Comments on: Not Improbable http://asymptotia.com/2007/01/27/not-improbable/ Fri, 21 Nov 2008 07:16:53 +0000 http://wordpress.org/?v=2.5.1 By: Clifford http://asymptotia.com/2007/01/27/not-improbable/#comment-28714 Clifford Sun, 18 Feb 2007 23:11:26 +0000 http://asymptotia.com/2007/01/27/not-improbable/#comment-28714 Brian. Thanks!! -cvj Brian. Thanks!!

-cvj

]]> By: Brian Head http://asymptotia.com/2007/01/27/not-improbable/#comment-28550 Brian Head Sun, 18 Feb 2007 05:14:09 +0000 http://asymptotia.com/2007/01/27/not-improbable/#comment-28550 As the offending musician, let me first say how inspired I am by everyone's interest and nimble mathematical minds. I think this attitude is what I enjoyed most about math. My approximation was based simply on how quickly the results converged even after only 5 seats - at which point I was running out of extra room on my program to run through the possibilities. The results oscillated and were zeroing in fast on 63%. I couldn't imagine a reason that the pattern would change. Frankly I was more interested in the various patterns of frustration in the seating arrangements themselves. Chains of duos each seating in each other's chair; trios doing the same; or everyone sitting exactly one to the right of their proper place, etc. I do carry the sentence of a B.S. in Mathematics (Univ. Md, many years ago), though what skills I may have had have long faded into the musical ether. Nevertheless, I kept my books, and I consulted Sheldon Ross' A First Course in Probability after the concert, (certainly outside the rules of this forum I'm sure). In there I found the so-called Matching Problem explained and elegantly proven. Since I'm not sure how to input mathematical notation here, I'll spare you the details, but the problem was this: A group of people throw their apparently identical hats into the center of a room. The hats are mixed up, and then each person picks up a hat. What are the chances that no one selects his own hat? The author actually finds it simpler to work out the odds of at least one man selecting his own hat (basically the question that we posed that night about the tickets), and uses a nifty inductive result stating that the probability of the union of n events equals the sum of the probabilities of these events taken one at a time, minus the sum of the probabilities of the events taken two at a time, plus the sum of the probabilities of the events taken three at a time, and so on. (For the simple case of two events, of course, the probability of the union of two events is the sum of their probabilities minus the probability of their intersection.) To me, this is a tremendous extension of that idea. From that Ross derives the lovely result that (for our ticket version) the odds of at least one person sitting in the proper seat is: 1 - 1/2! + 1/3! - 1/4! + 1/5! and so on. This converges, of course, to 1 - 1/e = .63212... - even I almost remembered the power series for e to the x... Cheers, Brian As the offending musician, let me first say how inspired I am by everyone’s interest and nimble mathematical minds. I think this attitude is what I enjoyed most about math.

My approximation was based simply on how quickly the results converged even after only 5 seats - at which point I was running out of extra room on my program to run through the possibilities. The results oscillated and were zeroing in fast on 63%. I couldn’t imagine a reason that the pattern would change. Frankly I was more interested in the various patterns of frustration in the seating arrangements themselves. Chains of duos each seating in each other’s chair; trios doing the same; or everyone sitting exactly one to the right of their proper place, etc.

I do carry the sentence of a B.S. in Mathematics (Univ. Md, many years ago), though what skills I may have had have long faded into the musical ether. Nevertheless, I kept my books, and I consulted Sheldon Ross’ A First Course in Probability after the concert, (certainly outside the rules of this forum I’m sure). In there I found the so-called Matching Problem explained and elegantly proven. Since I’m not sure how to input mathematical notation here, I’ll spare you the details, but the problem was this:

A group of people throw their apparently identical hats into the center of a room. The hats are mixed up, and then each person picks up a hat. What are the chances that no one selects his own hat?

The author actually finds it simpler to work out the odds of at least one man selecting his own hat (basically the question that we posed that night about the tickets), and uses a nifty inductive result stating that the probability of the union of n events equals the sum of the probabilities of these events taken one at a time, minus the sum of the probabilities of the events taken two at a time, plus the sum of the probabilities of the events taken three at a time, and so on. (For the simple case of two events, of course, the probability of the union of two events is the sum of their probabilities minus the probability of their intersection.) To me, this is a tremendous extension of that idea.

