So I didn’t *win* Monday Math Madness. For those of you not in the know, the blog WildAboutMath is having a rotating math problem contest with another website, and I put my hat in the ring. And I got the answer right. Sadly, the random number generator did not love me (click link for a few different solutions).
This week’s problem was:
A popular blog has just three categories: brilliant, insightful, and clever. Every blog post belongs to exactly one of the three categories and the category for each post is selected at random. What is the probability of reading at least one post from each category if a reader reads exactly five posts?
After solving it, I tried thinking of natural extensions. Clearly, one is: “if you have p posts and n tags: what is the probability that a reader reads at least one post from each category if a reader reads exactly p posts?” Or one could think about the more difficult question, where the frequency of each tag is different (not equal chance of stumbling across each post).
My submitted solution after the fold. You can see how hard it is to explain math clearly in an email.
Here’s my entry for Monday Math Madness…
I think the answer is: 150/243 (simplified: 50/81 or about 61.7%)
I arrived at the solution through a brute force counting method. Since the numbers were so small, I figured it would be the quickest way to do it.
First, because I’m incredibly boring and unimaginative, I let A stand for the 1st tag, B stand for the 2nd tag, and C stand for the 3rd tag. I saw there were a few possibilities, which I enumerated below. The number in the parentheses afterwards represents how many different ways you can arrange the posts to get that combination of tags (e.g. for AAAAB (5), that represents five posts which are tagged: AAAAB, AAABA, AABAA, ABAAA, BAAAA.)
Only ONE tag represented in 5 posts:
AAAAA (1) BBBBB (1) CCCCC (1)
Exactly TWO tags represented in 5 posts:
AAAAB (5) BBBBC (5) AAAAC (5)
AAABB (10) BBBCC (10) AAACC (10)
AABBB (10) BBCCC (10) AACCC (10)
ABBBB (5) BCCCC (5) ACCCC (5)
All THREE tags represented in 5 posts:
AAABC (20) BBBAC (20) CCCAB (20)
AABBC (30) AACCB (30) BBCCA (30)
So the final result is: (20+20+20+30+30+30)/243 = 150/243.
Of course, there are some things to note and clarify.
1. We didn’t need to enumerate anything but the last category (“all three tags represented in 5 posts”) because it is clear that there are going to be 243 total possibilities. (If there are 3 tag choices, and 5 posts, then there are going to be 3^5 different combinations of posts.) Or you could enumerate everything except the last category. But I like counting things; it’s cathartic. So I did them all.
2. The question that any mathlete would ask is: how did you get the numbers in the parentheses? The answer is:
For “one tag in 5 posts”, we see that we are chosing only 1 tag, and the rest have to be the same. So it is 3C1 (3 choose 1). That gives us a total of 3 different possibilities [3 columns above].
So that total is 3.
For “two tags in 5 posts”, we see that we are choosing 2 tags from 3, so we have 3C2 (3 choose 2). That gives us 3 possibilities for tags combinations [3 columns above; posts with tags A and B, posts with tags A and C, and posts with tags B and C].
However, let’s say that we have chosen the tags A and B. How many ways can you arrange them in 5 posts? The answer is 5C1+5C2+5C3+5C4 [5+10+10+5]. This is because the first number (5C1) represents the number of ways you can have one post with tag B with all the rest As. The second number (5C2) represents the number of ways you can have two posts with tag B with all the rest As. The third number (5C3) represents the number of ways you can have three posts with tag B with all the rest As. And the fourth number (5C4) represents the number of ways you can have four posts with tag B with the remaining one A.
So the total number is 3*[5+10+10+5]=90
And since we did all that work, and we know the total different possible number of posting those 5 blogs is 243, we have (243-90-3)/243=150/243.
(Or you could just count up the different ways to have all 3 categories, as I did above. You get 150 that way too.)