Another Beauty argument

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Wei-Hwa Huang

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Jul 7, 1999, 3:00:00 AM7/7/99
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Suppose we didn't erase Beauty's memory, and she knows this.
Everything else remains the same as in the previous problem --
on heads, she gets wakened on Monday; on tails, she gets
wakened on Monday and Tuesday.

Obviously, when we ask her the question on a wakening, she will
either think:
(1) "Well, I don't remember being wakened, so it must be Monday. This
means that the probability of the coin being heads was 1/2."
(2) "Well, I remember being awakened yesterday, so today is Tuesday
and the probability of the coin being heads is 0."

Now, suppose we DO erase her memory but don't tell her. Then she will
always think it is Monday, and she will always answer 1/2.

And finally, if tell her we're going to erase her memory, we revert to
the original problem (and note that it doesn't matter if we actually
erase her memory or not): A halfer SB will think:
"My memory's been erased, so I don't know what today is for sure. But
by such-and-such an argument, the probability is 1/2."
And a thirder will think:
"My memory's been erased, so I don't know what today is for sure. But
by such-and-such an argument, the probability is 1/3."

Obviously the answer is different between the two problems because of the
added memory wipe and SB's knowledge of it.

The halfer, then, thinks that Sleeping Beauty's knowledge of the erasure
is irrelevant, while the thirder seems to think it is.

--
Wei-Hwa Huang, whu...@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
"You may outrank me, but a bath will change that."

Nick Wedd

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Jul 7, 1999, 3:00:00 AM7/7/99
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In article <7m0bp2$h...@gap.cco.caltech.edu>, Wei-Hwa Huang
<whu...@ugcs.caltech.edu> writes

>The halfer, then, thinks that Sleeping Beauty's knowledge of the erasure
>is irrelevant, while the thirder seems to think it is.

"seems"? We _do_ think it is relevant.

Nick
--
Nick Wedd ni...@maproom.co.uk

Julius K Yako

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Jul 8, 1999, 3:00:00 AM7/8/99
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what is your point?

Excerpts from netnews.rec.puzzles: 7-Jul-99 Another Beauty argument by
Wei-Hwa Hu...@ugcs.calte

> Suppose we didn't erase Beauty's memory, and she knows this.
> Everything else remains the same as in the previous problem --
> on heads, she gets wakened on Monday; on tails, she gets
> wakened on Monday and Tuesday.
>
> Obviously, when we ask her the question on a wakening, she will
> either think:
> (1) "Well, I don't remember being wakened, so it must be Monday. This
> means that the probability of the coin being heads was 1/2."
> (2) "Well, I remember being awakened yesterday, so today is Tuesday
> and the probability of the coin being heads is 0."
>
> Now, suppose we DO erase her memory but don't tell her. Then she will
> always think it is Monday, and she will always answer 1/2.
>
> And finally, if tell her we're going to erase her memory, we revert to
> the original problem (and note that it doesn't matter if we actually
> erase her memory or not): A halfer SB will think:
> "My memory's been erased, so I don't know what today is for sure. But
> by such-and-such an argument, the probability is 1/2."
> And a thirder will think:
> "My memory's been erased, so I don't know what today is for sure. But
> by such-and-such an argument, the probability is 1/3."
>
> Obviously the answer is different between the two problems because of the
> added memory wipe and SB's knowledge of it.
>

> The halfer, then, thinks that Sleeping Beauty's knowledge of the erasure
> is irrelevant, while the thirder seems to think it is.
>

Wei-Hwa Huang

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Jul 8, 1999, 3:00:00 AM7/8/99
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Nick Wedd <Ni...@maproom.co.uk> writes:
>In article <7m0bp2$h...@gap.cco.caltech.edu>, Wei-Hwa Huang
><whu...@ugcs.caltech.edu> writes

>>The halfer, then, thinks that Sleeping Beauty's knowledge of the erasure


>>is irrelevant, while the thirder seems to think it is.

>"seems"? We _do_ think it is relevant.

You do?? No wonder it seems that way, then! :-)

Jamie Dreier

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Jul 8, 1999, 3:00:00 AM7/8/99
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> >>The halfer, then, thinks that Sleeping Beauty's knowledge of the erasure
> >>is irrelevant, while the thirder seems to think it is.
>
> >"seems"? We _do_ think it is relevant.
>
> You do?? No wonder it seems that way, then! :-)


A few weeks ago I was walking down the street with a friend, and we saw a
man with an odd-looking hat. I said, "That hat looks funny." My friend
replied, "Maybe that's because it *is* funny."

-Jamie

--
SpamGard: For real return address replace "DOT" with "."

Wei-Hwa Huang

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Jul 9, 1999, 3:00:00 AM7/9/99
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Julius K Yako <ya...@andrew.cmu.edu> writes:
>what is your point?

That the key to the paradox is the question:
"Should Beauty's knowledge of the mind erasure change her assessment of the
probabilities?"

James Smith

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Jul 10, 1999, 3:00:00 AM7/10/99
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Wei-Hwa Huang wrote:

> That the key to the paradox is the question:
> "Should Beauty's knowledge of the mind erasure change her assessment of the
> probabilities?"

Actually, the key to the paradox is that thirders leave out a piece
of the probability equation. This is sort of like starting a proof
with "Given that 1 + 1 = 3, ...".

Jim

jon wild

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Jul 11, 1999, 3:00:00 AM7/11/99
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Hi - sorry to invent another situation perhaps analogous to the SB puzzle.
It's all in the name of trying to figure out what I think about the
original. Could you (anyone who's thought about SB) say whether you think
everyone should answer this puzzle the same way they answer SB:

A man stops you on the street and says "this morning I decided to flip a
coin, and tell one stranger about it if it came up heads, and two if came
up tails. Have a nice day." What is your credence that the coin came up
heads?

If you think factors like what time of day the stranger approaches you, or
how crowded the street is, might affect your credence, then do you think
replacing "one" and "two" with n and 2n makes the puzzle more
straight-forward?

That's it - sorry if someone's talked about something similar already, I
missed a few weeks in the middle of the discussion.

Cheers - Jon

David A Karr

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Jul 11, 1999, 3:00:00 AM7/11/99
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It only looks that way if you have a mental condition that makes the
numeral "2" look like a "3" to you. Sort of like the way King Arthur
in Monty Python and the Holy Grail kept counting, "1, 2, 5 ... no, 3!"

