October 24, 2000 Volume 29, No.7

This Week's Colloquium
Attributable risk, a simple but underused concept, is defined as the percent of cases of a disease which can be "attributed" to a risk factor.  The deceptive simplicity of the concept provides fertile ground for a discussion of its mathematical and statistical underpinnings.  Not surprisingly, in the complex world of multiple diseases with multiple, interrelated causes, implementation of the concept proves to be not at all simple. This presentation discusses the development of the concept from its simple phase to a complex generalized model of attributable risk which, by permitting the simultaneous consideration of multiple causation and confounding factors, may come acceptably close to the "Real World." A software package will be discussed briefly which employs computer intensive resampling methods (Bootstrap, Jackknife) to estimate the distribution of the generalized estimate. The utility of the concept will be illustrated using data from a large study designed to explicate risk factors for ischemic stroke. This presentation will combine issues of medicine, public health, mathematics and statistics in a meaningful and understandable fashion.

This Week's Speaker
W. Michael O'Fallon grew up in Litchfield, MN, received an undergraduate degree in Mathematics at St. Johns University and a master's degree from Vanderbilt University. He subsequently taught Mathematics at St. Johns for three years before going to North Carolina as an NDEA Fellow at UNC. He later earned a Ph.D. in Mathematical Statistics from UNC and has specialized in the application of statistical principles to medical and epidemiological research. He has published over 400 manuscripts, most in collaboration with medical colleagues from Duke University Medical School and the Mayo Clinic. He is currently the President of the American Statistical Association, as well as an ex-soccer referee and wrestler, though, unlike our governor, the latter was not professional.

Statistic Students:
Mike O'Fallon, a statistician at the Mayo Clinic will be here Oct. 26 and Bruce Craig from the statistics dept at Purdue University will be here Nov 2.  Both are happy to chat with anyone who might be interested in career opportunities or graduate school in statistics/biostatistics. If you are interested in talking with  either Prof. O'Fallon or Prof. Craig please contact Michael Kahn.

Best Wishes
Our best wishes go to Prof. Richard Allen who is at home recovering from surgery. Take it easy Rich, we hope to see you soon.

Last Week's Solutions
Problem: Suppose we have a 36-story building and two eggs. What is the least number of droppings required that is guaranteed to determine, in all cases, the lowest floor that causes eggs to break?

Solution by Eric Johnson, Laura Johnson, and Jerad Parish: If an egg drops from a window and breaks, all floors leading up to it must be verified; e.g. if it does not break at 12 and it does at 18, we must check 13,14,15,16 and 17.  Thus, partition the floors as follows: 8,15,21,26,30,33,35,36.

Drop an egg at each floor in the sequence; if an egg breaks, check all floors from the last one that worked.  As this series increases by 8-7-6-5-4-3-2-1, the number of floors needed to check plus the number of drops remains constant at 8.  In all cases, the largest number of floors that need checking is 8.

To show that 7 will not work, partition by a series that increases by 7-6-5-4-3-2-1. This series sums to 28, so floors 29-36 cannot be checked except by extra drops, which invalidates 7.  Thus, the minimum number of drops is 8.

*Problem also solved by Bob Breid, Jason Grimm, Robert Hillard, Erik Johnson, Jonathon Kuipers and Michael Zahniser.

Problem of the Week
David Molnar, the associate problem guru, suggested the following: 40 thieves, all of different ages, steal a huge pile of identical gold coins and must decide how to divide them up. They settle on the following procedure: the youngest divides them among the thieves however s/he wishes, then all 40 thieves vote on whether they are satisfied with the division. If a majority (21) vote NO, the youngest is killed, and the next youngest gets to try to divide the booty among the remaining 39 thieves (including him/herself), with the same penalty if a majority vote goes against the division. And so on.

Now, each of the 40 thieves is completely rational, a perfect logician, and knows that this is true of the others as well. Further, each thief always acts in her/his own interest, ignoring (possibly) the interest of the group and plain old fairness. Given all this, how should the youngest of the 40 thieves divide the loot?

**Please submit all solutions to Cliff Corzatt (corzatt@stolaf.edu) by noon on Friday.

To subscribe to the Math Mess, please contact Donna Brakke at brakke@stolaf.edu.

Editor-in-Chief: Jill Dietz
Associate Editor: Jennifer Beilfuss
Problems Editor: Cliff Corzatt
MM Czar:   Donna Brakke
mathmess@stolaf.edu