Conference rusure::math

2034.0. "Pillow Problems" by NETCAD::ROLKE (Tune in, turn on, fail over) Thu Mar 21 1996 10:24

This is a follow-on to problems 2030 and 2031.
I just love the solution given in 2030.4 -- there's something very
appealing about a clean answer.  After looking over the code I
flip the page saying, "Hey! Where is it?".  But the other side
is blank because all the code fits on one side.  Good job.  I
like problems like this.

"Pillow-Problems (Thought Out During Wakeful Hours)",
by Charles L. Dodgson, M.A., 1893, is an interesting book
Apparently Dodgson, or Lewis Carroll, thought up this whole book of
questions, derived the answers and solutions while putting nothing
on paper.  What a guy.  Anyway, here's some examples of his stuff
with respect to combination and probability.

Note: Sorry for any transcription errors; this stuff is hard to get from
the original text with the wacky grammar, punctuation and typeface.   
I even used a looking glass to figure out what the book said!  

Chuck

5. A bag contains one counter, known to be either white or black.
A white counter is put in, the bag shaken, and a counter drawn out,
which proves to be white.  What is now the chance of drawing a 
white counter?


16. There are two bags, one containing a counter, known to be 
either white or black; the other containing 1 white and 2 black.
A white is put into the first, the bag shaken, and a counter drawn
out, which proves to be white.  Which course will now give the
best chance of drawing a white -- to draw from one of the two bags
without knowing which it is, or to empty one bag into the other
and then draw?


19. There are three bags; one containing a white counter and a 
black one, another two white and a black, and the third 3 white
and a black.  It is not known in what order the bags are placed.
A white counter is drawn from one of them, and a black from 
another. What is the chance of drawing a white counter from the 
remaining bag?


27. There are 3 bags, each containing 6 counters; one contains
5 white and one black; another, 4 white and 2 black; the third,
3 white and 3 black. From two bags (it is not known which) 2
counters are drawn, and prove to be black and white.  What is the
chance of drawing a white counter from the remaining bag?


41. My friend brings me a bag containing four counters, each of
which is either black or white.  He bids me draw two, both of which
prove to be white. He then says, "I meant to tell you, before you
began, that there was at least one white counter in the bag.
However, you know it now, without my telling you. Draw again."
(1) What is now my chance of drawing white?
(2) What would it have been, if he had not spoken?

Answers follow...then solutions...

Answers.

5. Two-thirds.

16. The first course gives chance = 1/2; the second, 5/12.
Hence the first is best.

19. Eleven-seventeenths.

27. Seventeen-twentyfifths.

41. (1) Seven-twelfths. (2) One-half.

Solutions.

5. At first sight, it would appear that, as the state of the bag,
after the operation, is necessarily identical with its state before it,
the chance is just what it then was, vis. 1/2. This, however, is an error.

The chances, before the addition, that the bag contains (a) 1 white 
(b) 1 black, are (a) 1/2 (b) 1/2. Hence the chances after the addition,
that it contains (a) 2 white (b) 1 white, 1 black, are the same, vis.
(a) 1/2 (b) 1/2.  Now the probabilities, which these 2 states give to the
observed event, of drawing a white counter , are (a) certainty (b) 1/2.
Hence the chances, after drawing the white counter, that the bag, before
drawing, contained (a) 2 white, (b) 1 white, 1 black, are proportional to
(a) 1/2 * 1 (b) 1/2 * 1/2; i.e. (a) 1/2 (b) 1/4; i.e. (a) 2 (b) 1.  Hence
the chances are (a) 2/3 (b) 1/3.  Hence, after the removal of a white
counter, the chances, that the bag now contains (a) 1 white (b) 1 black,
are for (a) 2/3 and for (b) 1/3.

Thus the chance, of now drawing a white counter, is 2/3.
                                        Q. E. F.


16. The 'a priori' chances of possible states of first bag are 'W, 1/2;
B, 1/2'. Hence chances, after putting W in, are 'WW, 1/2; WB, 1/2'. The
chances, which these give to the 'observed event', are 1, 1/2.  Hence
chances of possible states 'W, B', after the event are proportional to
1, 1/2; i.e. to 2, 1; i.e. their actual values are 2/3, 1/3.

Now, in first course, chance of drawing W is 1/2 * 2/3 + 1/2 * 1/3;
i.e. 1/2.

And, in second course, chances of possible states 'WWBB, WBBB' are 2/3, 
1/3; hence chance of drawing W is 2/3 * 1/2 + 1/3 * 1/4; i.e. 5/12.

Hence first course gives best chance.
                                        Q. E. F.


19. Call the bags A, B, C; so that A contains a white counter and a black
one; &c.

The chances of the orders ABC, ACB, BAC, BCA, CAB, CBA are, a priori, 1/6
each. Since they are equal, we may, instead of multiplying each by the 
probability it gives to the observed event, simply assume those probabilities
as being proportional to the chances after the observed event.