Wei-Hwa's thesis rings truer and truer each time it comes around.

--
David A. Karr "Groups of guitars are on the way out, Mr. Epstein."
ka...@shore.net --Decca executive Dick Rowe, 1962

Gerry Quinn

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Jul 11, 1999, 3:00:00 AM7/11/99
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Thirders (like me) will see the analogy as valid. Halfers have a
problem with the fact that two different people are approached instead
of one with a bad memory. (Hmmm, I feel a version coming on in which
there are two twin Beauties who are subjected to a brain swap on Monday
night...)

I did post a similar alternative in which Beauty parachutes repeatedly
into one of two large fields (based on the toss of a coin), one of which
has one daisy and one of which has two. Finally, she lands by a daisy.
What are the odds that the last toss was heads?

In fact, this may be encapsulate the real difference. If you think
"What are the odds?" is equivalent to "What credence do you have?", you
are a thirder.

- Gerry Quinn

Nick Wedd

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Jul 12, 1999, 3:00:00 AM7/12/99
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In article <37874B4C...@cc.newcastle.edu.au>, James Smith
<en...@cc.newcastle.edu.au> writes

>Actually, the key to the paradox is that thirders leave out a piece
>of the probability equation. This is sort of like starting a proof
>with "Given that 1 + 1 = 3, ...".

What piece do you think it is that they leave out? And how would you
answer Jon Wild's question, elsewhere in this thread?

James Smith

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Jul 12, 1999, 3:00:00 AM7/12/99
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Nick Wedd wrote:

> >Actually, the key to the paradox is that thirders leave out a piece
> >of the probability equation. This is sort of like starting a proof
> >with "Given that 1 + 1 = 3, ...".
>
> What piece do you think it is that they leave out?

Thirders solve the equations

P(heads,Monday) + P(tails,Monday) + P(tails,Tuesday) = 1
P(heads,Monday) = P(tails,Monday) = P(tails,Tuesday)

to get P(heads) = 1/3 when they should be solving the equations

P(heads,Monday) + P(tails,Monday) + P(tails,Tuesday) - P(tails,Monday
AND Tuesday) = 1
P(heads,Monday) = P(tails,Monday) = P(tails,Tuesday) = P(tails,Monday
AND Tuesday)

to get P(heads) = 1/2. So the bit they leave out is
P(tails,Monday AND Tuesday).

> And how would you
> answer Jon Wild's question, elsewhere in this thread?

Haven't seen it yet.

Jim

James Smith

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Jul 12, 1999, 3:00:00 AM7/12/99
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Gerry Quinn wrote:

> >A man stops you on the street and says "this morning I decided to flip a
> >coin, and tell one stranger about it if it came up heads, and two if came
> >up tails. Have a nice day." What is your credence that the coin came up
> >heads?
> >
> >If you think factors like what time of day the stranger approaches you, or
> >how crowded the street is, might affect your credence, then do you think
> >replacing "one" and "two" with n and 2n makes the puzzle more
> >straight-forward?
> >
> >That's it - sorry if someone's talked about something similar already, I
> >missed a few weeks in the middle of the discussion.
>
> Thirders (like me) will see the analogy as valid. Halfers have a
> problem with the fact that two different people are approached instead
> of one with a bad memory. (Hmmm, I feel a version coming on in which
> there are two twin Beauties who are subjected to a brain swap on Monday
> night...)

No, the analogy is the same and the answer is the same. It doesn't
matter
how many people are approached, for either side, it can't change the
probability.

> I did post a similar alternative in which Beauty parachutes repeatedly
> into one of two large fields (based on the toss of a coin), one of which
> has one daisy and one of which has two. Finally, she lands by a daisy.
> What are the odds that the last toss was heads?

She counts up the daisies in the field and knows what the last toss was.

Jim

Michael Hochster

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Jul 12, 1999, 3:00:00 AM7/12/99
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I think you are confused about what thirders are actually
saying. Let me restate it as clearly as I can. The
events MONDAY and TUESDAY are shorthand for
"Today is Monday" or "Today is Tuesday." Thus,
the event (MONDAY and TUESDAY) is empty.

Now, let's back up.

(1) if Beauty knows it's Tuesday, the probability the
coin came up heads is 0.

(2) If she knows it's Monday, she is in EXACTLY the same
position as another person who is awakened only once on
Monday with no memory wiping and asked what is the
probability that the coin is heads. Therefore, if she knows
it's Monday, the probability is 1/2.

Since she does not know what day it is and there is
some chance it's Tuesday, the probability must be
STRICTLY less than 1/2. Let's not quibble about the
exact value.

**** Posted from RemarQ - http://www.remarq.com - Discussions Start Here (tm) ****

r.e.s.

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Jul 12, 1999, 3:00:00 AM7/12/99
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Michael Hochster <anon...@web.remarq.com> wrote ...
[snip]
:
: (1) if Beauty knows it's Tuesday, the probability the

: coin came up heads is 0.
:
: (2) If she knows it's Monday, she is in EXACTLY the same
: position as another person who is awakened only once on
: Monday with no memory wiping and asked what is the
: probability that the coin is heads. Therefore, if she knows
: it's Monday, the probability is 1/2.
:
: Since she does not know what day it is and there is
: some chance it's Tuesday, the probability must be
: STRICTLY less than 1/2. Let's not quibble about the
: exact value.

