superprismatic

My last post(#15) was awkward and ill-worded. Here's a better one: Suppose many two-child households each filled out a questionnaire truthfully. The questionnaire contained two questions: "Do you have a daughter?" and "Do you have a son?". If you take all of the questionnaires with a yes answer to "Do you have a daughter?", what portion of these would have a no answer to "Do you have a son?" (this is the same as saying that both are daughters)? To simulate this, just use two tosses of a fair fair coin to decide the genders of the two children. Use this to fill out each questionnaire. If you then take those with a yes answer to "Do you have a daughter?", you will find that 1/3 of those have a no answer to "Do you have a son?". Try it! I just did it quickly for 12 questionnaires and I got 10 with a yes to "Do you have a daughter?". Of these 10, 4 (that's 4/10 = .4) had a no answer to"Do you have a son?". The more questionnaires you do, the closer this gets to 1/3.

Look closely, the Ω has a zero coefficient in each of the equations!

It may help to think of the problem in terms of how this could arise in the real world. Imagine that I am looking at census records. In this census there were two questions of interest to us: "Do you have two children?" and "Do you have a daughter?". So, we imagine that we make a pile of records for which both of these questions have a "yes" answer. Call this pile X. Suppose that another question on the census is "Do you have a son?" What portion of the records in pile X would you expect to have a "no" answer to the question about a son? We can use a coin flip to make fake census records of this type. Each record would require three coin tosses to make -- one for each of the three questions. Then we can actually make pile X and count what portion has a "no" response to the question about a son. I suggest Twinhelix try this and see what happens!

No, I simply took the original problem and performed the following substitutions: $25 becomes $1, $5 becomes $29, $2 becomes $26, $29 becomes $53, and $1 becomes $23 with an appropriate change to the wording of the last sentence.

I'm glad this little puzzle tickled as much interest that it did! It really seems counter intuitive. Congratulations to EventHorizon for his quick correct response. For a problem where accuracy is paramount, he got the height right on the nose. Thank you to everyone who worked on it.

Suppose the earth were a perfectly smooth sphere with radius 6,371,000 meters. Now, there's an old problem which supposes that a rope girdles the earth along a great circle. An extra meter of rope is then spliced into it. You then assume that the rope is raised the same amount above the earth's surface all the way around. You are asked to determine what the amount of the rise is. It's pretty easy to see that the rise is independent of the radius of the earth and is 1/(2π) meters. But we consider a different problem here: Suppose the earth were a perfectly smooth sphere with radius 6,371,000 meters. Now, suppose a rope girdles the earth along a great circle. An extra meter of rope is then spliced into it. Now, grab the rope in one spot and pull it taut directly away from the center of the earth. How far above the surface of the earth will it be able to be pulled? Give the answer to the nearest meter.

BobbyGo's explanation doesn't take into account some dependencies, For example, if the first 4 kings are drawn in the first 12 draws, we are guaranteed never to get a match with a king. So the factor of 48/52 should only be used when we face a king in the case where no king has been drawn yet. Here's a small example with a 4-long deck: 1,2,3,4. Now, using BobbyGo's method, we get a probability of not matching of (3/4)4 which is approximately .32. But, actually listing all 24 ways that the cards could be shuffled, we can see that 10 of these do not produce a match. So, the actual probability is 10/24 which is about .42, a bit bigger than .32. So, an exact calculation requires one to consider all dependencies. I can't figure out how to do this yet, but it might be able to be done with a recursive calculation.

Make 0246891357 from 0123456789 using only two permutations P1 and P2 where P1 is a left circular shift and P2 is flipping the elements in the first two pairs. Specifically, P1 will transform ABCDEFGHIJ into BCDEFGHIJA, and P2 will transform ABCDEFGHIJ into BADCEFGHIJ. Use the fewest possible number of moves where a move consists of an application of either P1 or P2. Here's an example which shows a way to make 0817923456 from 0123456789: 0123456789 Starting position 7890123456 P1 applied 7 times 8709123456 P2 7091234568 P1 0719234568 P2 8071923456 P1 applied 9 times 0817923456 P2 [/code]for a total of 20 moves.

Dart!

