Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.


Paul
Posts:
676
Registered:
7/12/10


Re: Maximisation problem
Posted:
Jul 27, 2013 5:33 AM


On Saturday, July 27, 2013 10:14:46 AM UTC+1, quasi wrote: > pepstein5 wrote: > > >Peter Percival wrote: > > >> Paul wrote: > > >> > > >> >Let n be a fixed integer > 1. n logicians walk into a bar. > > >> >The barwoman says "Do all of you want a beer?" The first > > >> >logician says "I don't know." The second logician says > > >> >"I don't know." ... The n1st logician says "I don't know." > > >> >The nth logician says "Yes please." > > >> > > >> That doesn't answer the question correctly. > > >... > > > > > >Please could you state your objection? Everyone is narrowly > > >focusing on the yes/no question: "Do all of the n logicians > > >want a beer?" > > > > It's not exactly a yes/no question since some of the answers > > have been "I don't know". > > > > >Everyone saying "I don't know" clearly wants a beer because, > > >if they didn't want a beer, they would know that not everyone > > >wants a beer and would answer "no" instead of "I don't know." > > > > The joke has the implicit assumption that each logician would > > answer yes or no if asked individually as whether or not > > they want a beer. With that assumption, together with > > infinitely many levels of recursion about that assumption, > > the logic of the joke works. > > > > To eliminate that issue, the joke could be stated as follows. > > > > BEGIN JOKE > > > > Let n be a fixed integer > 1. n logicians walk into a bar. > > The barwoman says "Do all of you want a beer?" > > > > Assume that > > > > (A1) Each logician either wants a beer or doesn't want a beer. > > > > (A2) Each logician knows that each of the others either wants > > a beer or doesn't want a beer. > > > > (A3) Each logician knows that each logician knows that ... > > > > and so on, for infinitely many levels. > > > > (B1) Also assume that each logician will answer either > > "Yes","No", or "I don't know", and will only answer "I don't > > know" if they can't deduce the preferences of the others. > > > > (B2) Each logician knows that each logician will answer > > either "Yes","No", or "I don't know", and will only answer > > "I don't know" if they can't deduce the preferences of the > > others. > > > > (B3) Each logician knows that each logician knows that > > each logician will answer ... > > > > and so on, for infinitely many levels. > > > > With those assumptions, > > > > The first logician says "I don't know." The second logician > > says "I don't know." ... The n1st logician says "I don't > > know." The nth logician says "Yes please". > > > > END JOKE > > > > Of course, the joke should be left as it is, so as not to > > ruin it. The goal is humor, not precision. >
quasi, I agree with everything in your posting. Would you be so kind as to answer the question I opened the thread with? Which value of n do you think makes the joke work best? The reasonable choices seem to be n = 2, n = 3 or just leaving n as a variable. n >= 4 just seems to make the joke unnecessarily longwinded.
My preference is n = 2 though you would then say "both" instead of "all" as has been pointed out.
What do you think?
Thanks,
Paul