Of course a halfer would have no trouble refuting the
above "intuitive" claim that p(H|MON)=1/2:

(2*) Clearly p(MON|H)=1, because if SB knows that the
coin is Heads, then she knows that "Today is Monday".
Also, it's clear that p(MON|T)<1 (strictly), because
if SB knows that the coin is Tails, then she knows
that "Today might be either Monday or Tuesday". Hence,
p(H|MON)
= p(MON|H)*p(H)/(p(MON|H)*p(H) + p(MON|T)*p(T)
= 1 * (1/2)/( 1 *(1/2) + p(MON|T)*(1/2))
which must be STRICTLY greater than 1/2. But let's
not quibble about the exact value. ;o) ;o)

Both (2) and (2*) are defensible when H, MON, etc are
treated as events, as though a legitimate probability
model has been constructed. (It hasn't been, imo.)
--
r.e.s.
XXr...@ix.netcom.com (Spam-block=XX)


Michael Hochster

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Jul 13, 1999, 3:00:00 AM7/13/99
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r.e.s. wrote:

>Of course a halfer would have no trouble refuting the
>above "intuitive" claim that p(H|MON)=1/2:

>(2*) Clearly p(MON|H)=1, because if SB knows that the coin
>is Heads, then she knows that "Today is Monday". Also,
>it's clear that p(MON|T)<1 (strictly), because if SB knows
>that the coin is Tails, then she knows that "Today might
>be either Monday or Tuesday". Hence, p(H|MON) =
>
>p(MON|H)*p(H)/(p(MON|H)*p(H) + p(MON|T)*p(T)

>1 * (1/2)/( 1 *(1/2) + p(MON|T)*(1/2))


Good one! Except: I think your p(H) is actually the
answer to the interview question (i.e., Beauty's
personal probability of heads, not knowing what day
it is). If you want to assume that quantity is 1/2,
why bother with any further analysis? :)

Mike

Michael Hochster

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Jul 13, 1999, 3:00:00 AM7/13/99
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>The problem with this argument is that it is assuming that
>(tails,Monday) and (tails,Tuesday) are mutually exclusive
>events; when they are actually mutually inclusive (if
>there is such a term). To say that Beauty knows that it is
>Monday or Tuesday but not both is to say that subjectively
>Beauty is awoken on either Monday or Tuesday but not both,
>and that changes the problem.

>Jim

I repeat: the event (tails, Monday) is shorthand for
"the coin landed tails and TODAY is Monday," and
similarly for tails. Today cannot be both Monday and
Tuesday. Therefore (tails, Monday) and
(tails, Tuesday) ARE exclusive.

David A Karr

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Jul 13, 1999, 3:00:00 AM7/13/99
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James Smith <en...@cc.newcastle.edu.au> wrote:
>To say that Beauty knows that it is Monday or Tuesday but not both

As I sit here and respond to a Usenet post, I know that it is Monday
or Tuesday but not both. I also know what English speakers typically
mean when they say, "It is Monday or Tuesday but not both." Sure
enough, if you assign these words a different meaning, the truth of
the statement may indeed change.

r.e.s.

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Jul 13, 1999, 3:00:00 AM7/13/99
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Michael Hochster <anon...@web.remarq.com> wrote ...
:

: r.e.s. wrote:
:
: >Of course a halfer would have no trouble refuting the
: >above "intuitive" claim that p(H|MON)=1/2:
:
: >(2*) Clearly p(MON|H)=1, because if SB knows that the coin
: >is Heads, then she knows that "Today is Monday". Also,
: >it's clear that p(MON|T)<1 (strictly), because if SB knows
: >that the coin is Tails, then she knows that "Today might
: >be either Monday or Tuesday". Hence, p(H|MON) =
: >
: >p(MON|H)*p(H)/(p(MON|H)*p(H) + p(MON|T)*p(T)
: >1 * (1/2)/( 1 *(1/2) + p(MON|T)*(1/2))
:
:
: Good one! Except: I think your p(H) is actually the
: answer to the interview question (i.e., Beauty's
: personal probability of heads, not knowing what day
: it is). If you want to assume that quantity is 1/2,
: why bother with any further analysis? :)

That isn't being assumed, though. As I used it, p(H) is
not conditional on SB being awake for interview, but is
the "a priori" probability of Heads, given only the game
rules (which state that it's a fair coin), i.e.
p(H|G)=1/2 -- which I was abbreviating as p(H).

Also, since the thirder argument (2) was worded in terms
of SB "knowing that it's Monday" etc, I used MON to mean
"Today is Monday, and I (Beauty) know it".

So the same equation (just Bayes' Rule), written in more
complete notation, might look like this, bearing in mind
that Beauty never forgets the rules of the game (G):

p(H|G.MON)
=p(MON|G.H)*p(H|G)/(p(MON|G.H)*p(H|G) + p(MON|G.T)*p(T|G)
1 * (1/2)/( 1 * (1/2) + p(MON|G.T)*(1/2))
> 1/2 (strictly)
because, as before, p(MON|G.T)<1/2 (strictly).

--
r.e.s.
XXr...@ix.netcom.com (Spam-block=XX)


David A Karr

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Jul 13, 1999, 3:00:00 AM7/13/99
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r.e.s. <XXr...@ix.netcom.com> wrote:
>Michael Hochster <anon...@web.remarq.com> wrote ...
>: Good one! Except: I think your p(H) is actually the
>: answer to the interview question (i.e., Beauty's
>: personal probability of heads, not knowing what day
>: it is). If you want to assume that quantity is 1/2,
>: why bother with any further analysis? :)
>
>That isn't being assumed, though. As I used it, p(H) is
>not conditional on SB being awake for interview, but is
>the "a priori" probability of Heads, given only the game
>rules (which state that it's a fair coin), i.e.
>p(H|G)=1/2 -- which I was abbreviating as p(H).
>[...]

>p(H|G.MON)
>=p(MON|G.H)*p(H|G)/(p(MON|G.H)*p(H|G) + p(MON|G.T)*p(T|G)
> 1 * (1/2)/( 1 * (1/2) + p(MON|G.T)*(1/2))

But it can certainly occur that the coin is heads (H) and
Beauty knows the rules of the game (G) on Sunday (NOT MON),
so why don't you write p(MON|G.H) < 1 ?

What I would like you to write (and I think Michael would too) is
p(MON|G.H.W) = 1, where G is the rules of the game, H is the
proposition that the coin came up heads, and W is the proposition that
Beauty is in one of the game awakenings _right_ _now_. W is a strange
kind of proposition: it is a function of time, that is, it is false at
one time, true at a later time, and false again at an even later time.

Jamie Dreier doesn't like this, so he says W cannot be used in any
non-trivial fashion in his probability model; for similar reasons, he
will also reject any non-trivial term of the form
p("Today is Monday" | X). Jamie's approach gives us a consistent
halfer probability model. In this model, as Wei-Hwa postulated, the
memory wiping is not relevant; the reason for its irrelevance is that
it affects only inadmissible "propositions" such as "I've had this
experience earlier than _right_ _now_."

