# Reframing as a fundamental tool of creativity – lessons from a game of backgammon

Reframing a situation remains one of the best ways to see a new solution in strategy. As I said a few weeks ago, because strategy occurs in the domain of “wickedness”, the way you choose to describe the strategic problem you’re facing will drive the solution you come up with.

But reframing is a fundamental tool in creativity as well. I reckon this can best be seen in the world of mathematics. I know this is counter-intuitive, but I’ll keep saying it – a good mathematician is highly creative. All elegant proofs have some great “creative move” where the problem is “reframed” from another perspective. That reframing then makes the problem solving easy.

#### Reframing in backgammon

I was playing backgammon with my 14 year old this weekend. Zoe was trying to work out the odds of being hit if she left a counter exposed one place from mine. Remember that in backgammon you roll two dice and you can move your counter the number of spaces shown on the first die, and another counter (or the same one) the number of spaces shown on the second die.

So, if I am one space away from Zoe’s exposed counter, what are the chances of my hitting her? Well, I have to roll a “1” on one of my dice. What are the chances of that? Well “1 in 6”. So that seems like good odds. It might be worth the risk for Zoe to leave her counter exposed.

But of course I get two die rolls. I get two attempts at the “1 in 6” chance. What does that do to the odds?

Well, it was getting really confusing as Zoe and I talked back and forth about this. So, instead, what we did was “reframe” the problem.

Zoe is familiar with the X-Y cartesian plane from mathematics. We decided to call one die the “X die” and one die the “Y die”. So, when you roll the two dice you now get an “ordered pair”, a coordinate on the cartesian plane. At this point, it becomes really easy to work out the odds of being hit:

Clearly there are 36 possible points on this plane (6 x 6). And the dark circles show every point with either an X-coordinate of “1” or a Y-coordinate of “1”. There are 11 of these dark circles. The chances of being hit are 11/36, or just under “1 in 3”. Not good odds.

Look what we did. A simple reframe allowed us to see the whole problem in an entirely new way. It then became almost trivial to work out the odds. A classic mathematical move.

#### Archimedes’ classic reframe

One of the most famous reframes in the history of mathematics is Archimedes’ reframe to work out the surface area of a sphere. The classic definition of a sphere is “The locus of points at a given distance from a given point”. But Archimedes came up with a much more useful definition of a sphere. He saw it as “The surface described by a circle revolving around a diameter” (I am quoting from Polya’s 1945 book “How to solve it“).

Archimedes reframed the problem and chose to see the sphere as a circle spinning on its axis in space. By seeing the sphere as a circle spinning in three dimensions Archimedes can then imagine a regular polygon, inside the sphere, spinning with it. And in this way he is able to work out the sphere’s surface area in an era before calculus. There’s a series of videos here, describing the exact details of this proof if you have twenty minutes and the interest!

You can get a good sense of the approach to the problem by watching the first minute of this video:

#### Nathan’s classic reframe in “X+Y”

A couple of months ago, I described Nathan’s reframe in the movie “X+Y”. He is faced with a problem as follows. You have a sequence of cards, all face down. A permitted move on this sequence is as follows. You can turn one face down card, face up. But you must also turn over the card immediately to the right of it (if that card was face up, it’s now face down, and vice versa). Show, that if you keep applying this move to the sequence that eventually you will have to stop.

Nathan makes the “reframe move” of seeing the cards, not as cards, but as a sequence of 1s and 0s. He then uses the principles of binary arithmetic to show that the sequence will stop eventually.

The clip is worth watching here:

#### Over to you

Where have you reframed a problem to see a totally new solution? Leave a comment below. As always, if you are enjoying these posts, please subscribe by pressing the button in the top right corner.