Discover magazine has error that is "Not Even Wrong"

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1. Mr Mean:

   Dr Lasky,

   As stated in your post, you’ve computed the median. This is one possible interpretation of the word “average.” I would argue, though, that a more common interpretation of the word “average” is “mean.” For example, what if I asked you for the average of the following numbers: 1, 2, 3, 4, 100? Would you answer 3, which is the median? Or, would you answer 22, which is the mean?

   Dickenson has, in fact, correctly stated the mean (or expected value). To compute the expected value of a random variable, we multiply each possible outcome by the probability
   of that outcome and take the sum of the products. For example, if we have a population in which 20% of the people make $10/hr, 70% make $20/hr, and 10% make $30/hr, the mean
   income for the population is:
   0.2*10 + 0.7*20 + 0.1*30

   So to compute the mean age at which someone would die, we have to figure out all of the possible ages at which they could die, multiply each age by the probability of dying
   at that age, and add up the products.
   Year 1 is easy enough. There’s a 1/1743 chance that the person dies, so we multiply that
   by 1 and get:
   1/1743 * 1
   Year 2 isn’t so bad. To die in year 2, first they have to survive year 1. The probability of that is 1742/1743. Then, they have a 1/1743 chance of dying. So for year 2, we have
   (1742/1743) * (1/1743) * 2
   For year 3, they have to survive years 1 and 2, and die in year 3. So we have:
   (1742/1743)^2 * (1/743) * 3
   And in year 4:
   (1742/1743)^3 * (1/1743) * 4
   In general, the probability of dying in year y is:
   (1742/1743)^(y-1) * 1/1743
   multiplying by the year gives us:
   (1742/1743)^(y-1) * 1/1743 * y

   So, now we just have to add all of products. This gets a little ugly, because there are an infinite number of things to add. We have:
   sum[y=1->inf] ( (1742/1743)^(y-1) * (1/1743) * y )

   Assuming you don’t do binomial expansions every day, it might be easier to estimate the solution to this rather than solve it exactly. Here is some perl code do do that. It only considers the people who live less than 1000000 years. Since only 0.<247 zeros go here>580511580314418 percent of the people are still alive at 1000000 years, this is a pretty good estimate.

   ————————————- #!/usr/bin/perl
   $tot=1000000;
   $base=1743;

   $avg=0;
   $stillalive=1;

   for($i=1; $i<$tot; $i++) { $avg+=(1/$base)*$stillalive*$i; $stillalive*=($base-1)/$base;
   }
   print “avg $avgn”;
   ————————————-

   For the sake of completeness, we can use the following:
   1/(1-1/x)^2=1+2/x+3/x^2+4/x^3+5/x^4+...
   which I found here:
   http://www.jimloy.com/algebra/series.htm
   to solve the equation exactly.

   So we rewrite:
   sum[y=1->inf] ( (1742/1743)^(y-1) * (1/1743) * y )
   as
   (1/1743) * ( sum[y=1->inf] ( (1742/1743)^(y-1) * y ) ) )
   let x=1743/1742:
   (1/1743) * ( sum[y=1->inf] ( y/x^(y-1) ) )
   or
   (1/1743) * (1+2/x+3/x^2….)

   The equation from jimloy.com tells us the sum of the series, so we have:
   (1/1743) * 1/(1-1/x)^2
   plugging 1743/1742 back in for x:
   (1/1743) * 1/(1-1/(1743/1742))^2
   (1/1743) * 1/(1-1742/1743)^2
   (1/1743) * 1/(1/1743)^2
   (1/1743) * 1743^2
   1743

   I was going to comment that this seems like a bit more work than your statement “It appears he just inverted the 1/1743.” implies. However, if we went back through the math and used 1/p instead of 1/1743, we’d get a mean of p. So, in this type of problem, you can just invert the probability to get the mean. You won’t get many points for showing your work, though.

   So, assuming Dickinson means mean, then the solution is not only not not even wrong, but also not wrong.

2. Mr Mean:

   It looks like I am a failure at pasting. This line of his perl code:
   print “avg $avgn”;
   is missing a character. If you make it:
   print “avg $avgn”;
   the script will work a lot better.

3. Ron Lasky:

   Mr. Mean,

   You are correct. I solve the problem folling your lead for 50,000 years using Excel.

   I will print a correction soon, after I clean the egg off of my face and let my recent posting have time to be read.

   Thanks for your time and effort. My assumption that the author simply took the inverse of 1/1743 was wrong and unfair to him. I also assumed (wrongly) that the median and mean would be close in value.

   Well, my hope is that we all learned something.

   Thanks again,

   Dr. Ron

4. Mr Mean:

   I noticed this in the post:
   “Some time ago I pointed out an error in Discover Magazine and even one from Marilyn Vos Savant, who is claimed to have the world’s highest IQ.”

   Actually, I believe both the earlier Discover article, and Marilyn Vos Savant are correct. I posted comments at:

   http://www.indium.com/drlasky/entry.php?mode=add&id=506

   and

   http://www.indium.com/drlasky/entry.php?mode=add&id=510

   describing why I believe this.

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