Isn't it 1 + (9 choose 2) (to choose either 0 or 1 pairs of questions to switch) out of 9! possible orders? --Joshua Zucker On 7/20/06, Robert Baillie <rjbaillie@frii.com> wrote:

Hypothetical situation: On July 7, a reporter (let's call her Eileen Byrne) asks Donald Rumsfeld questions on 9 topics, in this order: 1. North Korea 2. William Perry's op-ed on striking NK 3. diplomatic efforts on NK 4. stability of Iraq 5. political pressure in U.S. for withdrawal 6. U.S. troops' morale 7. American journalists giving away state secrets 8. Iraq War is being fought at home, in the media 9. Hamdan decision

On July 8, a different reporter (let's call her Monica Crowley) asks Donald Rumsfeld questions on the same 9 topics, in the same order, except with items 6 and 9 reversed.

Let's assume that there are only these 9 topics to choose from. What is the probability that these "reporters" chose the order of their questions independently? (That is, either the same order, or at most one switch of two questions?)

Bob Baillie

All Math-Fun members, especially newcomers, are reminded that the mailing list rules strongly discourage political discussion. The reason for this is practical: People signed up for math discussion, and the heat of politics will easily swamp the list. (Old-timers will recall that I'm arbitrary, harsh, unfair, and have an itchy trigger finger wrt enforcement.) That said, let's address the math question. If we assume the nine questions are tossed in a hat and drawn at random, the question is already answered. If we posit some logical structure to the questions, so that some should precede others, then we get the problem of specifying the partial order - presumably Q1 should precede Q2 - and the puzzle: Given a partial order, how many ways can it map into a total order? Knuth discusses this in Volume 3, Searching and Sorting, when he recounts the problem of sorting 12 things with 29 compares. After you've done a few compares, your data can be summarized as a partial order; planning your next compare, what the results could be, and how well the two possible updated partial orders split the universe of remaining possibilities. IIRC, 2^29 > 12!, but there's no way of doing the compares that always wins. In contrast, 2^7 > 5!, and you *can* sort 5 things with 7 compares. The problem of just doing the counting, given a partial order, turns out to be difficult. I'm not sure where it fits in the P...NP (or #P) ranking. Rich PS: Feel free to harangue me in private about policy, but don't impose on the group. -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com on behalf of Robert Baillie Sent: Thu 7/20/2006 3:36 PM To: math-fun Subject: [math-fun] How to Compute This Probability? Hypothetical situation: On July 7, a reporter (let's call her Eileen Byrne) asks Donald Rumsfeld questions on 9 topics, in this order: 1. North Korea 2. William Perry's op-ed on striking NK 3. diplomatic efforts on NK 4. stability of Iraq 5. political pressure in U.S. for withdrawal 6. U.S. troops' morale 7. American journalists giving away state secrets 8. Iraq War is being fought at home, in the media 9. Hamdan decision On July 8, a different reporter (let's call her Monica Crowley) asks Donald Rumsfeld questions on the same 9 topics, in the same order, except with items 6 and 9 reversed. Let's assume that there are only these 9 topics to choose from. What is the probability that these "reporters" chose the order of their questions independently? (That is, either the same order, or at most one switch of two questions?) Bob Baillie _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun