# 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)

# 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 🙂