From that Ross derives the lovely result that (for our ticket version) the odds of at least one person sitting in the proper seat is: 1 - 1/2! + 1/3! - 1/4! + 1/5! and so on. This converges, of course, to 1 - 1/e = .63212… - even I almost remembered the power series for e to the x… Cheers, Brian

]]> By: Carl Brannen http://asymptotia.com/2007/01/27/not-improbable/#comment-25861 Carl Brannen Sun, 28 Jan 2007 09:43:02 +0000 http://asymptotia.com/2007/01/27/not-improbable/#comment-25861 Before writing down the "solution" to the problem, I want to claim that most of my evening has been spent reading Massie's book on the naval side of the Great War, "Castles of Steel", and dealing with primitive idempotency equations in 6x6 matrices of products of Pauli algebra projection operators. That said, one can rework [tex]A_n[/tex] to be in the form: [tex]A_n = (n-1)!\left(1 + \sum_{k=2}^{n-2}\frac{A_k}{k!}\right)[/tex] This suggests dividing both sides by n! giving: [tex]\frac{A_n}{n!} = \frac{1}{n}\left(1 + \sum_{k=2}^{n-2}\frac{A_k}{k!}\right)[/tex] Replacing [tex]A_n/n![/tex] by [tex]B_n[/tex], one obtains a readily soluble formula: [tex]B_n = \frac{1}{n}\left(1 + \sum_{k=2}^{n-2}B_k\right)[/tex] To solve this, define [tex]B(x) = \sum_{n=0}^{n=\infty}x^n B(x)[/tex] with suitable choices for B(1) and B(0). This can be solved by the usual methods of combinatorial mathematics. I.e. one obtains a differential equation something like: [tex]B(x) = \int x \frac{dB}{dx} \;dx + x^2B(x)[/tex] which one solves for the solution that meets the definition of B. I leave the rest as an exercise for the reader. (I.e. I've been working on discrete degrees of freedom QM calculations (or qubits) for 3 years, and now am so unused to continuous degrees of freedom that I probably can't differentiate sin(x) twice and get the same result.) Before writing down the “solution” to the problem, I want to claim that most of my evening has been spent reading Massie’s book on the naval side of the Great War, “Castles of Steel”, and dealing with primitive idempotency equations in 6×6 matrices of products of Pauli algebra projection operators.

That said, one can rework A_n to be in the form:
A_n = (n-1)!\left(1 + \sum_{k=2}^{n-2}\frac{A_k}{k!}\right)

This suggests dividing both sides by n! giving:

\frac{A_n}{n!} = \frac{1}{n}\left(1 + \sum_{k=2}^{n-2}\frac{A_k}{k!}\right)

Replacing A_n/n! by B_n, one obtains a readily soluble formula:

B_n = \frac{1}{n}\left(1 + \sum_{k=2}^{n-2}B_k\right)

To solve this, define

B(x) = \sum_{n=0}^{n=\infty}x^n B(x)

with suitable choices for B(1) and B(0). This can be solved by the usual methods of combinatorial mathematics. I.e. one obtains a differential equation something like:

B(x) = \int x \frac{dB}{dx} \;dx + x^2B(x)

which one solves for the solution that meets the definition of B. I leave the rest as an exercise for the reader. (I.e. I’ve been working on discrete degrees of freedom QM calculations (or qubits) for 3 years, and now am so unused to continuous degrees of freedom that I probably can’t differentiate sin(x) twice and get the same result.)

]]> By: anon http://asymptotia.com/2007/01/27/not-improbable/#comment-25826 anon Sun, 28 Jan 2007 05:21:19 +0000 http://asymptotia.com/2007/01/27/not-improbable/#comment-25826 <i>I’d think that, given the number of physicists here, the number 63% would enable people to guess the correct answer in the large n limit</i> Well, <i>now</i> I feel stupid. Should have seen that, given the numerical result. :-) Anyhow, 'derangement' is a fun word. I’d think that, given the number of physicists here, the number 63% would enable people to guess the correct answer in the large n limit

Well, now I feel stupid. Should have seen that, given the numerical result. :-) Anyhow, ‘derangement’ is a fun word.

