Tangled up in red

The much anticipated day back at work didn't happen. A slight relapse is going to mean a trip to the doctor and some drugs. Hopefully that'll sort me out.

It's an ill wind, and all that, and the extra day off has given me time to look at something that I only found out about relatively recently and think is probably worthy of an enrichment lesson. But more on that later, as it didn't take long before I was once again caught up in Entanglement.

Despite a good few hours of my day wasted invested playing the game, I don't seem to be making much progress. My top score so far is a paltry 348, which does at least put me 78th on the leader board. For today. So far. I'm trying to set a target of scoring at least 200 points, a target I have met. Once. Which is starting to make me question whether the act of setting targets alone is enough to improve performance. Who knew?

The main problem I'm having is which strategy to use. Okay, the main problem is that I'm rubbish, but the main problem that I can do something about is which strategy to use. I've tried several: inside to out, outside to in, prioritising making long chains and trying to attach to them, and prioritising joining up loose ends.* But all of those seem inferior to my current "try something different and see if you can stumble upon a better strategy". Well, I say inferior... the current strategy hasn't actually produced any results yet. But I'm optimistic.

All of this makes me wonder whether we should be taught more Euclidean geometry in school. I remember lots of algebra and statistics, but not many geometric proofs. Which reminds me of an interesting probability proof that uses similar triangles. Oh, if only Blogger would let me include PDF files I could show you! Maybe that's a good thing... you could try it for yourself. Here's the problem:

You are playing a game which involves spinning two coins. If they are both the same, you win; if they are different, your friend wins. If the coins are fair, then so is the game. But what if the coins aren't? For the sake of simplicity, let's say that they have the same bias, and in favour of the same side (hmmm... maybe I should have started by saying there was one coin, which we spin twice). Does the game now favour me or my friend?

There's a fairly simple algebraic proof which the absence of a superscript button on the formatting toolbar prevents me from demonstrating (Wow! Now I know just how Pierre de Fermat felt!) but can you come up with a geometric one? Go on... give it a try!

* This might make more sense if you've played the game.