From the Greek meaning 'heavy with wine'
A blog devoted to science and reason
Written after a glass or two of Pinot Noir.

Sunday, October 23, 2011

Don't Ask Marilyn

In Parade magazine - that free magazine that appears in your Sunday paper - has a column written by Marilyn Vos Savant who is in the Guiness Book of World Records for supposedly having the world's highest IQ.

Here is today's column:
I’m a math instructor and I think you’re wrong about this question: “Say you plan to roll a die 20 times. Which result is more likely: (a) 11111111111111111111; or (b) 66234441536125563152?” You said they’re equally likely because both specify the number for each of the 20 tosses. I agree so far. However, you added, “But let’s say you rolled a die out of my view and then said the results were one of those series. Which is more likely? It’s (b) because the roll has already occurred. It was far more likely to have been that mix than a series of ones.” I disagree. Each of the results is equally likely—or unlikely. This is true even if you are not looking at the result. —George Alland, Woodbury, Minn. 
My answer was correct. To convince doubting readers, I have, in fact, rolled a die 20 times and noted the result, digit by digit. It was either: (a) 11111111111111111111; or (b) 63335643331622221214.

 Do you still believe that the two series are equally likely to be what I rolled? Probably not! One of them is handwritten on a slip of paper in front of me, and I’m sure readers know that (b) was the result.

The same goes for the first scenario: A person rolled a die out of my view and then informed me the result was one of these series: (a) 11111111111111111111; or (b) 66234441536125563152. It was far more likely to be (b), a jumble of numbers.
Having the highest IQ does not make one immune from being wrong.  The math teacher is right.  

Here's why.  The probability of throwing any number 1 through 6 on a fair die is 1/6.  So throwing a 1 has a probability of 1/6.  Throwing two 1s in a row is 1/6 x 1/6.  Throwing three 1s is 1/6 x 1/6 x 1/6.  Etc.  Throwing twenty 1s has a probability of 2.7 x 10^-16.  Not very likely is it?

Let's look at the other sequence.  The probability of throwing a 6 is 1/6.  The probability of throwing a 6, and then another 6 is 1/6 x 1/6.  The probability of throwing a 6, 6, and a 2 is 1/6 x 1/6 x 1/6.  The probability of throwing a 6, 6, 2, and a 3 is 1/6 x 1/6 x 1/6 x 1/6. And so on.  Then the probability of 66234441536125563152 is also 2.7 x 10^-16.  It doesn't make any difference if she plans on rolling the dice or she does it out of sight.

So what's going on, because Marilyn's answer does make common sense, even if her mathematics is off.  To understand what really makes the answer (b) requires some understanding of entropy - this is what she refers to as jumbled.

Look at the first sequence.  It is twenty 1s.  Now examine the second sequence except don't pay attention to the order.  That roll has three 1s, three 2s, three 3s, three 4s, four 5s, four 6s.  The probability of throwing  three 1s, three 2s, three 3s, three 4s, four 5s, four 6s is 0.239.  24%  That's pretty likely.  The reason this is so likely is that there are a very large number of ways to throw three 1s, three 2s, three 3s, three 4s, four 5s, four 6s.  There's only one way to throw twenty 1s.

Marilyn's first answer is wrong.  Her second answer (the "jumble of numbers" is more likely) is correct, but she makes no attempt to explain.  Not so smart, in my opinion.

1 comment:

  1. As quite often happens, Marilyn is right, but for a question that is different from what she published. She answered "Of the two sequences (a) 11111111111111111111 and (b) 66234441536125563152, one was rolled on a six-sided die and one was fabricated for unstated purposes. Which is more likely to have been actually rolled?" To answer this question, you need to know both the probability of rolling each (1/6^20) and the probability of fabricating each. Marilyn tacitly implies that we should assume the probability of fabricating (a) is far greater than (b). That is probably true since the question demonstrates the properties of random appearances, but her answer can't be given without making that assumption explicit.