]]> By: Aaron Bergman http://asymptotia.com/2007/01/27/not-improbable/#comment-25819 Aaron Bergman Sun, 28 Jan 2007 04:24:00 +0000 http://asymptotia.com/2007/01/27/not-improbable/#comment-25819 Permutations that leave nothing invariant are often called 'derangements'. A lot of computations along these lines are done in Graham, Knuth and Patashnik's <i>Concrete Mathematics</i>, a great book, BTW. I'd think that, given the number of physicists here, the number 63% would enable people to guess the correct answer in the large n limit :). I ran across JCD's problem a while ago and solved it by deriving a recurrence and solving it. The solution's simple enough that there must be an easy way to see why it's true. Anyone know such an argument argument? Permutations that leave nothing invariant are often called ‘derangements’. A lot of computations along these lines are done in Graham, Knuth and Patashnik’s Concrete Mathematics, a great book, BTW. I’d think that, given the number of physicists here, the number 63% would enable people to guess the correct answer in the large n limit :).

I ran across JCD’s problem a while ago and solved it by deriving a recurrence and solving it. The solution’s simple enough that there must be an easy way to see why it’s true. Anyone know such an argument argument?

]]> By: andy http://asymptotia.com/2007/01/27/not-improbable/#comment-25816 andy Sun, 28 Jan 2007 03:53:04 +0000 http://asymptotia.com/2007/01/27/not-improbable/#comment-25816 For real details on the use of group theory in music, I suggest you check out John Baez's Week 234 and links therein. Andy For real details on the use of group theory in music, I suggest you check out John Baez’s Week 234 and links therein.

Andy

]]> By: Clifford http://asymptotia.com/2007/01/27/not-improbable/#comment-25801 Clifford Sun, 28 Jan 2007 00:36:11 +0000 http://asymptotia.com/2007/01/27/not-improbable/#comment-25801 Interestingly, Brian, on his way out between pieces (he had to leave early) hurriedly whispered "about 63%" to myself and Tameem.... It seems he agrees with your numerical experiments Anon and your formula, Carl Brennan. Perhaps Brian will tell us how his computation worked. Excellent... -cvj Interestingly, Brian, on his way out between pieces (he had to leave early) hurriedly whispered “about 63%” to myself and Tameem…. It seems he agrees with your numerical experiments Anon and your formula, Carl Brennan.

Perhaps Brian will tell us how his computation worked.

Excellent…

-cvj

]]> By: Carl Brannen http://asymptotia.com/2007/01/27/not-improbable/#comment-25798 Carl Brannen Sun, 28 Jan 2007 00:13:34 +0000 http://asymptotia.com/2007/01/27/not-improbable/#comment-25798 Ooops. extra factor of (k-1)! slipped in. Try: [tex]A_n = (n-1)! + \sum_{k=2}^{n-2}\frac{(n-1)!}{(n-k)!}\;A_{n-k}[/tex] Ooops. extra factor of (k-1)! slipped in. Try:

A_n = (n-1)! + \sum_{k=2}^{n-2}\frac{(n-1)!}{(n-k)!}\;A_{n-k}

]]> By: Carl Brannen http://asymptotia.com/2007/01/27/not-improbable/#comment-25796 Carl Brannen Sun, 28 Jan 2007 00:08:03 +0000 http://asymptotia.com/2007/01/27/not-improbable/#comment-25796 A better recursion. Let [tex]A_n[/tex] be the number of permutations of n elements with no fixed points. Then [tex]A_n = (n-1)! + \sum_{k=2}^{n-2} \frac{(n-1)!(k-1)!}{(n-k)!}A_{n-k}[/tex] The first term is the number of cycles of length n. The second term is a sum over all the ways that the first element can be in a cycle of length k. It consists of the number of ways of picking that cycle (given that the cycle's first element is given), multiplied by the number of ways of writing the remaining n-k elements as a permutation with no fixed points. A better recursion. Let A_n be the number of permutations of n elements with no fixed points. Then

A_n = (n-1)! + \sum_{k=2}^{n-2} \frac{(n-1)!(k-1)!}{(n-k)!}A_{n-k}

The first term is the number of cycles of length n. The second term is a sum over all the ways that the first element can be in a cycle of length k. It consists of the number of ways of picking that cycle (given that the cycle’s first element is given), multiplied by the number of ways of writing the remaining n-k elements as a permutation with no fixed points.

]]> By: Carl Brannen http://asymptotia.com/2007/01/27/not-improbable/#comment-25791 Carl Brannen Sat, 27 Jan 2007 23:37:55 +0000 http://asymptotia.com/2007/01/27/not-improbable/#comment-25791 Oooops. Left off a / in that recursion relation, oh well. Should be [tex]A_{N,M} = A_{N-M,0}\; N!/(M!\;(N-M)!)[/tex] Some first few results, 1/1, 1/2, 4/6, 12/24, 76/120, 460/720, counted without the above formula. Oooops. Left off a / in that recursion relation, oh well. Should be
A_{N,M} = A_{N-M,0}\; N!/(M!\;(N-M)!)

