The problem is explained in a number of books and web sites, the following is taken from the Wikipedia page:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? (Assuming you want to win a new car and not a goat).


The problem is not really intuitive, many people think that eliminating one of the choices reduces the choice to a 50:50 choice, i.e. by eliminating one of the losing doors leaves a winning door and a losing door. So there is no advantage in swapping. This thinking is incorrect, by swapping the contestant gives himself/herself a 67% chance of winning. To convince myself of the probability I wrote the below Java program. When running the program 100 times for no swap there were 35 wins and running 100 times where the door was swapped produced 64 wins.


The program:



/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */

package montyhall;
import java.io.FileWriter;
import java.io.IOException;

/**
 *
 * @author Paul
 */
public class Main {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        String result = "";
         try
            {
                FileWriter writer = new FileWriter("C:\\Users\\Paul\\Desktop\\output.csv");
                for (int i = 0; i < 100; i++){

                    result = "";

                    Box bx1 = new Box();
                    Box bx2 = new Box();
                    Box bx3 = new Box();

                    bx1.setId("1");
                    bx2.setId("2");
                    bx3.setId("3");

                    bx1.setPrize(1, 3);
                    bx2.setPrize(1, 3);
                    while (bx2.getPrize() == bx1.getPrize()){
                        bx2.setPrize(1, 3);
                    }
                    if (bx1.getPrize() + bx2.getPrize() == 3){
                        bx3.specialSetPrize(3);
                    }
                    else if(bx1.getPrize() + bx2.getPrize() == 4){
                        bx3.specialSetPrize(2);
                    }
                    else bx3.specialSetPrize(1);

                    Contestant cont = new Contestant();
                    cont.setBoxID(1, 3);
                    if (cont.getBoxID() == 1){
                        cont.setPrize(bx1.getPrize());
                    }
                    else if (cont.getBoxID() == 2){
                        cont.setPrize(bx2.getPrize());
                    }
                    else cont.setPrize(bx3.getPrize());
                    //******************************************************************

                    //in the case of swap -------------------comment out as necessary
                    if (cont.getPrize() == 3){
                        result = "lose";
                    }
                    else result = "win";


                    //in the case of not swap ----------------comment out as necessary
                    /*if (cont.getPrize() == 3){
                        result = "win";
                    }
                    else result = "lose";*/

                    //******************************************************************
           
                    writer.append(result + "\n");
                }
                writer.flush();
                writer.close();
        }
        catch(IOException e)
        {
            e.printStackTrace();
        }
    }
}


package montyhall;
import java.util.Random;
/**
 *
 * @author Paul
 */
public class Box {
    private String id;
    int prize;

    public Box() {
        this.id = "";
        this.prize = 0;
    }

    /**
     * @return the id
     */
    public String getId() {
        return id;
    }

    /**
     * @param id the id to set
     */
    public void setId(String id) {
        this.id = id;
    }

    /**
     * @return the prize
     */
    public int getPrize() {
        return prize;
    }

    /**
     * @param prize the prize to set
     */
    public void setPrize(int min, int max) {
        Random rand = new Random();
        int randomNum = rand.nextInt((max - min) + 1) + min;
        this.prize = randomNum;
    }

    public void specialSetPrize(int prize){
        this.prize = prize;
    }
}


package montyhall;
   import java.util.Random;
/**
 *
 * @author Paul
 */
public class Contestant {

    private int prize;
    private int boxID;

    public Contestant() {
        this.prize = 0;
        this.boxID = 0;
    }


    /**
     * @return the prize
     */
    public int getPrize() {
        return prize;
    }

    /**
     * @param prize the prize to set
     */
    public void setPrize(int prize) {
        this.prize = prize;
    }

    /**
     * @return the boxID
     */
    public int getBoxID() {
        return boxID;
    }

    /**
     * @param boxID the boxID to set
     */
    public void setBoxID(int min, int max) {
        int rn = chooseABox(min, max);
        this.boxID = rn;
    }

    public int chooseABox(int min, int max){
        Random rand = new Random();
        int randomNum = rand.nextInt((max - min) + 1) + min;
        return randomNum;
    }

