Le mar. 18 févr. 2025, 03:42, Bruce Kellett <[email protected]> a
écrit :

> On Tue, Feb 18, 2025 at 12:00 PM Quentin Anciaux <[email protected]>
> wrote:
>
>> Bruce,
>>
>> Consider the following thought experiment, which directly parallels MWI
>> and illustrates why your argument assumes what it tries to prove.
>>
>> Imagine we have a machine that can perfectly duplicate an observer, just
>> as MWI implies happens during quantum measurements. The experiment works as
>> follows:
>>
>> 1. The observer enters a sealed box.
>>
>> 2. Inside the box, they are duplicated into 10 copies.
>>
>> 3. Each duplicate is placed in an identical room with only one visible
>> difference:
>>
>> One of them sees a 0 written on the wall.
>>
>> Nine of them see a 1 written on the wall.
>>
>> 4. The observer, upon exiting the box, can only report what they
>> personally experienced.
>>
>
> It has occurred to me what is wrong with this example. Instead of
> considering a two-outcome experiment, where we get either a zero or a one,
> you have considered a ten-outcome experiment, with one zero and nine ones.
> This is not equivalent to the binary case under consideration. In the
> binary case, we get 2^N possible sequences after N trials, (not the 10^N
> sequences as in your example). Because there are only two possible outcomes
> in my example, the majority of sequences will have approximately a 50/50
> split of zeros and ones. The majority of observers are then going to use
> their data to estimate a probability of 0.5 for getting a zero. Now this
> bears absolutely no relation to the actual Born probability, which is
> a^2.The majority of observers estimate p = 0.5, whatever the value of a.
> This is because there are only 2^N possible binary sequences of length N,
> and we get the same 2^N sequences whatever the values of a and b. That is
> why I say the amplitudes have no effect.
>
> Bruce
>

There are only two possible outcome in my example, seeing 1 or seeing 0.

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