# All 5 entries tagged St220

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## October 08, 2017

### The media has a problem with uncertainty

Writing about web page https://fivethirtyeight.com/features/the-media-has-a-probability-problem/

Not just the media, but it's a fair point. Have a look at what Nate Silver has to say.

## 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?

1. 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.
2. Let X0 be the vector encoding the target image (the butterfly above). Generate entirely random independent vectors X1, X2, ..., Xn-1 and distribute the corresponding white noise images to the first n-1 students.
3. Student n is given an image corresponding to the vector Xn obtained by multiplying (coordinate-wise) all the other vectors:
Xn = X0* X1 * X2 * ... * Xn-1
where "*" denotes coordinate-wise multiplication.
4. It is simple arithmetic that X0 = Xn* X1 * X2 * ... * Xn-1. So all students working together possess the information to recover the butterfly image.
5. On the other hand one can use elementary probability to show that, if one selects any subset of size n-1 of the vectors X1, X2, ..., Xn, 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-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 Xn be the result of the nth throw, so the nth running average is Yn=(X1+...+Xn)/n. I will read out the numbers Y10, Y9, ..., Y1 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 Yn then with probability Yn/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 Y1/5. Otherwise you buy me a pint.''

You calculate rapidly: E[Y1/5]=E[X1/5]=(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!

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## Most recent comments

• The paper includes a nice example of application of a log–normal distribution, which is used to mode… by Wilfrid Kendall on this entry
• See also their webapp https://045.medsci.ox.ac.uk/ by Wilfrid Kendall on this entry