MicroPosts

• Convert RDA to CSV with a script

  Throw the following into rda2csv.r:

  #!/usr/bin/env Rscript
  
  argv <- commandArgs(TRUE)
  inFile <- toString(argv[1])
  print(paste("Reading:", inFile))
  
  outFile <- gsub(".rda$", ".csv", inFile)
  print(paste("Writing:", outFile))
  
  inData <- get(load(inFile))
  write.csv(inData, file=outFile)
  

  Make it executable: chmod +x rda2csv.r

  And run it on an RDA file: ./rda2csv.r path/to/file.rda

• Risking millions on coin flips

  Using the Binomial Distribution to tilt the odds in your favor.

  I found this pearl in Naked Statistics (chapter 5):

  In 1981, the Joseph Schlitz Brewing Company spent $1.7 million for what appeared to be a shockingly bold and risky marketing campaign for its flagging brand, Schlitz. At halftime of the Super Bowl, in front of 100 million people around the world, the company broadcast a live taste test pitting Schlitz Beer against a key competitor, Michelob. Bolder yet, the company did not pick random beer drinkers to evaluate the two beers; it picked 100 Michelob drinkers.

  It turns out, Schlitz's marketing department was not a bunch of idiots. And here's why.

  If you toss a coin twice you can get either one of the following:

  1. heads, heads
  2. tails, tails
  3. heads, tails
  4. tails, heads

  Therefore, the probability of getting one tails out of two flips is 2/4 (ie, case 3. and 4.).

  That is described by the Binomial Distribution where n is the number of trials (coins) and k is the number of successes (tails):

  p = n! / [k! * (n - k)!] * (1/2)^n
      2! / [1! * (2 - 1)!] * (1/2)^2
      2  / 1               *  1/4
      1/2 (same as 2/4 from above)
  

  The statisticians at Schlitz considered that:

  • commercial beers taste the same
  • in a blind tasting, a Michelob drinker would say that Schlitz tastes better 50% of the times (it's a coin flip)
  • getting 40% or more of Michelob drinkers to say Schlitz is better is worth $1.7 million

  But what's the probability?

  If you take 10 Michelob drinkers, you need at least 4 successes. In other words, you want to sum the probability of 4, 5, 6, 7, 8, 9, and 10 tails with coin flips:

  p = 10! / [4! * (10 - 4)!] * (1/2)^10 +
      10! / [5! * (10 - 5)!] * (1/2)^10 +
      ...
      10! / [10! * (10 - 10)!] * (1/2)^10
    = 0,82
  

  So there's a 82% chance of succeeding.

  And if you raise the number of drinkers to 100, you get to a shocking 98%!