> The problem is that it is wrong to neglect rare sequences of coin flips.
> A series of 6 heads in a row has a probability of only 1/64, but OTOH
> it produces 64 people who have all seen 6 heads, which largely compensates
> for the low probability.
Even though the frequency approach seems to support my answer, I don't
think it works in general. As a counter example, suppose the experiment
was modified so that any clone made during the experiment is killed at the
end of it. This should not change any of the probabilities mentioned
before, but it radically alters the long-run frequencies if the experiment
is repeated.
I will offer my own tentative analysis. Suppose in a different experiment
a subject is given a sleeping pill without being told what will happen
during his sleep. When he wakes up, the experimenter tells him this:
"You are not a clone. However during your sleep a fair coin was tossed and
if it landed head up an exact duplicate was made of you. What is the
probability that the coin landed head up?"
Compare that question to this one:
"You are not a clone. However during your sleep a fair coin was tossed and
if it landed head up a brain scan was made of you and stored in a
computer. What is the probability that the coin landed head up?"
or this one:
"You are not a clone. However during your sleep a fair coin was tossed and
if it landed head up a glass of water was boiled. What is the probability
that the coin landed head up?"
Now I think the correct answer to all these questions is 1/2. Once you
know you are not a clone, the knowledge that a coin flip outcome may
have caused some unobservable external material to be rearranged should
not change the distribution of that outcome.
But notice that the state of knowledge of the subject after the first
question is the same as the state of knowledge of the subject of the
original experiment after question 3, and therefore their answers must be
the same.
Let C stand for "I am a clone", H stand for "the coin landed head up", and
K stand for the state of knowledge of the subject of the original
experiment immediately after he wakes up. We have P(H | not C and K) =
1/2. Also, it is obvious that P(C | H and K) = 1/2. Now we can deduce P(C
| K) and P(H | K) which are the answers to question 1 and 2.
P(H and not C | K) = P(H | not C and K) P(not C | K) = P(not C | H and K) P(H | K)
= 1/2 * P(not C | K) = 1/2 * P(H | K)
which implies P(not C | K) = P(H | K)
P(H and C | K) = P(C | H and K) P(H | K) = P(H | C and K) P(C | K)
= 1/2 * P(H | K) = 1 * P(C | K)
Therefore 1 - P(C | K) = P(H | K) = 2 * P(C | K), and so P(C | K) = 1/3
and P(H | K) = 2/3.
This set of answers may seem counter-intuitive because you will observe
the subject say "the probability that the coin will land head up is 1/2"
before he goes to sleep and say "the probability that the coin landed head
up is 2/3" immediately after he wakes up. However the alternative is
equally counter-intuitive. If answer set A is correct, and suppose the
copying process is slightly defective so that the original can tell that
he is not a clone when he wakes up, then you will observe the subject say
"the probability that the coin will land head up is 1/2" before he goes to
sleep and say "the probability that the coin landed head up is 1/3"
immediately after he wakes up.