The equation you've written, however, is a mix-and-match affair. As
you yourself pointed out, there is no coherent probability model
underlying it. A coherent model (based on time-dependent probability
functions) does, I think, exist, but you would have to rewrite your
equations above to fit it, for example by replacing p(H|G) by p(H|G.W),
where p(H|G.W) is not known "a priori" to be 1/2.

Michael Hochster

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Jul 13, 1999, 3:00:00 AM7/13/99
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r.e.s. wrote:
p(H|G.MON) =p(MON|G.H)*p(H|G)/(p(MON|G.H)*p(H|G) + p
(MON|G.T)*p(T|G)

>1 * (1/2)/( 1 * (1/2) + p(MON|G.T)*(1/2)) > 1/2 (strictly)


>because, as before, p(MON|G.T)<1/2 (strictly).

p(MON|G.H) = 1 does not hold from Beauty's point of
view unless the experiment is in progress. In order
for your calculations to be consistent, all probability
statements including p(H|G) should also be conditional on
the experiment being in progress. But p(H|G, experiment in
progress) is the quantity in question.

r.e.s.

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Jul 13, 1999, 3:00:00 AM7/13/99
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David A Karr <ka...@shore.net> wrote ...

: r.e.s. <XXr...@ix.netcom.com> wrote:
: >Michael Hochster <anon...@web.remarq.com> wrote ...
: >: Good one! Except: I think your p(H) is actually the
: >: answer to the interview question (i.e., Beauty's
: >: personal probability of heads, not knowing what day
: >: it is). If you want to assume that quantity is 1/2,
: >: why bother with any further analysis? :)
: >
: >That isn't being assumed, though. As I used it, p(H) is
: >not conditional on SB being awake for interview, but is
: >the "a priori" probability of Heads, given only the game
: >rules (which state that it's a fair coin), i.e.
: >p(H|G)=1/2 -- which I was abbreviating as p(H).
: >[...]
: >p(H|G.MON)
: >=p(MON|G.H)*p(H|G)/(p(MON|G.H)*p(H|G) + p(MON|G.T)*p(T|G)

: > 1 * (1/2)/( 1 * (1/2) + p(MON|G.T)*(1/2))
:
: But it can certainly occur that the coin is heads (H) and

: Beauty knows the rules of the game (G) on Sunday (NOT MON),
: so why don't you write p(MON|G.H) < 1 ?

Good point.
I tried to avoid (additional) references to suspect
"events" of the kind you call W below, and got it
wrong in the effort. Something like p(H|G.W) is indeed
involved in (2*), and that does beg the question.

However, MON is already of this suspect kind, since
the truth value of "Today is Monday" varies with the
actual day of the week. (I pointed this out, and my
qualms about it, months ago, btw.) So both arguments,
(2) and (2*), are suspect, because they both refer
to such events.

: What I would like you to write (and I think Michael would too) is


: p(MON|G.H.W) = 1, where G is the rules of the game, H is the
: proposition that the coin came up heads, and W is the proposition that
: Beauty is in one of the game awakenings _right_ _now_. W is a strange
: kind of proposition: it is a function of time, that is, it is false at
: one time, true at a later time, and false again at an even later time.

Of course I didn't want to write that, knowing
that W is, like MON, a "suspect event" -- I was
trying to avoid referring to yet another of them,
but failed. ;o(

: Jamie Dreier doesn't like this, so he says W cannot be used in any


: non-trivial fashion in his probability model; for similar reasons, he
: will also reject any non-trivial term of the form
: p("Today is Monday" | X). Jamie's approach gives us a consistent
: halfer probability model.

Maybe. I still have nagging doubts about whether
memory-loss is being (or can be?) properly modelled.

--
r.e.s.
XXr...@ix.netcom.com (Spam-block=XX)


r.e.s.

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Jul 13, 1999, 3:00:00 AM7/13/99
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Michael Hochster <hoch...@rocketmail.com> wrote ...

: r.e.s. wrote:
: p(H|G.MON) =p(MON|G.H)*p(H|G)/(p(MON|G.H)*p(H|G) + p
: (MON|G.T)*p(T|G)
:
: >1 * (1/2)/( 1 * (1/2) + p(MON|G.T)*(1/2)) > 1/2 (strictly)

: >because, as before, p(MON|G.T)<1/2 (strictly).
:
: p(MON|G.H) = 1 does not hold from Beauty's point of
: view unless the experiment is in progress. In order
: for your calculations to be consistent, all probability
: statements including p(H|G) should also be conditional on
: the experiment being in progress. But p(H|G, experiment in
: progress) is the quantity in question.

Good catch. Your point is well-taken, and
I think that the argument (2*) would indeed
require something like p(H|G, experiment in
progress) in the role of what I called p(H),
and that would be begging the question.
--
r.e.s.
XXr...@ix.netcom.com (Spam-block=XX)


James Smith

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Jul 14, 1999, 3:00:00 AM7/14/99
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Michael Hochster wrote:

> I think you are confused about what thirders are actually
> saying. Let me restate it as clearly as I can. The
> events MONDAY and TUESDAY are shorthand for
> "Today is Monday" or "Today is Tuesday." Thus,
> the event (MONDAY and TUESDAY) is empty.
>
> Now, let's back up.
>

> (1) if Beauty knows it's Tuesday, the probability the
> coin came up heads is 0.
>
> (2) If she knows it's Monday, she is in EXACTLY the same
> position as another person who is awakened only once on
> Monday with no memory wiping and asked what is the
> probability that the coin is heads. Therefore, if she knows
> it's Monday, the probability is 1/2.
>
> Since she does not know what day it is and there is
> some chance it's Tuesday, the probability must be
> STRICTLY less than 1/2. Let's not quibble about the
> exact value.

The problem with this argument is that it is assuming that


(tails,Monday) and (tails,Tuesday) are mutually exclusive events;
when they are actually mutually inclusive (if there is such a term).

To say that Beauty knows that it is Monday or Tuesday but not both

is to say that subjectively Beauty is awoken on either Monday

James Smith

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Jul 14, 1999, 3:00:00 AM7/14/99
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Michael Hochster wrote:
>
> >The problem with this argument is that it is assuming that
> >(tails,Monday) and (tails,Tuesday) are mutually exclusive
> >events; when they are actually mutually inclusive (if
> >there is such a term). To say that Beauty knows that it is
> >Monday or Tuesday but not both is to say that subjectively
> >Beauty is awoken on either Monday or Tuesday but not both,
> >and that changes the problem.

