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Author Topic: Riddle Me This! - Series  (Read 741 times)

katsumoto

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Riddle Me This! - Series
« Reply #120 on: March 22, 2004, 09:26:00 AM »

beerchug.gif

EPISODE XXIV

You're sitting at a bar, an explosively-dank hole in Funky Town USA.
You have an award that says, most likely to be the character in a riddle . . . looking around your attention finds a small chair with a mobile entity covering it with it's lesser half . . . It speaks, you hear "Glass of water please . . . " You see the bartender examine the seated man for a moment. He then reaches under the counter and pulls out a shotgun point-blank in the man's face. The seated man speaks again "Thank You" he says, and walks out of the bar . . . why did the man get what he needed?

Good luck!   biggrin.gif
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LepPpeR

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Riddle Me This! - Series
« Reply #121 on: March 22, 2004, 01:49:00 PM »

Well a ShotGun would certainly scare the Hiccups right out of me.


So my guess is that he had the hiccups and the shot gun got rid of them but the water might have helped as well.
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katsumoto

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Riddle Me This! - Series
« Reply #122 on: April 01, 2004, 06:40:00 AM »

biggrin.gif
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SKoT

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« Reply #123 on: April 01, 2004, 07:13:00 AM »

Saturday: 684/4800
Sunday: 687/4800
Monday: 685/4800
Tuesday: 685/4800
Wednesday: 687/4800
Thursday: 684/4800
Friday: 688/4800


Lol, this math class is boring btw
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LepPpeR

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Riddle Me This! - Series
« Reply #124 on: April 01, 2004, 11:57:00 AM »

In a period of 400 years, the cycle of weekdays makes one complete cycle through the years (it cannot be a shorter period because of the odd leap year rule; it does not have to be a longer period because a period of 400 years has 400 × 365 = 146000 normal days plus 100 - 3 = 97 leap days, which makes 146097 days in total, which is divisible by 7). So, we only have to look at the probability in a period of 400 years, for example the years 2001 up to 2400.
For a normal year, the distribution of the 13th days over the days of the week is as follows, if January 13th falls on weekday x:


   x   x+1  x+2  x+3  x+4  x+5  x+6
   2    1    1    3    1    2    2

For a leap year, the distribution of the 13th days over the days of the week is as follows, if January 13th falls on weekday x:


   x   x+1  x+2  x+3  x+4  x+5  x+6
   3    1    1    2    2    1    2

For subsequent years, these distributions shift cyclic 1 place to the right for normal years, and 2 places to the right for leap years.

For three normal years followed by one leap year, we can now calculate the distribution of the 13th days:


   x   x+1  x+2  x+3  x+4  x+5  x+6
   2    1    1    3    1    2    2   (normal year, not shifted)
   2    2    1    1    3    1    2   (normal year, shifted 1 place to the right)
   2    2    2    1    1    3    1   (normal year, shifted 2 places to the right)
+  2    1    2    3    1    1    2   (leap year, shifted 3 places to the right)
 -----------------------------------
   8    6    6    8    6    7    7   (3 normal years and 1 leap year, not shifted)

For four normal years, we find:


   x   x+1  x+2  x+3  x+4  x+5  x+6
   2    1    1    3    1    2    2   (normal year, not shifted)
   2    2    1    1    3    1    2   (normal year, shifted 1 place to the right)
   2    2    2    1    1    3    1   (normal year, shifted 2 places to the right)
+  1    2    2    2    1    1    3   (normal year, shifted 3 places to the right)
 -----------------------------------
   7    7    6    7    6    7    8   (4 normal years, not shifted)

In four years, the latter two distributions shift 5 places cyclic to the right. So for a period of 16 years with 4 leap years, we find:


   x   x+1  x+2  x+3  x+4  x+5  x+6
   8    6    6    8    6    7    7   (3 normal years and 1 leap year, not shifted)
   6    8    6    7    7    8    6   (3 normal years and 1 leap year, shifted 5 places to the right)
   6    7    7    8    6    6    8   (3 normal years and 1 leap year, shifted 10 places to the right)
+  7    8    6    6    8    6    7   (3 normal years and 1 leap year, shifted 15 places to the right)
 -----------------------------------
  27   29   25   29   27   27   28   (16 years with 4 leap years, not shifted)

For a period of 100 years with 24 leap years (the years 2001-2100, 2101-2200, and 2201-2299), we can now calculate the distribution of the 13th days:


   x   x+1  x+2  x+3  x+4  x+5  x+6
  27   29   25   29   27   27   28   (16 years with 4 leap years, not shifted)
  29   25   29   27   27   28   27   (16 years with 4 leap years, shifted 20 places to the right)
  25   29   27   27   28   27   29   (16 years with 4 leap years, shifted 40 places to the right)
  29   27   27   28   27   29   25   (16 years with 4 leap years, shifted 60 places to the right)
  27   27   28   27   29   25   29   (16 years with 4 leap years, shifted 80 places to the right)
  27   28   27   29   25   29   27   (16 years with 4 leap years, shifted 100 places to the right)
+  8    7    7    6    7    6    7   (4 normal years, shifted 120 places to the right)
 -----------------------------------
 172  172  170  173  170  171  172   (100 years with 24 leap years, not shifted)

For a period of 100 years with 25 leap years (the years 2301-2400), we can now calculate the distribution of the 13th days:


   x   x+1  x+2  x+3  x+4  x+5  x+6
  27   29   25   29   27   27   28   (16 years with 4 leap years, not shifted)
  29   25   29   27   27   28   27   (16 years with 4 leap years, shifted 20 places to the right)
  25   29   27   27   28   27   29   (16 years with 4 leap years, shifted 40 places to the right)
  29   27   27   28   27   29   25   (16 years with 4 leap years, shifted 60 places to the right)
  27   27   28   27   29   25   29   (16 years with 4 leap years, shifted 80 places to the right)
  27   28   27   29   25   29   27   (16 years with 4 leap years, shifted 100 places to the right)
+  7    8    6    6    8    6    7   (3 normal years and 1 leap year, shifted 120 places to the right)
 -----------------------------------
 171  173  169  173  171  171  172   (100 years with 25 leap years, not shifted)

Now we can calculate the distribution of the 13th days for the period 2001 up to 2400:


   x   x+1  x+2  x+3  x+4  x+5  x+6
  172  172  170  173  170  171  172   (100 years with 24 leap years, not shifted)
  170  173  170  171  172  172  172   (100 years with 24 leap years, shifted 124 places to the right)
  170  171  172  172  172  170  173   (100 years with 24 leap years, shifted 248 places to the right)
+ 172  171  173  169  173  171  171   (100 years with 25 leap years, shifted 372 places to the right)
 ------------------------------------
  684  687  685  685  687  684  688   (400 years, not shifted)

Since January 13th, 2001 (x in the distribution) is a Saturday, we get the following probability for the 13th days over the days of the week:

Saturday: 684/4800
Sunday: 687/4800
Monday: 685/4800
Tuesday: 685/4800
Wednesday: 687/4800
Thursday: 684/4800
Friday: 688/4800


Conclusion: the probability that the 13th of a certain month in a certain year is a Friday, is the highest.



Have a nice day...... I do not condone the cheating on homework.
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