- Contemporary messages sorted: [ by date ] [ by thread ] [ by subject ] [ by author ] [ by messages with attachments ]

From: Wei Dai <weidai.domain.name.hidden>

Date: Thu, 4 Jun 1998 17:21:36 -0700

I just realized I made a couple of mistakes in my previous post. The

measures for a,b,c,d,e,f should be 2/7,1/7,1/7,1/7,1/7,1/7 respectively.

On Thu, Jun 04, 1998 at 04:58:47PM -0700, Wei Dai wrote:

*> According to the above definitions, P(I will see a heads-up coin and clock
*

*> reading 2|I currently see clock reading 0) = 1/2, whereas P(I see a
*

*> heads-up coin|I see clock reading 2) = 2/3. In general under this
*

*> definition P(I currently observe X|I observed Y) <> P(I will observe X|I
*

*> currently observe Y) whenever there is some possibility of copying or
*

*> death between observations X and Y.
*

Actually with the setup I gave, P(I will see a heads-up coin and clock

*> reading 2|I currently see clock reading 0) should be 1. If there were
*

two observer-instants at time 0, a0 and a1 each with measure 1/7, such

that a0 gives rise to b and a1 gives rise to c, that probability would be

1/2. So there is a crucial difference between observer splitting and

observer differentiation under my proposed definition.

Received on Thu Jun 04 1998 - 17:22:49 PDT

Date: Thu, 4 Jun 1998 17:21:36 -0700

I just realized I made a couple of mistakes in my previous post. The

measures for a,b,c,d,e,f should be 2/7,1/7,1/7,1/7,1/7,1/7 respectively.

On Thu, Jun 04, 1998 at 04:58:47PM -0700, Wei Dai wrote:

Actually with the setup I gave, P(I will see a heads-up coin and clock

two observer-instants at time 0, a0 and a1 each with measure 1/7, such

that a0 gives rise to b and a1 gives rise to c, that probability would be

1/2. So there is a crucial difference between observer splitting and

observer differentiation under my proposed definition.

Received on Thu Jun 04 1998 - 17:22:49 PDT

*
This archive was generated by hypermail 2.3.0
: Fri Feb 16 2018 - 13:20:06 PST
*