> I repeat: the event (tails, Monday) is shorthand for


> "the coin landed tails and TODAY is Monday," and
> similarly for tails. Today cannot be both Monday and
> Tuesday. Therefore (tails, Monday) and
> (tails, Tuesday) ARE exclusive.

So therefore in the original problem P(heads) = P(tails) = 1/2
and P(tails,Monday) = P(tails,Tuesday) = 1/4.

So P(heads) = P(Monday) * P(heads,Monday) + P(Tuesday) *
P(heads,Tuesday)
= 3/4 * 2/3 + 1/4 * 0
= 1/2

Jim

Michael Hochster

unread,
Jul 14, 1999, 3:00:00 AM7/14/99
to

James Smith wrote:

So therefore in the original problem P(heads) = P(tails) =
1/2 and P(tails,Monday) = P(tails,Tuesday) = 1/4.
So P(heads) = P(Monday) * P(heads,Monday) + P(Tuesday) * P
(heads,Tuesday)

3/4 * 2/3 + 1/4 * 0
1/2


OK, I am about ready to give up here. You have not
responded to what I wrote before at all. Let me just
enumerate everything that I see that is wrong with
the calculation above, and you can work out whether
you still agree with yourself.

First, you state P(heads) = 1/2. Then you do some
analysis to compute P(heads). Is this the same
quantity? If so, it is not surprising you reach the
answer you have assumed at the outset. If you
are computing a conditional probability, what are
you conditioning on?

Second, you have expressed the law of total probability
incorrectly. You should write:

P(heads) = P(heads, Monday) + P(heads, Tuesday)
Where the "," in the usual notation means "and."
The right hand side can be rewritten:
= P(heads | Monday) P(Mon) + P(heads | Tuesday)*P(Tue)
where P(A | B) means the probability of A GIVEN B is
known (the usual notation).

[All these probability statements are implicitly
conditional on the experiment being in progress,
an important point! This is why it is not obvious
straightaway that P(heads) = 1/2]

We agree P(heads | Tuesday) = 0,
so focus on the other term. What probability of heads
should Beauty say given that she knows it's Monday?
If you think this is something other than 1/2, try
expressing clearly the difference between her position
and the position of another person who is awakened
only on Monday (and knows this) and asked for the
probability of heads. If you can be convinced
P(heads |Mon) = 1/2, the only way P(heads) can be
1/2 is if P(Mon) =1, which it clearly is not.

David A Karr

unread,
Jul 14, 1999, 3:00:00 AM7/14/99
to
r.e.s. <XXr...@ix.netcom.com> wrote:
>David A Karr <ka...@shore.net> wrote ...
>: What I would like you to write [...] is p(MON|G.H.W) = 1, [...]

>
>Of course I didn't want to write that, knowing
>that W is, like MON, a "suspect event" -- I was
>trying to avoid referring to yet another of them,
>but failed. ;o(

In for a penny, in for a pound; I think once you've admitted suspect
"events" such as "Today is Monday," you have to admit all the events
of that kind, and W is such an event (like Monday, it is a simple
Boolean function of time).

>: Jamie's approach gives us a consistent


>: halfer probability model.
>
>Maybe. I still have nagging doubts about whether
>memory-loss is being (or can be?) properly modelled.

I think the answer is "cannot be modeled," since the memory that
Beauty loses in this example consists entirely of her ability to
assess statements such as "Today is Monday." So Jamie would say that
no _relevant_ memory is lost. Beauty's answers are based on the same
facts that she would have used before she went to sleep, and she still
knows those facts.

I still think it is sometimes reasonable (and in particular, is
reasonable in Sleeping Beauty) to assign weights of credence to
statements of the form "Today is Monday" when the truth is not known
for sure. And this leads me to a thirder model. But if I had to stop
assigning probabilities to those statements then I would agree with
Jamie.

Note the following behaviors if you subject the following people to
this protocol: you follow the standard SB protocol, except that after
they answer your question about credence in heads, you tell them what
day it is and ask again about credence in heads. Different people
will behave as follows on Monday:

Thirders like me:
Initial credence that coin is heads: 1/3
Credence after you say "Today is Monday": 1/2

Halfers who reject entirely any meaning for P(Today is Monday):
Initial credence that coin is heads: 1/2
Credence after you say "Today is Monday": 1/2

Halfers who compute with values such as P(Today is Monday):
Initial credence that coin is heads: 1/2
Credence after you say "Today is Monday": 2/3

Whatever my problems with either thirder or Jamie-Dreier-halfer
analyses, they are nothing in comparison to my problems with any
analysis that forces you to say heads have probability 2/3 when you
come into possession of the same facts held by someone who was never
drugged at all.

James Smith

unread,
Jul 15, 1999, 3:00:00 AM7/15/99
to
Michael Hochster wrote:

> OK, I am about ready to give up here. You have not
> responded to what I wrote before at all. Let me just
> enumerate everything that I see that is wrong with
> the calculation above, and you can work out whether
> you still agree with yourself.

Fair enough. We could well be arguing about different things.

> First, you state P(heads) = 1/2. Then you do some
> analysis to compute P(heads). Is this the same
> quantity? If so, it is not surprising you reach the
> answer you have assumed at the outset. If you
> are computing a conditional probability, what are
> you conditioning on?

The trouble is that there is no other way to compute
the probability in the first place. The only thing you have
in the original problem that you can compute the probability
from is that P(heads) = P(tails) = 1/2.

> Second, you have expressed the law of total probability
> incorrectly. You should write:
>
> P(heads) = P(heads, Monday) + P(heads, Tuesday)
> Where the "," in the usual notation means "and."
> The right hand side can be rewritten:
> = P(heads | Monday) P(Mon) + P(heads | Tuesday)*P(Tue)
> where P(A | B) means the probability of A GIVEN B is
> known (the usual notation).

How do you calculate P(Mon) and P(Tue) ? Without these
quantities what you have written is essentially meaningless.
The equation could well read

P(heads) = P(heads | Monday)* P(Mon) + P(heads | Tuesday)*P(Tue)
1/2 = 1/2 * 1 + 0 * 0

When you respond to this, note that you can't say

P(tails) = P(tails,Monday) + P(tails,Tuesday)

because this assumes that (tails,Monday) and (tails,Tuesday)
are mutually exclusive, and they aren't.

