# Vintage Cube Stats Puzzle

Supposing that Power 9 cards are always picked over non-Power 9 cards, what is the chance you never see a piece of power in the drafting phase of a Vintage Cube event?

This puzzle probably doesn’t make a whole lot of sense to people who don’t play Magic, and even then it requires a bit of knowledge about the Vintage Cube draft format.

That’s why I rewrote it in a different context and posted it here to the Math Riddles subreddit.

It’s more of problem than a riddle (and at the time of writing this entry the post is sitting on 3 upvotes, 2 downvotes, ouch) but I know that subreddit has a great community of problem-solvers so figured I’d post it there.

I have tried to solve this “by hand” but it seems to run into a bit of complexity, which I’ll briefly explain:

First of all, we know that if you have at least one piece of Power in any of your starting boosters then you’ll see one. That probability is easy enough to calculate. But we can’t just do 1 minus this probability to get the answer to the puzzle, because even if we don’t start with a piece of Power in our boosters, the person on our left or right might have two or more pieces in the starting booster they pass to us and so we would see one of those. We therefore need to also include the probability that one of those three boosters contains two or more pieces given that none of ours contained any pieces. That’s not too tricky…

…but then we need to think about the three starting boosters that are two passes away from us. If any of those have three or more pieces of Power in them then we’ll see one of those pieces. So we include the probability that one of those three boosters contains three or more pieces given that none of ours contained any pieces, and none of our neighbours’-starting-boosters-passed-directly-to-us had two or more pieces. That conditional should be just a case of (1 minus the two probabilities we already calculated) * (probability there are three or more in one of those packs). However, that “three or more” probability changes depending on whether our neighbours packs had one or zero pieces of power. We have multiple cases to consider.

Then when you think about three passes away from us and four or more pieces you run into even more cases to consider, and so on.

I think I could do a good job of this with dynamic programming, or perhaps an alternative inclusion/exclusion approach, but it’s actually where I decided to write a simulation instead to get an idea of the answer more quickly!

Here you can look at and execute the code I wrote. The output gives an estimate for the probability that you see at least one piece of power. So to answer to the original question subtract that from 1!

(Psst! It’s ≈0.24)

# John Polkinghorne’s Erdös Number

John Polkinghorne is a neighbour of my parents and, very tenuously, a relative.

He was active in Cambridge around the time of events in The Theory of Everything, and knew many of the people in it, including Hawking himself. Since getting to the cinema is a little hard for him now, a few months ago my parents invited John over to watch the DVD and eat Sunday lunch. I attended too, as chief video operator (!).

It was really fun, and interesting to hear John’s insights into the film and the people it featured.

The previous time I had spoken to John at a meal I had brought up Paul Erdös because I knew John had met Erdös, from an anecdote in John’s autobiography. Erdös was in Cambridge to receive an honorary degree and John was one of the people helping him. Erdös was expected downstairs by 9:30am but hadn’t appeared yet. John went to investigate and found him in his room still not quite ready. He asked John if he thought he should wear a tie, and John said that yes, he probably ought to, but this posed a problem because Erdös had never tied one before and didn’t know how! John had only tied his own ties so doing so on someone else was counterintuitive as everything would be mirrored. I guess you’ll have to read the autobiography to find out how the story ended.

Anyway…I was pretty sure that John was aware of the concept of Erdös numbers, but didn’t know his own. So I decided to find out what his Erdös number is. This proved quite difficult and I got in touch with Jerrold Grossman from The Erdös Number Project to help. He passed me on to Chris Fields, who was able to give me a chain of 5 co-authors. I did the googling to determine exactly which papers these were and made the following information sheet.

Click the image to enlarge.

I gave the sheet to John on the Sunday we watched The Theory of Everything, and he was really pleased with it. John indeed hadn’t known his own Erdös number! He even said “That’s jolly clever of you Jim to find that out”, which gave me cause for a wry smile.

# The Lonely Runner Conjecture

The Lonely Runner Conjecture states that if K runners set off at constant different speeds to run laps around a 1 mile circular track then for each runner there is some time when she is at least 1/K miles from all the other runners.

I made a little simulation to help visualise this.

Try it! (Click the image)

As the wikipedia page shows, the conjecture is proven for K up to 7 (which was shown in 2008), but beyond that we don’t know!

I posted the problem for K = 4 to the Math Riddles subreddit because that place is cool and I like it.

Unfortunately, I posted it slightly wrong by stating “more than” instead of “at least”. This was picked up in a particularly thorough answer 🙂

# Triangle Chop

• Take two random points inside a triangle.
• Join the points with a straight line and extend it to the triangle’s edge.
• Cut along the line.

How many pieces do you get?
…What shapes are they?
…On average, what is the ratio of their sizes?

Try it! (Click the image)

(Original problem)

# Wednesday Maths Problems 2013-03-13

Insert \$20 and a machine will generate random numbers between 0 and 1 until their total exceeds 1. It will then dispense \$7 for each number it generated.

Try it with fake money.

Should you play in real life?

You know the line.

Tomorrow is Pi Day in the US. When might a pedantic Englishman celebrate Pi Day?

# For what m and n can a regular m-gon be inscribed in a regular n-gon?

One day as an undergraduate I visited my supervisor with a problem that had crossed me in an otherwise unrelated assignment.

I showed him on the whiteboard how an equilateral triangle can be inscribed in a square and how a square can be inscribed in a regular pentagon. Both used the intermediate value theorem with given pairs of triangles or rectangles. You can probably see where I went with that. There may be much nicer ways to show it, but that was what I used at the time.

We casually discussed for a few minutes whether a regular pentagon could similarly be inscribed in a regular hexagon, a regular hexagon in a regular heptagon, and so on, coming up with very little. He showed me another problem involving inscribing and regular n-gons but his one had circles and a line and a constant that nobody knows. I left, my initial puzzle still a mystery.

The question has needled me over the years and become something I have thought about when waiting in queues, at bus stops, and on station platforms. The formulation has generalised in my mind to the set of values for which a regular m-gon can be inscribed in a regular n-gon, rather than my initial n and n+1.

The obvious results sprang out at me, for example in the case when m divides n, but never a fully comprehensive answer. Frequently I would search the internet, sure that there would be a complete solution with elementary techniques that I had simply overlooked.

Today I was going to post the problem on this newly-born blog, as a nice start to set the tone of the posts, but decided of course to have one more check online first.

Lo and behold: Solved, by an S. J. Dilworth of the University of South Carolina.

Perhaps it is not surprising I had not found this before, it was only published in August 2010.

I have spent a large portion of today reading the paper.