How many birthdays does the average man have?

[ Editor: replies from the September 1993 thread, in order.  The
  question numbers refer to the list in the problem. ]

I'm posting this mainly for the benefit of the rec.puzzles archive and
in a vain attempt to reduce bandwidth.

ST = standard trick answer
MA = my answer

1.  ST = one
    MA = depends on definition of birthday

2.  ST = all twelve
    MA = "have 28" generally *means* "have exactly 28"; or, consider
         alternative calendars

3.  ST = only dead people are buried
    MA = USA and Canada are disjoint regions (people can be buried
         alive after all)

4.  ST = MA = 6

5.  ST = no
    MA = yes; man X marries woman Y, woman Y dies, man X marries woman
         Y's sister (call her woman Z), man X dies; in the original
         marriage man X married his widow's (woman Z's) sister
         perfectly legally.

6.  ST = MA = they didn't play each other

7.  ST = 77
    MA = 30/(1/2) + 17.

8.  Already in rec.puzzles archive.

9.  ST = MA = 2

10. ST = nickel + half dollar
    MA = consider other countries that use dollars (e.g., a Hong Kong
         five-cent coin and a Hong Kong half-dollar coin)

11. ST = match
    MA = match unless one of the other items is self-lighting

12. ST = 1 hour
    MA = depends how much attention I pay to the doctor's instructions

13. ST = 9
    MA = 17

14. ST = meat
    MA = clerks often don't weigh meat

15. ST = MA = 12

16. ST = Clinton?
    MA = can quibble about referent of "the president"

-- Tim Chow <tycchow@math.mit.edu>

I thought that I would try to get these out of the way before too much
time goes by.  They are all "trick" questions, but many of them can be
turned around to "trick the trickster".

1.  The average man has one, which is celebrated multiple times.  Some
non-average people like George Washington had more than one birthday.
When the calendar changed in 1752, he chose to celebrate a day one year
after his last birthday, which meant having a different date.

2.  All of them have at least 28.  Only February (sometimes) has
exactly 28.

3.  Nothing at all prevents this.  The man would probably be very
unhappy, since he is still living.

4.  Usually 6, 3 per side.  Sometimes 3, as the bottom half of the
ninth inning is not played if the home team is already ahead.

5.  It is not legal in California for dead people to marry.  (even for
tax purposes)

6.  They did not play each other.

7.  77 or 1.714, depending on your operator precedence.

8.  White.  The house is supposed to be at the North Pole.  (I believe
some people have stated that no bears actually roam near the North
Pole)

9.  Two more than you had before.

10. A half dollar and a nickel.  The half dollar is not a nickel.  They
could also both be "not a nickel" if you use an older American coin
called the "half-dime", also worth 5 cents.

11. The match, assuming the others were not already lit.

12. You would have finished taking them 1 hour after you started.
Their period of effectiveness (another meaning of last) is unknown.

13. 9

14. Meat.

15. A dozen.

16. Our present president's name was William Clinton in 1960.  It was
not the name he was born with, but it was changed before 1960.  Viewed
in a different fashion, the president's name in 1960 was either
Eisenhower (the inauguration was in January) or Kennedy.

-- Rolf Wilson <rolf@geoserv.isgs.uiuc.edu>

1.  one, and it occurs on the day he is born.
2.  all of them.
3.  because they don't bury people who are alive.
4.  six, three for each team.
5.  no; if the man has a widow then he's dead and dead people can't
    marry.
6.  they are not playing each other.
7.  77.
8.  white, he is at the north pole.  (technically, there are no bears
    at the n.p.)
9.  two (reminds me of Bucky Banzai's statement "wherever you go, there
    you are!")
10. a half dollar and a nickel; the one that is not a nickel is a half
    dollar.
11. the match.
12. 1 hour.
13. 9.
14. meat.
15. a dozen what?
16. William Jefferson Blythe Rizenthaler Clinton

-- Dave Dodson <dodson@convex.COM>

In article <1993Sep24.165621.12004@galois.mit.edu>
tycchow@cayley.mit.edu (Timothy Y. Chow) writes:
>>3.  Why can't a man living in the USA be buried in Canada?
>ST = only dead people are buried
>MA = USA and Canada are disjoint regions (people can be buried alive
>     after all)

To me, "living in the USA" means that "the USA" is a truthful answer
to the question, "Where do you live?"  One is not "living in Canada"
if one is merely on vacation or business there.

