Puzzle 7

Problem 1 (Related to Dr. Rossmo's talk): Recently, a rare disease called "calculitis" has begun to afflict freshmen at universities and colleges across the United States. Fortunately, the disease can be treated if it is detected early. One way to determine whether a freshman has calculitis is to apply a chemical to his/her scalp; this chemical should cause the patient's hair roots to turn purple if he/she is infected. (This test is called the root test. The root test has a 98% chance of returning a correct result if the patient is not infected, and a 99.9% chance of returning a correct result if the patient is infected. Approximately 0.2% of college students become infected with calculitis during their first year of college. Answer the following questions, explaining each of your answers carefully:

(a) Suppose we choose a freshman at random and have him/her take the root test. What is the probability that he/she has calculitis, and the test determines that he/she has calculitis?
(b) Suppose we choose a freshman at random and have him/her take the root test. What is the probability that he/she does not have calculitis, but the test determines that he/she has calculitis?
(c) Suppose we choose a freshman at random and have him/her take the root test. Assuming that the test determines that he/she has calculitis, what is the probability that the student actually has calculitis?

Solution (Submitted by Rina Sadun):

a. There is a .001998 chance that a randomly selected freshman will both test and be positive for calculitis. The probability they will have it is .002, the probability that given that they have it, they will test positive is .999, and when multiplied (to get the probability that both will happen) the product is .001998.

b. There is a .01996 chance that a randomly selected freshman will falsely test positive for calculitis. The probability they will not have it is .998, the probability that given that they don't have it, they will test positive is .02, and when multiplied (to get the probability that both will happen) the product is .01996.

c. There is approximately a .09099 chance that a randomly selected positive-testing freshman will actually be positive for calculitis. There is a .001998 chance of both testing and being positive (see a), and using the same method one can find that the chance of testing but not being positive is .01996. The sum of these (the total probability of testing positive) is .021958. The chance of being positive given a positive test result is .001998/.021958, which is about .09099.

(Many thanks to Rina Sadun for allowing us to post her well-written solution on our website.)

Puzzle 5

Problem (also from Dr. Starbird's book, "The Heart of Mathematics") : Let S be the set of all real numbers between 0 and 1 with the property that their decimal expansions only have 0's and 7's. For example, the following numbers are elements of S:

0.7777007707070777707000....

0.00000000000070000077777770000007....

Is the cardinality of S equal to the cardinality of the set of natural numbers? Why or why not?

Solution: The answer is no, they do not have the same cardinality. Remember that not all infinities are the same! This problem is actually very similar to the Dodge Ball game we played near the end of the lecture.

The best way to think about it is in terms of one-to-one correspondence. If they have the same cardinality, then you should be able to match up each natural number with a number in the set S. Now let's imagine what would happen is there WAS a way to do this. For example:

0.0700777007770000000.......
0.0070000777777777700......
0.7777007070707700070....
0.7707000707777700070....
0.0707077770007007007....

etc...

So we are assuming that the list I just gave is complete, that is, every natural number is paired with a number in S and viceversa (for every element in S there is a natural number that corresponds to it).

Can you see what happens? I can make a number that is in S but is not on the list in the following way: Take the first decimal to be the "opposite" of the first decimal in the first row, so 7. Take the second decimal to be the "opposite" of the second decimal in the second row, so 7. Take the third decimal to be the "opposite" of the third decimal in the third row, so 0. And so on... I would get a number that looks like this for the first few decimals: 0.77007....

Now, imagine, because we are doing math, that I do this infinitely many times. Then the number I just constructed is different from all the other numbers in my list, since it's different from the first one in the first decimal place, from the second one in the second decimal place, etc.

But this means I was wrong! I didn't have a complete list in the first place, because I found a number that wasn't on that list. Now, the last step is very important, because all I did is show that in the example above this doesn't work. How can we know that there is no possible way to write a complete list?

Well, no matter what list you give me, the strategy I just described always works (just like in Dodge Ball player two always wins). So there can't be a one-to-one correspondence between S and the natural numbers, since I can always find a missing element in the pairing.

Puzzle 4

It's in the box: There are two boxes: one marked A and one marked B. Each box contains either $1 million or a deadly snake that will kill you instantly. You must open one box. On box A there is a sign that reads: "At least one of these boxes contains $1 million." On box B there is a sign that reads: "A deadly snake that will kill you instantly is in box A." You are told that either both signs are true or both are false. Which box do you open? Be careful, the wrong answer is fatal!

Solution: One important thing to notice is that the statement "Each box contains either a million dollars or a deadly snake" is not the same as "one box contains a million dollars and the other one contains a snake". There are two possibilities for the contents of each of the boxes, so there could be two snakes or two boxes with a million dollars, this is part of what you need to think about.

Clearly both signs being true makes sense, but there's no reason (yet) that they HAVE to be true. Now what if they're both false? That would mean that the sign on A is false, which means that neither box contains a million dollars. This in turn would mean that they both have snakes in them (remember there are only two choices for what could be inside). But the sing on B is also false, and so we would know that there isn't a snake in A. This doesn't make any sense, that is, it's impossible for both signs to be false, and so both signs have to be true. Now it's clear that you should open box B.

Puzzle 3

Magic Triangle Problem: When you place the numbers 1-9 in the given circles so that the sum of the four numbers along each side of the triangle is the same, what values are possible for this common sum?(There can be more than one answer) Justify.

Puzzle 2

Problem (by M. Gardner): My wife and I recently attended a party with four other married couples. At the beginning, various handshakes took place. No one shook hands with himself or herself, or with his or her spouse, or with somebody more than once. After all the handshakes were over, I asked each person, including my wife (a total of nine questions to nine people), how many hands he or she had shaken. Each gave a different answer. How many hands did my wife shake?

Solution: You should notice that the nine people that are not the narrator have to have shaken from 0 to 8 hands. Now you should think about how these should be paired up. That is, if someone shook exactly 8 hands, then they shook everyone's hand, and therefore their spouse has to have shaken none (can you see why?). So one couple has a person that shook 8 hands and a person who shook 0 hands. This way, you can figure out that the person who shook 7 has to be married to the person who shook 2 hands, etc. As you go down the possibilities, you realize that there is a couple in which both people shook 4 hands. Since the narrator said everyone (not including him) shook a different number of hands, you know that he and his wife have to be the couple who shook the same number of hands. That is, his wife shook four hands.

Challenge Problem: Suppose that you have a group of people in a room. Some have met each other before, some of them have never met. How many people should there be in the room so that you can be sure that either 3 of them know each other or four of them have never met? Explain your reasoning. (Hint: you can think of this problem in terms of pictures.)

Puzzle 1

Problem: How many sets (including overlapping ones) are possible in the deck of SET?
Solution: The first thing you need to think about is that there are 81 cards total in a deck of set. So if you were going to pick three cards from the deck so that they form a Set, this is how you would do it:

You can pick any card as the first card, so you have 81 choices.
For the second card there are really no restrictions either, since any two cards form a set with a unique third card. So there are 80 choices for your second card.
In step 2 we said that there could only be one third card, and so we only have one choice.

This way we get 81*80*1 possible sets. Now, it is important the we don't count the same set more than once. With the method described above, we are getting that ABC and CAB are different sets (since we are picking the cards in a specific order, but of course order doesn't matter here). There are 6 possible ways to change the order of three cards without changing the set. So our final answer must be 81*80*1/6=1080.