May 2006 - mathfun@mailman.xmission.com

[math-fun] Tower of Hanoi problem
by David Wilson 10 Jun '06

10 Jun '06

In the n-plate Towers of Hanoi problem, consider the fixed base of each peg to be plate 0, and number the largest through smallest movable plates 1 through n. In the solution of the problem, how many times is plate p placed on top of plate q (0 <= p,q <= n)?

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RE: [math-fun] Bridge probability puzzle
by Cordwell, William R 31 May '06

31 May '06

I get Joshua's and Mr. McCaughan's answer to 14 decimal places, but it differs beyond that. There are (4 C 2) = 6 ways to pick which two suits you have; this also fixes the suits of your partner. You have, from these two suits, (26 C 13) ways of getting your 13 cards; your partner also has (26 C 13) ways of getting his 13 cards from the other two suits. You then divide by (52 C 13) ways of getting your hand randomly, and divide also by (39 C 13) ways of your partner getting his hand randomly from the remaining 39 cards. Thus, P = 6 * (26 C 13) * (26 C 13) / [ (52 C 13) * (39 C 13) ] = 21850/173640942909 = (acc'ding to Mathematica)1.25834377733429657622523386340107748418239154034425 x 10^(-7)(forty places) I'm not convinced that this is right, but I'd like to know why, if not. What are the odds of getting the same answer to 14 digits, but differing otherwise? :-) Bill C. -----Original Message----- From: math-fun-bounces+cordwell=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com]On Behalf Of Joshua Zucker Sent: Tuesday, May 30, 2006 10:32 AM To: math-fun Cc: dasimov(a)earthlink.net Subject: Re: [math-fun] Bridge probability puzzle I'll try an alternative counting method and see if I get the same answer as one of these other folks. Their two answers agree to about 7 significant figures, 0.00000012583437773342733107186438969809670960643871... vs 0.00000012583439386511151359617331968028508948288906.... Anyway, here's my approach. First of all, you could have only one suit: the probability of that happening is 4 possible hands out of (52 C 13). If that's the case, then the probability that partner has only two of the other three suits is, let's see, three ways to choose which two suits, times (26 C 13) ways of choosing the cards, out of (39 C 13) possible hands. But that overcounts the ways where partner has only one suit, counting each twice (say, if you have only clubs and partner has only spades, that's counted as one of the ways with hearts and spades and as one of the ways with diamonds and spades). So subtract the ways that partner has only one suit, 3 ways. On the other hand, you might have two suits: (4 C 2) ways to choose which two suits, times (26 C 13) ways to choose your cards, minus the two ways where you only have one suit, out of (52 C 13). Then partner has to have the other two suits, (26 C 13) out of (39 C 13) being that probability. So my final calculation looks like 4 / (52 C 13) * (3 * (26 C 13) - 3)/(39 C 13) + (4 C 2) * ((26 C 13) - 2) / (52 C 13) * (26 C 13) / (39 C 13) 54086240179999/429820857815004205200 is then my answer. --Joshua Zucker _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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RE: [math-fun] Bridge probability puzzle
by Cordwell, William R 31 May '06

31 May '06

Sorry for the multiple messages; Mr. McCaughan had the correct reasoning; if I have all one suit and my partner has all another suit, this can occur in multiple ways (overcounting) in my scheme. For example, if I have all spades, and my partner all hearts, this case shows up as one of the possibilities in me = {S, D} but also as me = {S, C}. Thus, I have twice 4*3 special cases, overcounting by 12 out of the (52 C 13)*(39 C 13) total possibilities. Bill C. -----Original Message----- From: math-fun-bounces+cordwell=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com]On Behalf Of Cordwell, William R Sent: Wednesday, May 31, 2006 3:24 PM To: math-fun Subject: RE: [math-fun] Bridge probability puzzle Ah, now I see that Joshua has the correction of -12/(52 C 13)/(39 C 13) different from mine. I need to figure out why I've overcounted by this. Bill C. -----Original Message----- From: math-fun-bounces+cordwell=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com]On Behalf Of Cordwell, William R Sent: Wednesday, May 31, 2006 3:08 PM To: math-fun Subject: RE: [math-fun] Bridge probability puzzle I get Joshua's and Mr. McCaughan's answer to 14 decimal places, but it differs beyond that. There are (4 C 2) = 6 ways to pick which two suits you have; this also fixes the suits of your partner. You have, from these two suits, (26 C 13) ways of getting your 13 cards; your partner also has (26 C 13) ways of getting his 13 cards from the other two suits. You then divide by (52 C 13) ways of getting your hand randomly, and divide also by (39 C 13) ways of your partner getting his hand randomly from the remaining 39 cards. Thus, P = 6 * (26 C 13) * (26 C 13) / [ (52 C 13) * (39 C 13) ] = 21850/173640942909 = (acc'ding to Mathematica)1.25834377733429657622523386340107748418239154034425 x 10^(-7)(forty places) I'm not convinced that this is right, but I'd like to know why, if not. What are the odds of getting the same answer to 14 digits, but differing otherwise? :-) Bill C. -----Original Message----- From: math-fun-bounces+cordwell=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com]On Behalf Of Joshua Zucker Sent: Tuesday, May 30, 2006 10:32 AM To: math-fun Cc: dasimov(a)earthlink.net Subject: Re: [math-fun] Bridge probability puzzle I'll try an alternative counting method and see if I get the same answer as one of these other folks. Their two answers agree to about 7 significant figures, 0.00000012583437773342733107186438969809670960643871... vs 0.00000012583439386511151359617331968028508948288906.... Anyway, here's my approach. First of all, you could have only one suit: the probability of that happening is 4 possible hands out of (52 C 13). If that's the case, then the probability that partner has only two of the other three suits is, let's see, three ways to choose which two suits, times (26 C 13) ways of choosing the cards, out of (39 C 13) possible hands. But that overcounts the ways where partner has only one suit, counting each twice (say, if you have only clubs and partner has only spades, that's counted as one of the ways with hearts and spades and as one of the ways with diamonds and spades). So subtract the ways that partner has only one suit, 3 ways. On the other hand, you might have two suits: (4 C 2) ways to choose which two suits, times (26 C 13) ways to choose your cards, minus the two ways where you only have one suit, out of (52 C 13). Then partner has to have the other two suits, (26 C 13) out of (39 C 13) being that probability. So my final calculation looks like 4 / (52 C 13) * (3 * (26 C 13) - 3)/(39 C 13) + (4 C 2) * ((26 C 13) - 2) / (52 C 13) * (26 C 13) / (39 C 13) 54086240179999/429820857815004205200 is then my answer. --Joshua Zucker _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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RE: [math-fun] Bridge probability puzzle
by Cordwell, William R 31 May '06

31 May '06

Ah, now I see that Joshua has the correction of -12/(52 C 13)/(39 C 13) different from mine. I need to figure out why I've overcounted by this. Bill C. -----Original Message----- From: math-fun-bounces+cordwell=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com]On Behalf Of Cordwell, William R Sent: Wednesday, May 31, 2006 3:08 PM To: math-fun Subject: RE: [math-fun] Bridge probability puzzle I get Joshua's and Mr. McCaughan's answer to 14 decimal places, but it differs beyond that. There are (4 C 2) = 6 ways to pick which two suits you have; this also fixes the suits of your partner. You have, from these two suits, (26 C 13) ways of getting your 13 cards; your partner also has (26 C 13) ways of getting his 13 cards from the other two suits. You then divide by (52 C 13) ways of getting your hand randomly, and divide also by (39 C 13) ways of your partner getting his hand randomly from the remaining 39 cards. Thus, P = 6 * (26 C 13) * (26 C 13) / [ (52 C 13) * (39 C 13) ] = 21850/173640942909 = (acc'ding to Mathematica)1.25834377733429657622523386340107748418239154034425 x 10^(-7)(forty places) I'm not convinced that this is right, but I'd like to know why, if not. What are the odds of getting the same answer to 14 digits, but differing otherwise? :-) Bill C. -----Original Message----- From: math-fun-bounces+cordwell=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com]On Behalf Of Joshua Zucker Sent: Tuesday, May 30, 2006 10:32 AM To: math-fun Cc: dasimov(a)earthlink.net Subject: Re: [math-fun] Bridge probability puzzle I'll try an alternative counting method and see if I get the same answer as one of these other folks. Their two answers agree to about 7 significant figures, 0.00000012583437773342733107186438969809670960643871... vs 0.00000012583439386511151359617331968028508948288906.... Anyway, here's my approach. First of all, you could have only one suit: the probability of that happening is 4 possible hands out of (52 C 13). If that's the case, then the probability that partner has only two of the other three suits is, let's see, three ways to choose which two suits, times (26 C 13) ways of choosing the cards, out of (39 C 13) possible hands. But that overcounts the ways where partner has only one suit, counting each twice (say, if you have only clubs and partner has only spades, that's counted as one of the ways with hearts and spades and as one of the ways with diamonds and spades). So subtract the ways that partner has only one suit, 3 ways. On the other hand, you might have two suits: (4 C 2) ways to choose which two suits, times (26 C 13) ways to choose your cards, minus the two ways where you only have one suit, out of (52 C 13). Then partner has to have the other two suits, (26 C 13) out of (39 C 13) being that probability. So my final calculation looks like 4 / (52 C 13) * (3 * (26 C 13) - 3)/(39 C 13) + (4 C 2) * ((26 C 13) - 2) / (52 C 13) * (26 C 13) / (39 C 13) 54086240179999/429820857815004205200 is then my answer. --Joshua Zucker _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Non-stuttering codes
by Michael Kleber 30 May '06

