*Motivation*

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## September 12, 2017

### The secret of success?

This TED talk makes a lot of sense to me, and chimes in with more than 35 years of lecturing experience. Have a look and see what it suggests to you.

### Secret–sharing and independence

The following remarkable procedure is entirely feasible: if I have a class of *n* students then I can distribute *n* different binary images, one to each student, such that each student's image looks like white noise,

and yet if * all* images are combined together in the right way then a meaningful picture emerges.

What's more, I can arrange matters such that if any strict subset of the *n* students tried to collaborate, then all they would get would be more white noise, no matter how they manipulated their *n*-1 images!

So any strict subset of the students would possess no information at all about the butterfly picture, but combined all together they would be in a position to produce a perfect reproduction of the image.

How can this be done?

- Code each black pixel of each image as -1, each white pixel as +1, and view each distributed image as a long vector sequence of +/-1 values.
- Let
*X*_{0}be the vector encoding the target image (the butterfly above). Generate entirely random independent vectors*X*,_{1}*X*_{2}, ...,*X*_{n}_{-1}*n*-1 students.

- Student
*n*is given an image corresponding to the vector*X*obtained by multiplying (coordinate-wise) all the other vectors:_{n}

*X*_{n}=*X*_{0}**X**_{1}*X*_{2}* ... **X*_{n}_{-1 }where "*" denotes coordinate-wise multiplication. - It is simple arithmetic that
*X*_{0}*=**X*_{n}**X**_{1}*X*_{2}* ... **X*_{n}_{-1}. So all students working together possess the information to recover the butterfly image.

- On the other hand one can use elementary probability to show that, if one selects any subset of size
*n*-1 of the vectors*X*,_{1}*X*_{2}, ...,*X*, then this subset behaves as if it is a statistically independent collection of vectors corresponding to white-noise images. (It suffices to consider just one pixel at a time, and show that the corresponding sequence of_{n}*n*-1 random +/-1 values obey all possible multiplication laws.) So no strict subset of the students has any information at all about the butterfly image.

There are many other ways to implement secret-sharing (Google/Bing/DuckDuckGo the phrase "secret sharing"). But this one is nice for probabilists, because it provides a graphic example of why *pairwise independence* (independence of any two events taken from a larger collection of events) need not imply complete independence.

## July 04, 2017

### An old advertisement for the ST318 "Probability Theory" module

In your third year at Warwick you often have a drink with a friend who enjoys probability (yes, you make some strange friends here). On the Sunday evening before the start of Term 2 she challenges you to a curious game of chance. Hiding a fair six-sided die (numbered 0,1,2,3,4,5) from your observation, she throws it ten times and secretly notes each outcome on a convenient beer mat. Then she turns to you and announces:

"I shall now read you the running averages of the dice throws, but in reverse order.''

(You must have looked puzzled, because she breaks off to explain)

"Let *X*_{n} be the result of the *n*^{th} throw, so the *n*^{th} running average is *Y _{n}*=(

*X*

_{1}+...+

*X*)/

_{n}*n*. I will read out the numbers

*Y*

_{10},

*Y*

_{9}, ...,

*Y*

_{1}in that order.''

Clearly a flicker of comprehension must have somehow crossed your face, because she now continues:

"At any stage you can stop me reciting the sequence. If the last average I read out was *Y*_{n} then with probability *Y*_{n}/5 (arranged using that book of random numbers which I carry everywhere) I will buy you a pint and the game ends. If you don't stop me at all then clearly you get a pint with probability *Y*_{1}/5. Otherwise you buy me a pint.''

You calculate rapidly: **E**[*Y*_{1}/5]=**E**[*X*_{1}/5]=(1/6) (0+1+...+5)/5=1/2 even if you wait till the last number, and you get to choose when to stop. Must be to your advantage!

"Seems a reasonably good deal to me,'' you say.

"I'm glad you think so;'' she replies, "before we begin you can buy me a pint to compensate. Then we can play the game *K* times, where *K* is as large as you think it should be so that on average you come out ahead.''

How big should *K* be?

The very next day you hear about this course, ST318 *Probability Theory*; by the end of the course you'll have learned enough to be able to answer the above question instantly *without any calculation at all*. All this and 15 CATS credit too. Now that is what you call fair! How can you resist?

## June 03, 2017

### A puzzle about inference

Here is a puzzle which I often use as a starter for a course of lectures on probabilistic coupling. The puzzle arose during some research -- I devised it as a simple example to show myself why a particular idea would not work -- and I developed the solution using calculations. In a lunch queue, I asked James Norris of Cambridge whether he could think of a calculation-free answer, and he found one almost immediately.

- You go to Starbucks with a friend.

- You are the quiet type, and stay in the front room near the window, sipping a coffee slowly to make it last.

- She is extrovert, and has gone through to a further room where the louder people gather.

There she plays a game: toss*n*coins, win if*r*heads or more. - After the game should have finished, a third friend comes through and says,

**either:**"She won the first toss'';

**or:**"She won at least one toss''.

*(Why don't people speak more clearly?!)* - Which is better news for your friend? Prove your answer without calculation!