Two adults + two young kids = two rooms?!
Jun 25, 2013, 10:24 am
#16
FlyerTalk Evangelist, Ambassador: World of Hyatt
Join Date: Jul 2001
Location: NJ
Programs: Hyatt Globalist, Fairmont Lifetime Plat, UA Silver, dirt elsewhere
Posts: 46,919
It's not likely they will allow 4 people in a room. One person had success at the Madeline property getting a larger room, but Vendome isn't going to do it unless you pay for a suite.
Jun 25, 2013, 10:38 am
#17
Join Date: Jan 2012
Location: Osaka
Programs: Hyatt Explorist, Hilton Gold, UA
Posts: 3,158
Jun 25, 2013, 3:02 pm
#19
Join Date: Jul 2010
Location: Between AMS and BRU
Posts: 8,852
Exactly... considering what kind of suite you will get for such a discount (in Paris). Or book two rooms...
But don't make the hotel the bad guy's here....they still seem to do almost everything that's reasonably possible for those spending points or free nights.
But don't make the hotel the bad guy's here....they still seem to do almost everything that's reasonably possible for those spending points or free nights.
Jun 26, 2013, 12:14 am
#21
Join Date: Aug 2005
Location: BRU, BE
Programs: LH, United, Hyatt, Air France
Posts: 464
As a Diamond member, I booked my room using points for 2 nights. We are 2 adults and 2 kids (6 year old and 17 month old). I am expecting an upgrade upon check-in since I am a Diamond member.
But the concierge is demanding me to book another room which will cost 410 Euros per night. This is not what I expected for a Diamond member. They site legal reasons. My travel is next week and this is the email I received today. In the US/Canada, there is absolutely no problem with any Hyatts imposing this.
Is there any flexibility with this rule?
But the concierge is demanding me to book another room which will cost 410 Euros per night. This is not what I expected for a Diamond member. They site legal reasons. My travel is next week and this is the email I received today. In the US/Canada, there is absolutely no problem with any Hyatts imposing this.
Is there any flexibility with this rule?