Jim

Michael Hochster

unread,
Jul 15, 1999, 3:00:00 AM7/15/99
to
Mike wrote:

The right hand side can be rewritten:
= P(heads | Monday) P(Mon) + P(heads | Tuesday)*P(Tue)
where P(A | B) means the probability of A GIVEN B is known
(the usual notation).

Jim wrote:

How do you calculate P(Mon) and P(Tue) ? Without these
quantities what you have written is essentially
meaningless. The equation could well read
P(heads) = P(heads | Monday)* P(Mon) + P(heads | Tuesday)*P
(Tue)
1/2 = 1/2 * 1 + 0 * 0

----------------------

Absolutely wrong! Remember, this calculation is from
Beauty's perspective. "Mon" is simply the statement
"today is Monday." P(Mon) is Beauty's degree of
belief that today (i.e. the day of the interview) is
Monday. This is clearly less than 1, since she
is not certain it is Monday.

Jim wrote:

When you respond to this, note that you can't say
P(tails) = P(tails,Monday) + P(tails,Tuesday)
because this assumes that (tails,Monday) and
(tails,Tuesday) are mutually exclusive, and they aren't.

------------

For the tenth time, they ARE. "tails, Monday" simply
means "the coin landed tails and TODAY is Monday",
similarly for "tails, Tuesday." Today cannot
be both Monday and Tuesday, hence "tails, Monday" and
"tails, Tuesday" are exclusive. If you are willing
to accept that it is possible to assign a probability
to the event "today is Monday" (and there are arguments
that it is not possible), the "thirder" calculation
is correct.

Michael Hochster

unread,
Jul 15, 1999, 3:00:00 AM7/15/99
to

James Smith wrote in message <378E29AE...@cc.newcastle.edu.au>...
>
>So, if (tails,Monday) and (tails,Tuesday) are mutually
>exclusive, then...
>
>The scientist tosses a coin. It can land heads or tails. These events
>are mutually exclusive and cover all possibilities, so
> P(heads) + P(tails) = 1

> P(heads) = P(tails) = 1/2


So far, agreed. Assuming you mean the subjective probability of heads
*before* the experiment.

>If the coin lands tails, the scientist awakens Beauty on Monday
>or Tuesday. These events are mutually exclusive (as you have so
>ardently stated) and cover all possibilites that can occur on
>a toss of tails, so
> P(tails,Monday) + P(tails,Tuesday) = 1/2
> P(tails,Monday) = P(tails,Tuesday) = 1/4


You have already determined that P(tails) = P(heads) = 1/2,
where P(tails) means beauty's probability of tails before the
experiment. The calculations above and below just go in
a circle. You assert P(heads) = 1/2 at the outset, then
break it into two parts and show they add up to 1/2. But
these calculations do not correspond to beauty's position.

Her position is, she does not know what day it is. But,
she knows that if she were told it was Monday, she
would assign a proability of 1/2 to heads. (You never
responded before to my query about why this is wrong:
again, how does her position upon being told it is
Monday differ from someone who is never drugged?).
She also knows that, if she were told it was Tuesday,
she would assign a probability of 0 to heads. On this
I am sure we agree. Does it not follow that she should
assign a probability to heads somewhere between
0 and 1/2 ?

>Plug those values into your calculations and you will discover
>that there is no contradiction.
>
>P(heads) = P(heads|Monday) * P(Monday) + P(heads|Tuesday) * P(Tuesday)
> 1/2 = 2/3 * 3/4 + 0 * 1/4


Again, the 3/4 value here should alert you that something is wrong.
P(heads | Monday) means beauty's personal assessment of
the probability of heads upon being told it is monday. Upon
receiving this information, she knows no memory wiping has occurred,
and in fact she just like any other person who is told that a fair
coin has been flipped. Therefore, this value should be 1/2.

James Smith

unread,
Jul 16, 1999, 3:00:00 AM7/16/99
to
Michael Hochster wrote:

> When you respond to this, note that you can't say
> P(tails) = P(tails,Monday) + P(tails,Tuesday)
> because this assumes that (tails,Monday) and
> (tails,Tuesday) are mutually exclusive, and they aren't.
> ------------
>
> For the tenth time, they ARE. "tails, Monday" simply
> means "the coin landed tails and TODAY is Monday",
> similarly for "tails, Tuesday." Today cannot
> be both Monday and Tuesday, hence "tails, Monday" and
> "tails, Tuesday" are exclusive. If you are willing
> to accept that it is possible to assign a probability
> to the event "today is Monday" (and there are arguments
> that it is not possible), the "thirder" calculation
> is correct.

So, if (tails,Monday) and (tails,Tuesday) are mutually
exclusive, then...

The scientist tosses a coin. It can land heads or tails. These events
are mutually exclusive and cover all possibilities, so
P(heads) + P(tails) = 1
P(heads) = P(tails) = 1/2

If the coin lands tails, the scientist awakens Beauty on Monday
or Tuesday. These events are mutually exclusive (as you have so
ardently stated) and cover all possibilites that can occur on
a toss of tails, so
P(tails,Monday) + P(tails,Tuesday) = 1/2
P(tails,Monday) = P(tails,Tuesday) = 1/4

Plug those values into your calculations and you will discover


that there is no contradiction.

P(heads) = P(heads|Monday) * P(Monday) + P(heads|Tuesday) * P(Tuesday)
1/2 = 2/3 * 3/4 + 0 * 1/4

P(tails) = P(tails|Monday) * P(Monday) + P(tails|Tuesday) * P(Tuesday)
1/2 = 1/3 * 3/4 + 1 * 1/4

And so, from Beauty's point of view, P(heads) is 1/2.

If you now want to turn around and say that (tails,Monday) and
(tails,Tuesday) are not mutually exclusive, then you can't
say

P(tails) = P(tails|Monday) * P(Monday) + P(tails|Tuesday) * P(Tuesday)


Jim

James Smith

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Jul 16, 1999, 3:00:00 AM7/16/99
to
Michael Hochster wrote:

> You have already determined that P(tails) = P(heads) = 1/2,
> where P(tails) means beauty's probability of tails before the
> experiment. The calculations above and below just go in
> a circle.

