Re: Doctor Who S01E06 Rogue

Blueshirt <blueshirt@indigo.news> wrote:

The Doctor wrote:
 

In article <v4bj99$1i3mv$1@dont-email.me>,
Daniel70  <daniel47@nomail.afraid.org> wrote:

The Doctor wrote on 12/06/2024 12:53 am:

 
I do not use illegit stuff.

 
Sorry!! When YOU, Gobble-de-gook, watch 'Doctor Who', are
YOU, Gobble-de-gook, watching a Canadian T.V. station
Broadcast, a Disney+ Broadcast or a BBC Broadcast??
 
Simple question, Gobble-de-gook!!

 
BBC and a legit subscription.

 
No, Dave it's not a legitimate subscription.
 
Pops Yadallee might pay for an IPTV service that carries BBC1,
but it's not legitimate.
 

He wilfully refuses to understand it, and would rather lie and lie than
admit to using illicit means for viewing the show. It’s a Catch-22 for him,
he’s sinning twice over now because he can’t pay the CAN7.99 a month
(£4.55) to watch the show legally - (with ads tier - premium is CAD 149.99
for the year, £7.11 a month).

Actually, if all he’s really interested in is Who, he only needs CAD24 a
year - 2 dollars a month saved up from his pocket money - 2 months access
for the season and 1 month to see the Christmas special.

But he’s such a successful businessman that he can’t afford the cost of a
single cup of coffee every two months in order to watch his favourite show
legally.

The BBC does not broadcast or stream its channels to Canada - it cannot,
because it sells licenses to local channels for each programme separately
and that would never work if the BBC was available there. Any BBC access he
has is illicit.

In Canada and the USA (in fact everywhere in the world apart from the U.K.
and Ireland) the ONLY legit channel for Doctor Who 2024 is Disney+, just as
in earlier years it was Space in Canada, and BBCA and then HBO in the USA.
(BBC America was an independent entity and had to licence Doctor Who from
the BBC. In the opposite direction, Orphan Black was a BBCA original
production and the BBC had to pay to show it in the U.K.).



--
“Most of the Universe is knackered, babes.”  - The Doctor