If, when I was on my honeymoon in Canada, you had told my landlord at
the time that I was no longer living in the USA, I think he would have
been quite surprised and a bit concerned.  Needless to say, I had no
intention of being buried alive while there (or of dying, which would
have rendered me no longer living in the USA).

>>11. [...] which one would you light first?
>ST = match
>MA = match unless one of the other items is self-lighting

If there is enough light for you to find your way into this room
without the match already having been lit, I see no obvious reason why
you should light any of these objects.  I've certainly been in rooms
with these objects without bothering to light them.

>>16. What was the presidents name in 1960?
>ST = Clinton?
>MA = can quibble about referent of "the president"

Does the answer change if I ask what was the President's name in
1800?

-- David Karr <karr@cs.cornell.edu>

In article <27v9qo$rb6@vixen.cso.uiuc.edu> rolf@geoserv.isgs.uiuc.edu
writes:
>1.  The average man has one, which is celebrated multiple times. Some
>non-average people like George Washington had more than one birthday.
>When the calendar changed in 1752, he chose to celebrate a day one
>year after his last birthday, which meant having a different date.

Hmmm, I dispute that Washington had more than one birthday based on
information given in your answer.  I celebrate my birthday every May
3rd.  If I choose to celebrate it on May 16th, would I've gained
another birthday?  Why does Washington's choice to celebrate his
birthday on a different day than traditional give him 2?

>3.  Nothing at all prevents this. The man would probably be very
>unhappy, since he is still living.

Hmm, how can he be living in the USA and be buried (alive) in Canada?
In the Canadian Embassy (with loose definitions of in the USA and in
Canada)?

>10. A half dollar and a nickel. The half dollar is not a nickel. They
>could also both be "not a nickel" if you use an older American coin
>called the "half-dime", also worth 5 cents.

I could be wrong, but isn't "penny" and "nickel" both informal names
for the US one cent piece and US five cent piece respectively, and so
no US coin is officially a "nickel"?

>16. [...] Viewed in a different fashion, the president's name in 1960
>was either Eisenhower (the inauguration was in January) or Kennedy.

Again, I dispute you...  I say that in 1960, the then-current
president of the US was Eisenhower, who left office in January (as you
said), but of 1961.  In 1960, the ELECTION was in November.

What was Clinton's name before he changed it?

-- Buddha Buck <bbuck@nyx.cs.du.edu>

In article <1993Sep24.164956.3689@cs.cornell.edu> karr@cs.cornell.edu
(David Karr) writes:
}>5.  Is it legal for a man in California to marry his widow's sister?
}
}Of course---there's no law against it.  In fact, I know two sisters,
}Ann and Bertha.  Ann is Charles' widow, but Charles married Bertha in
}California.  (Yes, each of the names Ann, Bertha, and Charles refers
}to exactly one person in this narrative.)  How can this be?

Arghh.  Never mind my last answers.  Charles was married to Ann, and
performed Bertha's wedding.  He later died.

-- Matthew T. Russotto <russotto@eng.umd.edu>

>>4.  How many outs are there in an inning of baseball?
>ST = MA = 6

Others have said, 3, considering the 9th inning.  No-one has said, as
far as I can tell, that if you include the 9th, 4 and 5 are also
valid.

-- Claudia Mastroianni <mastr@husc7.harvard.edu>

1.  One.

2.  Three every four years, from now until the year 2100.

3.  He can, but he will be very upset about it, most likely.  Assuming
that the definition of "living in the USA" involves having one's
primary residence there, not actually being physically present.

4.  Six, with the possible exception of the last inning, where there
may be three.

5.  I don't think the dead can marry, but I might be wrong-- California
is a strange place.

6.  They weren't playing against each other.

7.  42.  :-)  Actually, this problem probably works better orally:  77.

