# Re: predictions

From: Christopher Maloney <dude.domain.name.hidden>
Date: Tue, 03 Aug 1999 22:46:29 -0400

Hi!

I'd like to revive an old thread, that has been bothering me a lot
lately. I hope you'll all agree that it's a fascinating puzzle.
Wei Dai posed this way back, in February of last year:
http://www.escribe.com/science/theory/index.html?mID=38.
I've read the entire thread, and I don't think the question was
ever resolved. Wei, Nick Bostrom, and Hal Finney were the main
contributors.

I'll present the problem similar to the way it came to be described
toward the end of the thread: spoze that a person (whom we'll call
Jane) agrees to an experiment. At time t0 she's sitting there.
Just before time t1, a coin is flipped. If the coin lands tails,
then nothing special will happen. But if the coin lands heads, then
just before time t2, Jane will be duplicated. We want to know, from
Jane's subjective perspective, what are the odds that she will see
heads? To make it very clear, here is a graphical representation:

t0 |
|
t1 T / \ H
/ \
t2 / / \
| | \
t3 Y R B

One addition to the setup that came at the end of the thread is that
just prior to time t3, Jane(s) is/are shown a colored card. The
experimenter knows ahead of time that if the coin lands tails, he
will show Jane a yellow card. If the coin lands heads, he will show
Jane1 a red card and Jane2 a blue card. The reason for doing this is
so that we can say with certainty that at time t3, Jane1 and Jane2
are different people. Now, Jane is not told which card she will see,
but she knows she will be shown one of those three cards. From her
perspective, which color she is shown should be entirely random.

Wei's initial point was that this poses a problem for Tegmark's
approach to determining subjective probabilities:

... for any mutually exclusive and collectively exhaustive set of
probabilities Bi, the probability of an event A is given by
P(A) = Sum-over-i( P(A|Bi) P(Bi) )

Now let's consider Jane's thought processes at time t0. She wants
to know the probability that she'll see heads. Let's assume for
now that she wants to calculate the probability at time t3 that
she will have seen heads. From her perspective, the color
of the card she sees is random, so P(Y) = P(R) = P(B) = 1/3.
Furthermore, she knows that of the three possible outcomes for her,
represented by the three cards, only one will have seen the coin
land tails. That is (again, from her perspective) given that she
has seen a yellow card, the odds that she will have seen heads is
2/3. So we get

P(H,t3) = P(H|Y)P(Y) + P(H|R)P(R) + P(H|B)P(B) = 2/3.

So she believes, at time t0, that at time t3 the odds are 2/3 that
she will have seen heads. This is also the result one gets from
the "strong self-sampling assumption" (SSSA) (Wei, please correct me
if this is wrong).

But there's an alternative way of computing this probability. She
believes that at time t1, the odds are 1/2 that she'll have seen
heads. She also knows that if, at time t1, she saw heads, then the
odds will be 1 that at all later times, she'll continue to remember
seeing heads. Likewise, if she saw tails at time t1, she'll
continue to remember seeing tails. So

P(H,t3) = P(H,t1) = 1/2.

So which is correct? I know which solution I prefer, but I'd like
to get some feedback first.

Note that one need not bring MWI into this at all. The only big
assumption is the existence of a copy machine. Instead of MWI, one
can think of the identical experiment being carried out on an
ensemble of, say, 100 hapless souls Albert, Bernard, Caroline, etc.
At time t1, some number close to 50 will have seen heads. At time
t2, there will be 150 people, 100 of whom remember seeing heads,
and 50 of whom remember seeing tails. From a bird perspective,
if you picked any person at random from this group, the chance that
they'll have seen heads is 2/3.

```--
Chris Maloney
http://www.chrismaloney.com
"Donuts are so sweet and tasty."
-- Homer Simpson
```
Received on Tue Aug 03 1999 - 19:54:41 PDT

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