So you say you know probability... Amazingly silly question yet very tricky...

[Mickey]

Three men - A, B and C - were in separate cells under sentence of
death when the governor decided to pardon one of them. He wrote their
names on three slips of paper, shook the slips in a hat, drew out one
of them and telephoned the warden, requesting that the name of the
lucky man be kept secret for several days. Rumor of this reached
prisoner A. When the warden made his morning rounds, A tried to
persuade the warden to tell him who had been pardoned. The warden
refused. "Then tell me," said A, "the name of one of the others who
will be executed. If B is to be pardoned, give me C's name. If C is to
be pardoned, give me B's name. And if I'm to be pardoned, flip a coin
to decide whether to name B or C." "But if you see me flip the coin,"
replied the wary warden, "you'll know that you're .the one pardoned.
And if you see that I don't flip a coin, you'll know it's either you
or the person I don't name." "Then don't tell me now," said A. "Tell
me tomorrow morning." The warden, who knew nothing about probability
theory, thought it over that night and decided that if he followed the
procedure suggested by A, it would give A no help whatever in
estimating his survival chances. So next morning he told A that B was
going to be executed. After the warden left, A smiled to himself at
the warden's stupidity. There were now only two equally probable
elements in what mathematicians like to call the "sample space" of the
problem. Either C would be pardoned or himself, so by all the laws of
conditional probability, his chances of survival had gone up from 1/3
to 1/2. The warden did not know that A could communicate with C, in an
adjacent cell, by tapping in code on a water pipe. This A proceeded to
do, explaining to C exactly what he had said to the warden and what
the warden had said to him. C was equally overjoyed with the news
because he figured, by the same reasoning used by A, that his own
survival chances had also risen to 1/2. Did the two men reason
correctly? If not, how should each have calculated his chances of
being pardoned?

Regards,
Jyoti

[Mickey]

Before you break your head or worse, my head, let me tell you the
correct answer: The probability that A will survive is 1/3 and the
probability that C will survive is 2/3. I immediately struck that
Baye's Theorem is applicable. When I was wiki`ing it I figured out
that this is a very famous problem, here are the links to identical
problems:
http://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
http://en.wikipedia.org/wiki/Monty_Hall_problem
http://en.wikipedia.org/wiki/Three_Prisoners_problem


I got this from an old book: The Second Book of Mathematical Puzzles
and Diversions 1961 ed.

Regards,
Jyoti