Probability
Over at Math Notations, Dave recently posed a variation on a cube dissection problem. Here’s another:
A cube is painted on the outside and then cut into 27 equal cubes by dividing each edge in three with four planes which are parallel to the faces of the original cube.
The cubelets are placed in a bag. One is removed at random and tossed. The face on top is unpainted.
If that same cubelet is tossed again, what is the probability that an unpainted face will appear on top again?
Last Friday Blinkdagger announced a winner for MMM #8. Here’s MMM #9:
Consider all of the 6-digit numbers that one can construct using each of the digits between 1 and 6 inclusively exactly one time each. 123456 is such a number as is 346125. 112345 is not such a number since 1 is repeated and 6 is not used.
How many of these 6-digit numbers are divisible by 8?
While you may use a computer program to verify your answer, show how to solve the problem without use of a computer.
On March 3rd Blinkdagger and I posted the first Monday Math Madness problem. On March 11th, after the first contest ended, I posted a couple of different solutions to the problem. Pat Ballew, even though he wasn’t picked as the random winner, impressed me with a very clever solution to the problem that generalizes very nicely. He uses an approach called Markov state matrices, which I had never heard of. It seems to me that this approach is pretty similar to the one I posted from Richard Berlin. Pat and I exchanged several emails where he explained the method and here is my attempt to explain what Pat explained to me.
This was the problem:
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?
Pat’s approach starts by creating a matrix that encodes the probabilities of going from one “state” to another as a new blog post is read. State just refers to whether 0, 1, 2, or 3 categories have been encountered after reading some number of blog posts. After one blog post has been read we are in state 1 (1 category has been read). After two posts have been read we may be in state 1 (if both blog posts are in the same category), or state 2 (if the two categories are different), but not in state 3 (you could not have encountered three categories after having read only two blog posts.)
For the very first Monday Math Madness contest we got 13 submissions. Of the 13, 6 were correct. For the record, I solved the problem by enumerating the various cases where 3 categories were represented and computing and adding their probabilities. I also verified my solution to the problem by writing a computer program to enumerate all 243 (3^5) permutations of 3 categories and 5 blog posts and count the ones were all 3 blog categories were represented. So, I’m pretty confident I got the right answer 
I teach combinatorics as a one-term elective to high school seniors (and a few juniors). We use Ivan Niven’s “Mathematics of Choice - How to Count Without Counting” for two marking periods. But we’ve reached a turning point. Two-thirds of the way through, we drop counting and turn our attention to probability, then games and expected value. We will finish with some analysis of rou- and I break the word up -ll- into three pieces -ette.
Since I have had nothing to post, but since I wrote a nice little probability worksheet, I thought I’d post that. To see the questions, click —->
