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The test is over. Hopefully that wasn't too traumatic. Let's see how you did.

The answer to question 1 is b, the family with 3 boys and 3 girls. This one should have been pretty easy. Assuming the probability of having a boy is 50% (and the probability of having a girl is 50%) and the results are independent, the probability of having 6 boys is 1.56% or (1/2)^6. The probability of having 3 boys and 3 girls is 31.25%, or 5/16. This means a family of 3 boys and 3 girls is 20 times more likely than a family of 6 boys. The math might have been a bit tricky, but the answer makes sense even if you don't have a calculator for a brain.

The answer to question 2 is c, they are equally likely to occur. This might be a bit surprising to those of you who are new to probability. It confused me for a while, too. Lets look at a simpler problem. Hopefully that will help us understand the more complex case. Instead of a family of 6 lets look at a family of 2. There are 4 possible configurations of children ordered youngest to oldest:

  1. Youngest is a girl, oldest is a girl
  2. Youngest is a girl, oldest is a boy
  3. Youngets is a boy, oldest is a girl
  4. Youngest is a boy, oldest is a boy

Hopefully you're still with me. Each of the above four possibilities is equally likely to occur. The probability of each case is 1/4. Therefore the chance of having two girls is 1/4, the chance of having two boys is 1/4, and the chance of having one boy and one girl in any order is 1/2 (1/4 + 1/4). However, the chance of having the youngest child be a girl and the oldest child be a boy is 1/4. This represents one specific ordering of children -- and every ordering is equally likely with the probability of 1/4. The problem is that our human brains don't easily differentiate between case 2 and case 3 because they are very similar.

As you have probably guessed, question 3 is a trick. There is no correct answer because all four outcomes are equally likely for the same reason as question 2.

That wasn't so bad now, was it? It was? Really? Ok, well the next section will be easier. I promise.

Expected Value

Pretend it is a warm summer day and you are visiting a carnival with your friend Jack who is a compulsive gambler and your friend Sue who is a statistician. You're an odd group, but are a real kick at parties. While strolling through the booths you came across a rather peculiar showman. After catching your interest with an entertaining parlour trick he offers to play a game. He shows you a wheel marked in 100 subdivisons. One subdivision is marked WIN and 99 are marked LOSE. He goes on to say that it costs one dollar to play. If the wheel lands on WIN he will pay you 100 dollars. Otherwise he takes your dollar.

Smelling a scam you decide not to play. You think it would be nice to win a hundred dollars, but you'd rather not risk losing a dollar. Your gambling friend Jack disagrees. He proceeds to gleefully lose 40 dollars and would probably lose more if he wouldn't have run out of cash. Your statistician friend Sue shrugs and plays the game a few times with the detached enthusiasm that only a mathematician can possess. She loses all three times but hardly seems to notice. So who was right? Should you play or not play?

The truth is that from a statistical perspective it doesn't really matter. You could play the game all day long and expect to finish with the same amount of money you started with. In other words the expected value of this game is $0. You pay $1 and expect to get $1 in return. This is slightly counter-intuitive. A smart person would play this game and expect to lose their dollar because that is the most likely outcome. Therefore one might be tempted to say that the expected value of this game is minus $1. But you would be wrong.

The problem lies with the word "expected." The mathematical meaning of expected value is very different than its meaning in English. The word "expected" has a different meaning in both contexts. Mixing definitions will only lead to confusion. For the rest of this article we will refer to the mathematical definition which goes something like this: "The arithmetic mean of a random variable." Basically we take all the possible outcomes, multiply them by their probability of occurring, then add them up.

In the previous example the expected earnings is calculated as follows: $100 * 1/100 + $0 * 99/100 = $1. Since the game costs $1 to play the expected value is $0. This can be roughly translated into English as follows: If 100 people play this game you would expect 99 people to lose $1 and one person to win $99 ($100 prize minus $1 fee). Hence the expected value is $0: On average the entire population of people will neither win money nor lose money.

You might have noticed an interesting property of expected value. In the previous game the expected value is $0 even though it is not one of the possible outcomes. This shows us that the expected value may be an unlikely or even impossible outcome. You should not "expect" (in colloquial usage) to earn the expected value by playing the game only once. If you play the game 1,000 times or 1,000,000 times or an infinite number of times then you would "expect" to earn the expected value when you average your results. So repeat with me 10 times:

  1. Expected value is not what you expect to earn.
  2. Expected value is not what you expect to earn.
  3. Expected value is not what you expect to earn.
  4. Expected value is not what you expect to earn.
  5. Expected value is not what you expect to earn.
  6. Expected value is not what you expect to earn.
  7. Expected value is not what you expect to earn.
  8. Expected value is not what you expect to earn.
  9. Expected value is not what you expect to earn.
  10. Expected value is not what you expect to earn.

Before we solve our investing problem lets take a look at one more carnival game. Since the showman didn't snare you with the first game he gives you the chance to play a second one. He will let you roll a 6 sided die and will give you $1 times the number rolled. If you roll a 3, you get $3. Roll a six and strike it rich with a payoff of $6. The cost to play this game is $3.75. Should you play? It's hard to know. If you roll a 1, 2 or 3 you will lose money. If you roll a 4, 5 or 6 you will make money. But is it enough to to offset your chance of losing? Is the showman's clever rhyming of six with rich enough to entice you to play?

To answer this question we must compute the expected value of the game. If it is positive (meaning you will make money each time you play, on average) then you should play. Otherwise you should avoid playing. Since each dice roll is equally likely (probability is 1/6) the expected earnings of this game is $1 * 1/6 + $2 * 1/6 + $3 * 1/6 + $4 * 1/6 + $5 * 1/6 + $6 * 1/6 = $3.5. Since it costs $3.75 to play, the expected value of this game is minus $0.25 ($3.5 earnings - $3.75 cost to play). This means on average you will lose 25 cents each game. Therefore I offer the following mathematically proven superior rhyme: Do not play, walk away.

Continue to page 3 to see the solution to our investing problem and a few more corny jokes.