Yay! 400th post

Suppose you want to determine the parity of the number of ones in the binary representation of a 32-bit integer. Given that all integer arithmetic, logical, and shift (circular or end around) operations each count as one operation, what is the fewest number of operations needed to determine the parity of the number of ones in a 32-bit integer?

And regarding the example quote you gave, I have to ask, is there truely such a thing as a perfecly readable portion of Benjamin Franklin's autobiography? lol Very nice! You got it very quickly. Benny Franklin is a real favorite of mine, but I'll grant you that the autobiography is a difficult read. Thanks for doing the puzzle.

Suppose, I place some random-looking text in a 26 by 26 square matrix, indexing the rows and columns by the alphabet thusly: ABCDEFGHIJKLMNOPQRSTUVWXYZ A ntdmolbyl r u nsgo.heeaac B slr hb mnotdtlwadmaeee o C srmigP so"i lrm",'fPsryge D sl ,ntnyyiwfeo adali nm E mtno oate iehmayln ye t cm F ePktboe wstslod.nh iahe G yllhiaetst; lcoCo o r"cein H he sg,pk oom eeoeoradbr pr I nw oti,rhaohsIf d dmeca J a iad, r45cpn h,0ea n0o K tt taan th, hm deah w Iie L oec s tfao oct liolhrsiJnw M sRbHt "nusu.otye t 'Biro N ssuadeoaed n w aihivtrl O fic r tyidfosycnrls aeeo h P rsypBnweir a usnns'a noak Q oiado dnsIf lmwcFohfr a R w cn uhkfrrhtst a aldoios S e tf ioa.uItvlelt ere ltms T e'aelt hsL rsuvc th P""rie U t iledn hnaalg e,trtdial V tc eae r fesevei gnehrottn W klsor cbnasm emaw lse'hope X Istdeineswl rwnva coo e Y ity .Meei't h fllabltrlsl Z yp ioslndmviokcnbi ,oieti [/code] But notice that, if you permute the rows, columns, and the indexing alphabet with the same appropriate permutation, you get: [code] QUENFCLPSBMGWJYOZIVKRHDXAT Q From a child I was fond of U reading, and all the litt E le money that came into my N hands was ever laid out i F n books. Pleased with the C "Pilgrim's Progress", my f L irst collection was of Joh P n Bunyan's works in separa S te little volumes. I after B ward sold them to enable m M e to buy R. Burton's "Hist G orical Collections"; they W were small chapmen's books J , and cheap, 40 or 50 in a Y ll. My father's little lib O rary consisted chiefly of Z books in polemic divinity, I most of which I read, and V have since often regrette K d that, at a time when I h R ad such a thirst for knowl H edge, more proper books ha D d not fallen in my way, si X nce it was now resolved I A should not be a clergyman. T Plutarch's "Lives" there This is a perfecly readable portion of Benjamin Franklin's autobiography. This puzzle asks you to do the same type of unscrambling to the following: ABCDEFGHIJKLMNOPQRSTUVWXYZ A p eat vcpoai ,oeheolenc n B sl cfihaoeo cf ia rskptte C edokcgnd cananaeti dutn,e D otarrteer acdng yute oht E darys'hcoear tme ieca, t h F olanehdoerwa tlp o pgg et G Orueawascbhot frtto: u i H apxssiao tgrntdnmo e,iip f I yva A e e ao.uolyihltss J w rodeanoBp awne t'.lpte K .ngntauaom a!nP u irtnhef L iht laaesfate 'tlrt ,le u M ednn feitpot osru anghdds N mamta atel wrslp ys sea i' O e.sThtn egt a daefdr hoild P ht erbyhtsi b toewou bagl Q tpi ids,sc ir roctfhtosaa R la dehmnyitnitci e,p erot S eG egruoarneiemr,b n rmNHt T m yr liahsb lt teoehet ed U oocmneamarilnm rta ,uip a V weoi srr .' gideshgGhnTeon W uw oehshr eodttn eyc frod X eltt f tfbeddilhe tiirosg Y pci rodeVi crt .satohot is Z rtatilisea aihiwh nlaigW [/code] The correctly permuted matrix contains an interesting historical quotation. Please ignore the coloring of the text. This is an artifact of the code tag that I used to keep everything aligned.

Oh! Thanks, fabpig, I would never have thought of that.

What's wrong with Riddle_Lover's answer? It looks good to me, but perhaps I don't understand the rules.