}

The Coursera R programming course is part of the Data Science specialisation offered by Coursera. Week two introduces functions in R. There are a number of practical problems concerning air pollution data which have to be solved. Three hundred and thirty two csv files are provided. Each file contains data from one of 332 air quality monitors across the US, named 001, 002, ...,332. Each file has a date, a nitrate level and a sulphate level. The week two assignment has three parts, I provided my solution to part 1 in a previous post. Below is my partial solution to part two:

complete <- function(id) {
  nm <- c("ID", "nobs")
  df <- as.data.frame(matrix(nrow = 0, ncol = 2,dimnames = list(NULL, nm)))
  filenames <- paste(sprintf("%03d", id), ".csv", sep="")
  for (file in filenames){
    nm <- c("ID", "nobs")
    tempdf <- as.data.frame(matrix(nrow = 0, ncol = 2,dimnames = list(NULL, nm)))
    tempdata <- read.csv(file)
    lastrow <- tail(tempdata, 1)
    rowcount <- nrow(na.omit(tempdata))
    id <- lastrow$ID
    tempdf <- cbind(ID=id, nobs=rowcount)
    df <- rbind(df, tempdf)
    rm(tempdata)
    rm(rowcount)
    rm(id)
    rm(tempdf)
  }
  print(df)
}

The function generates the expected results for the test data given in the course. I did not include functionality to set the working directory as it is against Coursera rules to publish complete solutions.

The map below shows the relative engagement people have with museums and galleries by London borough:

The map below shows median income for tax payers per borough.

Although not 100% there is perhaps some possible correlation between income and visiting museums and galleries, wealthier areas seem more likely to visit museums and galleries.
A scatter plot of median income against likelihood of visiting a library or museum is shown below:

The same can not be said for people visiting libraries:

And if the data is plotted as a scatter plot:

Any pattern is less convincing.

​I used Excel for the visualisations. 

The above map shows the approximate number of street trees per borough in London in 2011. The darker the green the more street trees there are in that borough. Not surprisingly the centre of London has the fewest trees and the suburbs have the highest numbers. All data and blank borough map are available from the London Data Store. Below is the same map with tree numbers in 2007.

Because it is difficult to see the difference between the above two maps I calculated the percentage change in street trees for each borough between 2007 and 2011 and put these figures on the below map. White indicates little or no change, pink indicates a small decrease in the number of street trees and red indicates a large decrease, green indicates an increase in the number of trees.

This makes it easier to see the change, there is still a potential problem with  accessibility for people with colour blindness.

One of the problems of data mining is trying to understand the reasons for patterns in the data. For example:
 
Boris bikes, proper title = Barclay's Bicycle Hire, is a bicycle hire scheme in London. It started in July 2010. The number of bicycles hired per day is shown below:

Ignoring the first year or so of data (because the scheme was still being established) there appears to be a definite seasonal pattern. The peak is in the summer months and the trough is in the winter. This can be seen more clearly by looking at monthly rather than daily figures:

Monthly visitor numbers to London also show this kind of seasonal pattern:

The data was only available for visitors per quarter but the seasonal pattern does show up. So there may be a link between tourists visiting London and the number of bicycles being hired. In other words tourists might account for a significant part of the total bicycles hired. So I seem to have arrived at a reasonable conclusion based on the data, i.e. that without the tourists the number of cycles hired out could be significantly less. However if you look at the data for cycle flows on the Transport for London Road Network - this is for all cyclists not just the cyclists on the cycle rental scheme - this seasonal pattern also shows up prior to the cycle scheme becoming strongly established.

So maybe the true picture is that more people cycle in London in the summer than in the winter and that more people visit London during the summer season but the only way to determine if tourists make a significant contribution to the number of 'Boris Bikes' being hired out is to get data on the people who are doing the hiring. This data is not included in the dataset available.

The moral of the story being don't assume a correlation implies a causation. The following site has numerous (and humorous) examples of apparent correlations between completely unrelated things.

All the data used above is available here.

In recent years there has been some interest in capturing energy from footfall to generate electricity. For example during the 2012 London Olympics the exit/entrance to the closest underground station had a a system fitted which used pizoelectric tiles embedded in the floor to generate electricity. As the people walk across them a small amount of electrical energy is generated. The amount of energy captured is dependent on the number of people that pass across the tiles. A similar kind of system was installed at a couple of the busiest JR stations in Tokyo. In that system each person generates about 0.1 watts of power. If such a system were installed at international airports how much power might be generated in a year? The graph below shows the top ten busiest airports in the world in 2013, based on data from Airports Council International.

All the above airports have about sixty million people or more passing through each year that's about six million watts i.e. 6 MW, using the figures for the JR train stations assuming it is possible to capture all the power generated. To put this in perspective: a modern large wind turbine might generate around 3 or 4 MW per year depending on size and average wind speed. So it is not irrelevant but by itself it only provides a small part of the airports power needs and the numbers above are a best case approximation, the actual number might be a small fraction of this.