were a sinner to leave.

The third morning, the head monk announced that there were still sinners left and put the marks back. For a third time, he asked for anyone who knew they were a sinner to leave.

The fourth morning, the head monk announced that all of the sinners had left.  How did the sinners know to leave?

Answer: A month ago, I posted the locker problem where student one opens every locker, student two closes every even locker, student three opens/closes every third locker, and so on up through 1000 students. The question was which lockers remain open.

The answers are:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961

There is a pattern among these 31 numbers: they are all square. The reason for this is that every non-square number doesn't just have factors, but they have pairs of factors. For instance, six has factors:

1 x 6
6 x 1
3 x 2
2 x 3

The six and the one can be written twice, giving it two factors. Therefore, six will get opened by 1, closed by 2, opened by 3, and closed by 6. However, a number like 9 has factors:

1 x 9
9 x 1
3 x 3

Here, the three cannot be written twice since it is paired with itself. So, locker nine will get opened by 1, closed by 3, and opened by 9. The same logic applies to all of the lockers.