These probabilities are: --
     for
        ABC, 1/2 * 1/3; i.e. 1/6.
        ACB, 1/2 * 1/4; i.e. 1/8.
        BAC, 2/3 * 1/2; i.e. 1/3.
        BCA, 2/3 * 1/4; i.e. 1/6.
        CAB, 3/4 * 1/2; i.e. 3/8.
        CBA, 3/4 * 1/3; i.e. 1/4.
        
Hence the chances are proportional to 4, 3, 8, 4, 9, 6; i.e. they
are these Nos. divided by 34.

Hence the chance, of drawing a white counter from the remaining bag; is
  1/34 * ( 4 * 3/4 + 3 * 3/2 + 8 * 3/4 + 4 * 1/2 + 9 * 2/3 + 6 * 1/2) ;
  i.e. 1/34 * ( 3 + 2 + 6 + 2 + 6 + 3); i.e. 22/34; i.e. 11/17.

27. Call the bags A, B, C.

If the remaining bag be A, chance of observed event = 1/2 chance of drawing
white from B and black from C + 1/2 chance of drawing black from B and white
from C;
 i.e. it = 1/2 * (2/3 * 1/2 + 1/3 * 1/2) = 1/4.
 
Similarly, if remaining bag be B, it is 1/2 * (5/6 * 1/2 + 1/6 * 1/2) = 1/4;
and if it be C, it is 1/2 * (5/6 * 1/3 + 1/6 * 2/3) = 7/36.

Therefore chances of remaining bag being A, B, or C, are as 1/4 to 1/4 to
7/36; i.e. as 9 to 9 to 7.  Therefore they are, in value, (9, 9, 7)/25.

Now, if remaining bag be A, chance of drawing white from it is 5/6;
therefore chance, on this issue, is 5/6 * 9/25 = 3/10; similarly, for B,
it is 2/3 * 9/25 = 6/25; and, for C, 1/2 * 7/25 = 7/50. And entire chance
of drawing white from the remaining bag is the sum of these; i.e.
(15+12+7)/50 = 34/50 = 17/25.

41. (1) As there was certainly at least one W in the bag at first, the
'a priori' chances for the various states of the bag, 'WWWW, WWWB, WWBB,
WBBB,' were '1/8, 3/8, 3/8, 1/8'.

Those would have given, to the observed event, the chances '1, 1/2, 1/6, 0'.

Hence the chances, after the event, for the various states, are proportional
to '1/8 * 1, 3/8 * 1/2, 3/8 * 1/6'; i.e. to '1/8, 3/16, 1/16'; i.e. to
'2, 3, 1'. Hence their actual values are '1/3, 1/2, 1/6'.

Hence the chance, of now drawing W, is '1/3 * 1 + 1/2 * 1/2'; i.e. it
is 7/12.
                                        Q. E. F.

(2) If he had not spoken, the 'a priori' chances for the states 'WWWW,
WWWB, WWBB, WBBB, BBBB', would have been '(1, 4, 6, 4, 1) / 16'.

These would have given, to the observed event, the chances '1, 1/2, 1/6,
0, 0'.

Hence the chances, after the event, for the various states, are proportional
to '1/16 * 1, 1/4 * 1/2, 1/6 * 3/8'; i.e. '1, 2, 1'. Hence their actual
values are '1/4, 1/2, 1/4'.

Hence the chance, of now drawing W, is '1/4 * 1 + 1/2 * 1/2'; i.e. it
is 1/2.
                                        Q. E. F.

Too easy, you say?  Try these...

56. Given the altitudes of a triangle : construct it.

or

72. A bag contains 2 counters, as to which nothing is known except
that each is either black or white.  Acertain the colours without
taking them out of the bag.�

�Where can I hire someone who thinks like this?!?  "My cache memory has
some data. Tell me what it is without reading it." or "My router has
a packet......."   <smiley face here>

>
> 5. A bag contains one counter, known to be either white or black.
>

Does this mean there's an even chance of the "counter" (whatever that is)
being white or black ?  It seems like this would matter.

/Eric

>Does this mean there's an even chance of the "counter" (whatever that is)
being white or black ?  It seems like this would matter.

Yes, it does.  Dodgson's answers depend on unstated a priori assumptions like
this which aren't necessarily sound.  They are (barely) acceptable as a
convention, but modern conventions would usually require more explicitness.

> 5. A bag contains one counter, known to be either white or black.
> A white counter is put in, the bag shaken, and a counter drawn out,
> which proves to be white.  What is now the chance of drawing a 
> white counter?

If I initially believed the probability of the bag holding a white counter to
be p, then we have the following cases:

p/2:     I draw the same counter I put in, and the remaining one is white.
(1-p)/2: I draw the same counter I put in, and the remaining one is black.
p/2:     I draw the original counter (white), and the one I put in remains.
(1-p)/2: I draw the original counter (black), and the one I put in remains.

The fourth case is eliminated, since the counter I drew proved to be white.
Of the remaining cases, the probability that a white counter remains in the
bag is
	2p / (1+p)

If p=1/2, then we have a 2/3 chance that a white counter remains, which makes
sense because the four cases above are all equally likely.
	-- Bill