We can make choices based on expected values and standard deviations. Suppose that you have two urns (jars) containing colored balls and that certain payoffs are associated with the colors. Suppose that yellow balls are worth $100, blue balls are worth $ 20, and red balls are worth -$ 50. Designed by Gary Simon, September 2007.

Y = $100, B = $20, R = -$50 If the urn has 40 yellow balls 30 blue balls 30 red balls the expected value is $31. You might be willing to spend $15 in development costs to get to select one ball from this urn.

It is possible to compute a standard deviation around the expected value of $31. Here’s how: Variance = E[ squared difference from $31 ] = ($100 - $31) 2 × ($ 20 - $31) 2 × ( (-$50) - $31) 2 × 0.30 = $ 2 3,909 Standard deviation = $  $62.52 Would you regard this as risky?

You can also compare two jars. Jar A has Jar B has For both jars, the expected value is $29. …but SD(Jar A) = $41.10 > SD(Jar B) = $29.14.

At times you might be willing to take the larger standard deviation, provided that you also get the larger mean! Jar C has expected value = $37, standard deviation = $46.05 Jar D has expected value = $41, standard deviation = $47.84

In some decisions, you just don’t know enough! Consider a new urn, with yellow, green, and purple balls. You know that there are 900 balls in all: 300 balls are yellow an unknown number are green an unknown number are purple (green + purple = 600)

300 yellow ; green + purple = 600 Now select one ball. Which payout scheme would you prefer? SchemeYellowGreenPurple A$6000 B0 0 Write down your answer before proceeding....

Let’s play the game again. Return the ball and mix up everything again. You may have an opinion about the number of green balls. After seeing the first draw, you should do a Bayesian updating on that opinion. Since the urn is large (900 balls) and since the information is minimal (one draw), we will skip this step just for simplicity. Now, which of C and D do you prefer? SchemeYellowGreenPurple C$600 D0 Write down your answer before proceeding....

Most people choose A over B and then choose D over C. Why? Let’s think what this means.

SchemeYellowGreenPurple A$6000 B yellow ; green + purple = 600 Choosing A over B seems to be invoking a belief that there are more yellow balls than green balls.

SchemeYellowGreenPurple C$600 D0 300 yellow ; green + purple = 600 Choosing D over C seems to be invoking a belief that there are more green balls than yellow balls. It is logically inconsistent to choose A over B and then choose D over C. A foolish consistency is the hobgoblin of little minds, adored by little statesmen and philosophers and divines. Ralph Waldo Emerson

Many explanations have been proposed as to why this happens. The choices of A and D are those that eliminate the uncertainty by not knowing the number of green balls. SchemeYellowGreenPurple A$6000 B0 0 SchemeYellowGreenPurple C$600 D0

The choices of A and D can be thought of as “maximin,” in that they guard against the worst thing that can happen. You cannot describe these choices as “eliminating uncertainty.” There are two sources of variation here: The variability regarding the number of green balls. The variability associated with random sampling.

There is an additional perspective that appears when the choices are displayed next to each other. SchemeYellowGreenPurple A$6000 B0 0 SchemeYellowGreenPurple C$600 D0 The games are identical, except that the second has been enriched by paying out $60 for the purple balls!

If you prefer A over B, logic demands that you prefer C over D. This is known now as Ellsberg’s paradox, There are many variations of it, and it has ignited many arguments.

Daniel Ellsberg is better known for his tangential role in the Watergate crisis of the 1970s than for this wonderful intellectual gem. Ellsberg’s psychiatrist was the victim of an office break-in. The intruders were suspected of looking for embarrassing information on Ellsberg.