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The beginning of a well known book has been encrypted below. Each letter of the alphabet has been replaced by another using a simple substitution technique. For example, A may always be represented by G, etc. Can you name the book?

Hint: The phrase 'jsql 1632' should read 'year 1632'.

Answer: Robinson Crusoe by Daniel Defoe.

The piece of text is from the original book and should read:

I was born in the year 1632, in the city of York, of a good family, though not of that country, my father being a foreigner of Bremen, who settled first at Hull. He got a good estate by merchandise, and leaving off his trade, lived afterwards at York, from whence he had married my mother, whose relations were named Robinson, a very good family in that country, and from whom I was called Robinson Kreutznaer; but, by the usual corruption of words in England, we are now called - nay we call ourselves and write our name - Crusoe; and so my companions always called me.

a has been represented by q.
b has been represented by r.
c has been represented by x.
d has been represented by y.
e has been represented by s.
f has been represented by u.
g has been represented by n.
h has been represented by z.
i has been represented by f.
j has been represented by a.
k has been represented by k.
l has been represented by p.
m has been represented by m.
n has been represented by w.
o has been represented by e.
p has been represented by h.
q has been represented by g.
r has been represented by l.
s has been represented by d.
t has been represented by c.
u has been represented by b.
v has been represented by o.
w has been represented by t.
x has been represented by i.
y has been represented by j.
z has been represented by v.

Puzzle 808

Which is more probable, to throw:

at least one six with 6 dice
at least two sixes with 12 dice
at least three sixes with 18 dice
at least four sixes with 24 dice

[Ref: ZMXP]

Hint:

Answer: 6 dice.

The probability for one six with 6 dice is .665, 2 from 12 = .619, 3 from 18 = .597 etc. To explain:

For one dice, the probability is simply the probability of not getting a six on any dice, subtracted from 1, so:
1 - ((5/6)^6) = 0.665 (3 dp)

For two sixes, we subtract the probability of no sixes on 12 dice and the probability of exactly one six on 12 dice from 1. The probability of one six is (5/6)^11 (for 11 non-sixes) times (1/6) (for one six) times 12 (for the 12 different arrangements of 1 six and 11 non-sixes - we have to count all these because (5/6)^11*(1/6) is the probability that dice 1 is six and the other 11 are not six, but dice 2, or dice 3, or any of the others could be the one that is the six), so:
1 - (5/6)^12 - (5/6)^11*(1/6)*12 = 0.619 (3 dp)

Similarly for three sixes:
1 - (5/6)^18 - (5/6)^17*(1/6)*18 - (5/6)^16*(1/6)^2*18*17/2 = 0.597 (3 dp).
Etc.

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