Indeed they do. That is because the basis for calculating probability
within the problem is the fairness of the coin and then the split
on tails into Monday or Tuesday. If you want to argue some other
set of probabilities you have to derive them in a different way,
i.e by not relying on the fact that the coin is fair.

> Her position is, she does not know what day it is. But,
> she knows that if she were told it was Monday, she
> would assign a proability of 1/2 to heads. (You never
> responded before to my query about why this is wrong:

Looking at the problem objectively P(heads|Monday) = 1/2. Looking at
it subjectively you don't know what it is, because you do not know
what the subjective probabilities are. All you know is that they are
not the same as the objective probabilities. So your proof is based
on an assumption that need not be and in fact isn't true.

> >Plug those values into your calculations and you will discover
> >that there is no contradiction.
> >
> >P(heads) = P(heads|Monday) * P(Monday) + P(heads|Tuesday) * P(Tuesday)
> > 1/2 = 2/3 * 3/4 + 0 * 1/4
>

> Again, the 3/4 value here should alert you that something is wrong.

As it is plain that the subjective probabilities are not the same
as the objective ones, no particular value here could alert me that
something was wrong, save perhaps a value outside [0,1].

> P(heads | Monday) means beauty's personal assessment of
> the probability of heads upon being told it is monday. Upon
> receiving this information, she knows no memory wiping has occurred,
> and in fact she just like any other person who is told that a fair
> coin has been flipped. Therefore, this value should be 1/2.

Objectively it's 1/2. Subjectively it differs, because subjectively
if it is Monday it is more likely that the coin will be heads.

If you want to work out the subjective probabilities you cannot just
assume they are the same as the objective ones. You have to derive them.
The way to do that is to start with the objective probabilities and then
calculate the subjective ones by applying the changes necessary to reach
the subjective viewpoint. I've done that here twice. You are welcome
to try and do it without starting with the assumption that the coin is
fair. I would like to see your working.

Jim

hoch...@rocketmail.com

unread,
Jul 16, 1999, 3:00:00 AM7/16/99
to

> > P(heads | Monday) means beauty's personal assessment of
> > the probability of heads upon being told it is monday. Upon
> > receiving this information, she knows no memory wiping has occurred,
> > and in fact she just like any other person who is told that a fair
> > coin has been flipped. Therefore, this value should be 1/2.
>
> Objectively it's 1/2. Subjectively it differs, because subjectively
> if it is Monday it is more likely that the coin will be heads.
>

You know this cannot be true! Once SB is told it's Monday,
she knows nothing strange is going on. She has not been woken
twice, drugged, memory wiped, or anything else. She just knows
that a fair coin was flipped yesterday. You are suggesting that
the fact that she could be awakened TOMORROW if the coin is
tails should make it more likely to her subjectively that the
coin is heads? I hope you can see the absurdity of this.

> If you want to work out the subjective probabilities you cannot just
> assume they are the same as the objective ones. You have to derive
them.
> The way to do that is to start with the objective probabilities and
then
> calculate the subjective ones by applying the changes necessary to
reach
> the subjective viewpoint. I've done that here twice. You are welcome
> to try and do it without starting with the assumption that the coin is
> fair. I would like to see your working.
>
> Jim
>

Of course you need to know that the coin is fair. But that's
not ALL you need. I'll work the problem from A to Z with all
the details spelled out. If you disagree with the analysis
I'm sure I will never convince you, and I'll stop trying.

Let E denote the event "the experiment is in progress."
What we want is P(heads | E). "|" means "given."
This is a subjective probability, i.e., Beauty's
personal assessment of the chances. This is to be
distinguished from P(heads), which is the probability
BEFORE the experiment, understood to be 1/2.

You have baldly stated that P(heads | E) = P(heads) = 1/2.
The arguments you have offered to support this statement
ASSUME it; that is, you break up P(heads) in various ways
and then at the end just say, P(heads) = 1/2, (and imply
without argument) that P(heads | E) = P(heads).

Now let's see why this is wrong. We can write:

P(heads | E) = P(heads | E, today is Monday) * P(Today is Monday |E)
+ P(heads | E, today is Tuesday) * P(Today is Tuesday)

Note that "today is Monday" and "today is Tuesday" are
mutually exclusive events. Let's not bark up that tree again.
We can ignore the second term since
P(heads |E, today is Tuesday) = 0. By the arguments I have
now given several times, P(heads | E, today is Monday)
must be 1/2. This leaves us with:

P(heads | E) = 1/2 * P(Today is Monday | E) (*)

Now, how can we work this out? Let's define p = P(heads | E).
We can write:

P(Today is Monday | E) = P(Today is Monday| heads, E) *
P(heads | E)
+ P(Today is Monday| tails, E) *
P(tails | E)
= p + P(Today is Monday | tails, E) * (1-p) (**)

Given the information that the coin landed tails, SB knows
it could be either Monday or Tuesday, and from her point
of view neither is more likely than the other. So
combining (*) and (**)

Gives p = 1/2 * (p + 1/2*(1-p))
Solving for p gives p = 1/3.

Mike

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Michael Hochster

unread,
Jul 17, 1999, 3:00:00 AM7/17/99
to

>
>If she knows it is Monday then the subjective and objective
>probabilities
>are the same. Similarly, they are the same if she knows it is Tuesday.
>If she doesn't know what day it is, then they just as obviously differ.
>You know that from the objective point of view the possibility of it
>being
>Monday is not mutually exclusive of the possibility of it being Tuesday.
>For these to be mutually exclusive, if one of them happens during the
>course of the experiment then the other must not. Yet if the coin
>lands tails both must happen. Subjectively you are arguing that they
>are mutually exclusive. That being the case, and given that the
>probability of neither event is zero, the probabilities within
>the subjective view must change. Given that you are trying to
>argue that subjectively the probability of the coin landing heads
>changes to 1/3 you have to agree with this.
>
>Now, if you don't know what the subjective probabilities are,
>how can you say that the subjective probability must lie between
>the two objective probabilities? You can't. The whole argument becomes
>meaningless.


I am arguing ONLY from the subjective point of view, and I have
given reasons (I think convincing ones) for all the subjective
probability assignments I have descibed.