30 May '06

I have no idea at all whether the following problem is known in the literature: I'm looking for codes where the code words are "non-stuttering", i.e. have no two consecutive identical letters. The specific parameters I care about are a 4-letter alphabet with code words at minimum distance 4, probably with words in the vicinity of 9-10 letters. (Yes, for the application I have in mind, the 4-letter alphabet is {A,C,G,T}.) I can think of all sorts of very bad constructions for codes with this obnoxious property, but not anything that seems clever. (The map from all strings on {1,2,3} to non-stuttering strings on {0,1,2,3}, taking successive sums, is frustratingly useless, since it is very badly behaved with respect to Hamming distance.) Can anyone offer insight or references? --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.

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[math-fun] Bridge probability puzzle
by dasimov＠earthlink.net 30 May '06

30 May '06

What is the probability there is some pair of suits your bridge hand does not include, and your partner's hand does not include the other pair? --Dan

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[math-fun] New chess record
by Guy Haworth 28 May '06

28 May '06

The Konoval-Bourzutschky team have once again raised the DTZ-record for chess endgames. DTZ == Depth to (move-count) Zeroing (move), i.e. a P-push/capture/mate It stood at (KRNKNN) maxDTZ = 243 for some years, after Stiller's work on the Connection Machine at M.I.T. YK-MB raised it to 290, (317), 333 and now ... 517. MB announcement is at http://216.25.93.108/forum/viewtopic.php?t=2860 Krabbe rendition at http://www.xs4all.nl/~timkr/chess2/diary.htm Item 317 for the 517-mover Item 311 for the 330-mover and the 317-mover Item 298 for the 290-mover MB's announcement also includes the find of the first known full-point mutual zugzwang position not involving Knights or Pawns. MB notes that maxDTZ has more than doubled with the addition of each man: KRK 16 - KBNK 33 - KNNKP 81 - KRNKNN 243 - KRBNKQN 517 Guy

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Re: [math-fun] Library of Alexandria
by dasimov＠earthlink.net 27 May '06

27 May '06

Emeric wrote: << . . . on May 25, Michael R. Bloomberg, Republican Mayor of New York, to the graduating students of John Hopkins University School of Medicine: Today we are seeing hundreds of years of scientific discovery being challenged by people who simply disregard facts that don't happen to agree with their agenda. Some call it pseudo-science, other call it faith-based science, but when you notice where this negligence tends to take place, you might as well call it 'political science'. >> Careful, the moderator may make you sit in the corner. --Dan

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[math-fun] Library of Alexandria
by Henry Baker 27 May '06

27 May '06

I heard a scientist on the radio say today (without proof) that he thought that mankind would have made it to the Moon 200 years earlier if the Library of Alexandria had not burned down. Over the years, I've wondered why science & technology & math did not get further in the ancient world, so I'm curious if others have had similar ideas. Clearly the Roman Empire had a pretty decent standard of living -- it wasn't matched again in the Western world until the Victorian times -- probably due to the "Rats, Lice & History" rationale -- you don't get sick nearly as much when you have a decent sewer system. The population of Rome fell from a height of well over 1 million at the height of the Roman Empire to less than 30,000 in some parts of the Middle Ages -- the water, sewer and street systems had fallen into such disrepair that it could not support a big population. Clearly the Romans had good techology -- their weapons were excellent, their ships were excellent, etc. The Chinese of the same time period already had a very advanced system for sifting the population to find the best & brightest for their bureaucracy. The astronomers of China, Persia & certain parts of the Roman Empire were superb. Most of the geometry required for Kepler's Laws was known to some of the ancient mathematicians. What happened? Or what didn't happen? Was the printing press part of the answer? But the Chinese had printing. Was the human race set back by its failure to backup important files?

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Re: [math-fun] batting averages
by Michael Reid 27 May '06

27 May '06

> SHOULDA > wrote: > At minimum, > (c74) block([keepfloat:true],linsolve((2*integer+1)/(5*integer+3)=0.3985)) > > (d74) [integer = 26.066] > > (c75) subst(ceiling(%),5*integer+3) > > (d75) 138 > Serves me right for not using continued fractions. > at bats. ok, that looks better. but how do you get the form (2n + 1) / (5n + 3) ? is there some way without continued fractions? btw, the c.f. method i described before can be cleaned up slightly. to find the "simplest" rational number in the closed interval [a, b] , compute the convergents for the cf's of a and b . suppose p/q and P/Q are the last two they have in common. then the next convergent for each will be of the form (p + Pn) / (q + Qn) for some positive integer n . if the simpler of these two is in the target interval, then that's the one. otherwise, replace n by n+1 in its expression as above, and that gives the simplest fraction in the interval. (this may require a and b to be positive, but it's easy to reduce to that case.) mike

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