Screening for Down's Syndrome

In our discussion on Bayes theorem in the seminar yesterday, I brought up a personal anecdote. During my wife's first pregnancy, she was offered the choice of taking an integrated test to screen for Down's syndrome in the fetus.

I looked up the numbers for the accuracy and false positive rates and found that they were about 95% and 5% respectively (somewhat of a coincidence that these numbers add up to 100%).

The baseline rate of the syndrome steadily increases with the age of the mother.

For a 25 year old mother, it is 0.0001 (1/1100).
For a 35 year old mother it is 0.004 (1/250).
For a 45 year old mother it is 0.05 (1/20).

You can run these numbers through one of the online calculators I wrote about yesterday.

If the test is positive, then the posterior probabilities are again a function of age:

For a 25 year old mother, it is 1.7%.
For a 35 year old mother it is 7.1%.
For a 45 year old mother it is 50%.

Thus, at that time I concluded that the taking the test would only have been meaningful if my spouse were around 45 years old.

For young mothers, even a positive test result is not of particularly great practical value.

Bayes Theorem: Interactive Modules

In our undergrad seminar, we have been reading about Bayes Theorem from a very nice post by Eliezer Yudkowsky. However, most of the Java applets on the page don't seem to work (or at least I couldn't not get them to work!).

Fortunately, thanks to Geogebra, there are multiple interactive HTML5 "applets" which work straight away in any modern browser. If you have Geogebra on your system, you can download and modify the applet as well.

Here is a link to "Exploring Bayes' Theorem"

If you like "tree" based descriptions better, here is another applet/worksheet.

Dicey puzzle

Your friend rolls either one, two, or three dice (n=1, n=2 or n=3). Each die is a normal cube with six sides, displaying a number between 1 and 6. She doesn't tell you what n is, but tells you that the sum of the numbers on the dice is 7.

For example, she could have rolled 4 and 3 with n=2; or perhaps 5, 1, and 1 with n=3 etc. Obviously, n cannot be equal to one.

What are the odds of n=2 v/s n=3 given that the sum is 7?

Answer coming up in a week, but this is an example of simple Bayesian analysis.


Credits: picture from http://www.pwcphoto.com/studio/studio-07.htm.