>> By the arguments I have
>> now given several times, P(heads | E, today is Monday)
>> must be 1/2.
>

>The rest of your working is fine. This is not. It is an assumption.
>What is your reason for assuming it?
>


I have explained this several times. I don't know if you have read
the explanations and disagree or if you have not read them. If
you have read them, you have not responded to them. But I'll
spell it out again.

P(heads | E, today is monday) is the subjective assignment of
the probability Beauty should make to the coin landing heads,
given that she knows the experiment is in progress and that
today is Monday. At this time, Beauty knows that yesterday,
on Sunday, a fair coin was flipped, that she went to sleep,
and woke up. Imagine she has a friend with her, Sleeping
Cutie, who went to sleep at the same time, woke up at the
same time, and will be asked the same question about the
same coin. The only difference is that Sleeping Cutie will not
be wakened on Tuesday if the coin comes up heads. Now,
Beauty and Cutie are in possession of EXACTLY the same
facts about what has happened so far, which is the only information
that is relevant to how the coin has landed (in particular, SB's
knowledge that she may be wakened TOMORROW if the
coin landed tails is not relevant). Therefore they
must have the same SUBJECTIVE probability -- two rational
beings in possession of the same facts should come to
the same conclusions. Since SC has no basis for assigning
a probability other than 1/2, neither does SB. And that is my
reason for the "assumption" that P(heads | E, monday) =1/2.

Mike

James Smith

unread,
Jul 18, 1999, 3:00:00 AM7/18/99
to
hoch...@rocketmail.com wrote:

> > Objectively it's 1/2. Subjectively it differs, because subjectively
> > if it is Monday it is more likely that the coin will be heads.
> >
>
> You know this cannot be true! Once SB is told it's Monday,
> she knows nothing strange is going on.

If she knows it is Monday then the subjective and objective


probabilities
are the same. Similarly, they are the same if she knows it is Tuesday.
If she doesn't know what day it is, then they just as obviously differ.
You know that from the objective point of view the possibility of it
being
Monday is not mutually exclusive of the possibility of it being Tuesday.
For these to be mutually exclusive, if one of them happens during the
course of the experiment then the other must not. Yet if the coin
lands tails both must happen. Subjectively you are arguing that they
are mutually exclusive. That being the case, and given that the
probability of neither event is zero, the probabilities within
the subjective view must change. Given that you are trying to
argue that subjectively the probability of the coin landing heads
changes to 1/3 you have to agree with this.

Now, if you don't know what the subjective probabilities are,
how can you say that the subjective probability must lie between
the two objective probabilities? You can't. The whole argument becomes
meaningless.

> You have baldly stated that P(heads | E) = P(heads) = 1/2.


> The arguments you have offered to support this statement
> ASSUME it; that is, you break up P(heads) in various ways
> and then at the end just say, P(heads) = 1/2, (and imply
> without argument) that P(heads | E) = P(heads).

You are looking at this arse-about-tit, if you will forgive the french.
I am saying that the only obvious way to calculate the probabilities
is to assume that P(heads) = 1/2. I then show that nothing in the
calculations that flow from this disproves it.

> By the arguments I have
> now given several times, P(heads | E, today is Monday)
> must be 1/2.

The rest of your working is fine. This is not. It is an assumption.
What is your reason for assuming it?

Jim

James Smith

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Jul 18, 1999, 3:00:00 AM7/18/99
to
Michael Hochster wrote:

> I am arguing ONLY from the subjective point of view, and I have
> given reasons (I think convincing ones) for all the subjective
> probability assignments I have descibed.

You have argued that if the subjective viewpoint differs from the
objective in that SB knows it's Monday, P(heads) = 1/2.
You have argued that if the subjective viewpoint differs from the
objective in that SB knows it's Tuesday, P(heads) = 1.
You then proceed to argue based on the subjective viewpoint that
the three events are mutually exclusive. But these are three different
subjective viewpoints. Why should the probabilities in one have any
relationship to those in any other?

> >> By the arguments I have
> >> now given several times, P(heads | E, today is Monday)
> >> must be 1/2.
> >
> >The rest of your working is fine. This is not. It is an assumption.
> >What is your reason for assuming it?
>

> I have explained this several times. I don't know if you have read
> the explanations and disagree or if you have not read them. If
> you have read them, you have not responded to them. But I'll
> spell it out again.
>
> P(heads | E, today is monday) is the subjective assignment of
> the probability Beauty should make to the coin landing heads,
> given that she knows the experiment is in progress and that
> today is Monday.

If she knows that it is Monday then you are arguing from a different
subjective viewpoint. Why should the probabilies in the two viewpoints
bear any relationship to each other?

Jim

Michael Hochster

unread,
Jul 18, 1999, 3:00:00 AM7/18/99
to
I give up. I've explained my position as clearly as I can.
I don't even understand your questions now, especially
the way you are using the terms "objective" and
"subjective." It seems further discussion will not be
fruitful, so let's just call it quits.

Mike

James Smith

unread,
Jul 19, 1999, 3:00:00 AM7/19/99
to
Michael Hochster wrote:
>
> I give up. I've explained my position as clearly as I can.
> I don't even understand your questions now, especially
> the way you are using the terms "objective" and
> "subjective." It seems further discussion will not be
> fruitful, so let's just call it quits.

We've been arguing about this for a while. I'd like to take a step
back and look at the broader view.

We both start with the premise that Beauty's subjective viewpoint
differs from the objective one in that (tails,Monday) and
(tails,Tuesday) become mutually exclusive events.

I make the observation that P(heads) = 1/2 and from this derive
P(tails,Tuesday) = 1/4. You have agreed that the maths for this
is correct and contains no inconsistencies.

You make the observation that P(tails,Tuesday) = 1/3 and from this
derive that P(heads) = 1/3. I agree that the maths is correct and
contains no inconsistencies apart from one step which I still think
is incorrect. But let that pass for the moment.

It is obvious that the two observations and the original premise
lead to contradictions. You are saying that my observation must be
wrong. I am saying that yours must be wrong. There is a third
possibility, that the original premise must be wrong. This being
the case, what we have really shown is that, from Beauty's
viewpoint, the (tails,Monday) and (tails,Tuesday) events are not
mutually exclusive, which in turn means that Beauty's subjective
view does not differ from the objective one.

Jim

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