8.  White, because there aren't any other color bears near the North
Pole.

9.  Two.

10. One is a fifty cent piece, and the other is a nickel.

11. The match.

12. Anywhere from an hour and a half to two hours, assuming you follow
his orders.

13. 9.

14. Meat.

15. 12

16. The president of what?  Do you mean the name of the current
president 33 years ago, or the name of the person who was president in
1960?

-- Matthew T. Russotto <russotto@eng.umd.edu>

1. 1 (He's only born once.)
2. All of 'em.
3. He isn't dead (but that's never stopped me before.)
4. 6
5. Necrophelia is allowed in certain parts of CA
6. They didn't play against each other
7. 77
8. White
9. 2. I took 2, I have 2.
10. One is not a nickel, one is. (1/2 dollar and a nickel.)
11. The lamp to see what I'm doing(?)
12. with me, about 5 sec.
13. 9
14. He weighs steaks, pork chops, etc.
15. 12, and you still can't mail a letter.
16. He was probably called Billy.

-- Jay Pascale <jpascale1@vaxa.hofstra.edu>

1.  One, I guess.
2.  All twelve of them.
3.  Normally you would not bury someone who is still alive.
4.  Six.
5.  A dead man has no need for a wife.
6.  They did not play against each other.
7.  77
8.  White, since it is a polar bear.  The house is at the North Pole.
9.  Three.
10. A half-dollar and a nickel (the half-dollar is not a nickel)
11. The match.
12. One hour.
13. Nine.  Or, 17 if you count the dead ones.
14. Meat.
15. Twelve.
16. Bill

-- Bob Marshall <marshall@NEBULA.SSD.LMSC.LOCKHEED.COM>

Just for general edification - I would claim that an inning could have
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, or 12 outs as follows:

0, 1, or 2:  if a game is rained out, the unfinished top half of an
inning is allowed to remain in the record if the home team is winning
in the sixth inning or beyond.  This is an interpretation of a rule (I
don't have it in front of me) and these are the only ones I'm not sure
of.

3, 4, or 5:  as has been previously pointed out, these are possible
when the home team either has the lead after 8 1/2 or if it wins in the
9th.

6: DUH!

7 or above:  It is legal for a defensive side to record 3+ legal outs
in a half-inning.  The situation in which this would occur is if a run
were to cross the plate on the same play in which the third out was
recorded (not a force play), but that runner had failed to touch a base
on the way around.  In this case, an appeal play can be made by the
defense at any point before the last infielder crosses the base path on
their way to their dugout.  If allowed, this appeal becomes a legal
out, and the run is nullified.

In an extreme case, picture the bases being loaded with two out.  The
batter hits the ball deep into the right-field corner, all three
runners cross home plate without touching it, and the batter is tagged
out at the plate for the third out.  The defense could then legally run
appeal plays on each of the three base runners, resulting in a total
of six legal outs in a half-inning.  Repeat in the bottom half, and you
can have up to 12 legal outs in an inning.  (Unlikely?  Yes, but think
of the poor hitters who tripled into quadruple plays...)

Finally, some people might claim that the answer would in fact be
infinite based on the fact that a dropped third strike where the batter
reaches base is still called a "strike out".  Baseball does not
consider this a legal out in their scoring, but there would be some
semantic justification for this point of view.

I guess this one was trickier than anyone realized...  :-}

-- Scott Bales <sbales@cisco.com>

mastr@husc7.harvard.edu (Claudia Mastroianni) writes:
>Others have said, 3, considering the 9th inning.  No-one has said, as
>far as I can tell, that if you include the 9th, 4 and 5 are also
>valid.

Oh, I hate having to follow up and improve my answer!  Claudia is
right, if the home team goes ahead in the bottom of the ninth, the game
ends.  Now, to complete the answer, I believe that 0, 1 and 2 are also
possible answers if you count games that are rained out, but have
already gone enough innings to be counted.  Or would you actually call
it "zero" if a batter had already taken a pitch when the heavens opened
up and washed everyone away?

-- Rolf Wilson <rolf@geoserv.isgs.uiuc.edu>