Some first few results, 1/1, 1/2, 4/6, 12/24, 76/120, 460/720, counted without the above formula.

]]> By: Carl Brannen http://asymptotia.com/2007/01/27/not-improbable/#comment-25789 Carl Brannen Sat, 27 Jan 2007 23:34:29 +0000 http://asymptotia.com/2007/01/27/not-improbable/#comment-25789 Oooooh!!! I love challenges like this. And it's interesting to see how people approach this. I always assume that there is an easy way of making a calculation, so I just jump right in. Then I make a mistake and have to redo it. So here goes... The number of ways of sitting N people is N!, the numbe of permutations of N objects. Of these, I wish to count the number of ways that have at least one fixed point. So let [tex]A_{N,M}[/tex] be the number of permutations of N objects that happens to have M fixed points. What is desired is [tex]1 - A_{N,0}/N![/tex]. Naturally, one writes down some recursion relations: [tex]A_{N,1} = A_{N-1,0} \; N [/tex] [tex]A_{N,2} = A_{N-2,0} \; N(N-1)/2[/tex] [tex]A_{N,3} = A_{N-3,0} \; N(N-1)(N-2)/6[/tex] ... [tex]A_{N,M} = A_{N-M,0} \;N!\;M!\;(N-M)![/tex] the [tex]A_{N-M,0}[/tex] counts the permutations with no fixed points, the Pascal's triangle counts the number of ways of putting down the fixed points. Well, that's a start. Oooooh!!! I love challenges like this. And it’s interesting to see how people approach this. I always assume that there is an easy way of making a calculation, so I just jump right in. Then I make a mistake and have to redo it. So here goes…

The number of ways of sitting N people is N!, the numbe of permutations of N objects. Of these, I wish to count the number of ways that have at least one fixed point. So let A_{N,M} be the number of permutations of N objects that happens to have M fixed points. What is desired is 1 - A_{N,0}/N!. Naturally, one writes down some recursion relations:

A_{N,1} = A_{N-1,0} \; N

A_{N,2} = A_{N-2,0} \; N(N-1)/2

A_{N,3} = A_{N-3,0} \; N(N-1)(N-2)/6

…

A_{N,M} = A_{N-M,0} \;N!\;M!\;(N-M)!

the A_{N-M,0} counts the permutations with no fixed points, the Pascal’s triangle counts the number of ways of putting down the fixed points.

Well, that’s a start.

]]> By: anon http://asymptotia.com/2007/01/27/not-improbable/#comment-25784 anon Sat, 27 Jan 2007 22:37:09 +0000 http://asymptotia.com/2007/01/27/not-improbable/#comment-25784 Taking yet another approach: Numerically, it seems that for large numbers of seats, the probability that <i>at least one</i> patron has their assigned seat is ~ 65% (+/- 3% or so?). (This was true for both 100 and 1000 seats, at least.) I'm not proficient enough with combinatorics to find an analytic answer, at least not on a Saturday. Taking yet another approach:

Numerically, it seems that for large numbers of seats, the probability that at least one patron has their assigned seat is ~ 65% (+/- 3% or so?). (This was true for both 100 and 1000 seats, at least.)

I’m not proficient enough with combinatorics to find an analytic answer, at least not on a Saturday.

]]> By: JCD http://asymptotia.com/2007/01/27/not-improbable/#comment-25772 JCD Sat, 27 Jan 2007 20:56:18 +0000 http://asymptotia.com/2007/01/27/not-improbable/#comment-25772 Reminds me of a cute problem I heard once... A flight has 100 seats, each numbered 1 through 100. There are 100 passengers, each assigned to a specific seat. They board in order of seat number. Instead of necessarily sitting in his assigned seat (#1), the first passenger sits in a seat chosen at random. Each of the following passengers sits in his/her assigned seat, if available; if that seat is taken, he/she sits in a seat chosen at random. What is the probability that the 100th passenger gets his assigned seat? Reminds me of a cute problem I heard once…

A flight has 100 seats, each numbered 1 through 100. There are 100 passengers, each assigned to a specific seat. They board in order of seat number. Instead of necessarily sitting in his assigned seat (#1), the first passenger sits in a seat chosen at random. Each of the following passengers sits in his/her assigned seat, if available; if that seat is taken, he/she sits in a seat chosen at random.

What is the probability that the 100th passenger gets his assigned seat?

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