If you had to pick a size of spiritualist, what would you go for?

The Art of Uncertainty, by David Spiegelhalter, someone who I have long admired. I watched an Oxford Mathematics Public Lecture Chance, luck, and ignorance on YouTube and really liked the content and the way he presented it.

Early in the talk he quotes an Italian statistician de Finetti who famously declared "probability does not exist". Spiegelhalter goes on the say that after 50 years of mulling things over he has decided that Finetti was right! I don't think he is claiming that we can't attach numbers to whether we think some event will happen or not, only that probability (with the possible exception of subatomic particles) probability is not a property of the world but something humans have made up, a "personal expression of our uncertainty".

By this point he had won me over and I bought the book. The lecture, which I watched right through was fascinating. Towards the end he gave an arresting example using a pack of playing cards as a prop.

Randomly shuffle a pack of cards*. Now the order of cards has almost certainly never, ever happened before in the entire history of humankind. If every person ever born had lived to be 100 and shuffled a pack of cards every second of those 100 years the chance of any of them having generated the same order as the pack you hold in your hands is infinitesimally small. Here are the sums:

Estimated number human who have ever lived 120 billion

Seconds in 100 years about 3 billion

So all those people shuffling away could have produced only(!) 120,000,000,000 x 3,000,000,000

= 360,000,000,000,000,000,000 = 3.6 x 10²⁰ shuffles. Let's call it 4 x 10²⁰ to simplify the arithmetic coming in a moment.

But the total number of ways to order the pack is 52 x 51 x 50 x ... x 3 x 2 x 1 = 52 factorial, which my calculator says is about 8 x 10⁶⁷.

That means the chance that one of those people generated the same order as you is

4 x 10²⁰ ÷ 8 x 10⁶⁷ which is 1 in 4 with 47 zeros after it

It's theoretically possible but the chances it will are almost unimaginable small. 

* To do this properly you need to do at least 7 successive riffle shuffles or 1 minute of "smooshing", see the fascinating Numberphile video The Best (and Worst) Ways to Shuffle Cards. Anything less than this the pack is probably not truly randomised.

I came across this crossword clue

enclose : indeterminate number of clothes

Previously I posted a well known problem which asks: suppose we are on a circular boating lake when fog descends. We know the size and shape of the lake and can travel any distance and in any direction we please, but we don't know where we are on the lake, so we are rowing blind. What is the shortest distance we must be prepared to row to be sure of escaping from the pond. and what path should we follow?

The answer is that you should row in the straight line until you hit the bank and may have to travel a distance equal to the diameter of the pond, and this cannot be improved on. My proof is here. I actually found this for myself some years ago but it turned out to have already been published 40 year before. so I was late to the party.

A follow up question was: what if we are on a pond in the shape of an equilateral triangle? The diameter of an equilateral triangle is just the length of one of its sides and this is the least escape distance if you keep to a straight line. But rather surprisingly (well I think it is) a zig-zag escape path exists which is shorter, at about 0.98198 of the side length.

Here are two pictures of the path in different positions and you can probably see how it manages to always reach the perimeter of the triangle even though it is shorter than the diameter.

neutron : Ronald the newt

electron : vote for Ron!

boson : ship's officer

quark : duck noise

photon : imitation ton

particle : average tickle

To reiterate this question, imagine we are rowing on a round pond 200 m across when suddenly fog descends and visibility is effectively zero. You are lost and unfortunately the fog fell too quickly for you to have any idea of your location relative to the pond's edge. What is the shortest distance you can row to guarantee reaching the edge of the pond and what path should you follow?

To begin with, any straight path of length 200 m must be a diameter of the pond. Therefore, if we choose an direction at random (we do not know where the shore lies relative to our location so one direction is as good as another) and travel in a straight line, we are certain to reach the bank after at most 200 m. So this is am escape path.

It's natural to wonder if some ingenious non-linear path might guarantee hitting the pond's edge in a shorter distance, but we can show this is impossible.

Here's the proof.

Suppose we can guarantee reach the shore in a distance of less than 200 m by following some path starting at A and ending at M and let M be the point half-way along the path, so the path length for M to A and to B are both less than 100 m.

Consider an arbitrary point X somewhere along the path. The distance along the path from M to X must be less than 100 m and the straight line distance from M to X must therefore also be less than 100 m. Since X was an arbitrary point it follows that every point on the path is less than 100 m from M and we can draw a circle with centre M and radius less than 100 m.

But this is smaller than the pond, which has a radius of 100 m! So it's possible for the whole of AB to lie within the interior of the pond and it therefore cannot form a guaranteed escape path.

In a comment, Steven McDonald already gave a nice solution to the puzzle here. My own solution is equivalent but framed a bit differently; I went for a more visual explanation.

Each of the rectangles we want to count is determined by some two rows and two columns of the 10x10 grid, as shown in this example:

For the rows we have 10 choices for the first, which leaves 9 choices for the second, but that will have double-counted each combination, because for example row 5 and row 9 represents the same pair as row 9 and row 5. So we must divide by 2 and there are

(10 x 9) / 2  = 45

distinct combinations.

Similar reasoning gives 45 of ways to choose two distinct columns and so we find there are 45 x 45 = 2025 different rectangles.