If invited out, is paying for own meal reasonable or not please

[MirandaGoshawk]

Opinions wanted! Friend was invited to a pub for her close friend's (evening) birthday meal. Bottles of wine were provided & there were 12 people in total. At the end of the meal, friend was surprised that they split the bill. She had expected the birthday girl to pay. Is she being unreasonable, or was the birthday girl being tight?

Spot on LyingWitch

I am 58 and I have never been invited to a birthday meal out where the birthday person pays for everyone. So I would always expect to pay my share and usually chip in for the birthday person. It is a given among all the people I know - family, friends, workmates. I just don't know anyone affluent enough to pay for everyone.

It depends on the group of friends.
If everyone always pay their own, then it's fair for all.
If the birthday person (btw, girl ) always pays for everyone then it's still fair for everyone and they can choose what they can afford individually.
Everyone still ends up paying for 12 meals over the year (for a group of 12 friends).

I actually think that the birthday person paying is fairer because they adjust the venue to their budget. Whereas if you have a lower income and pay each separate meal, you have to pay according to other people's choice and I suspect it will cost you more overall, or you have to miss some dinners.

So, you choose a venue with £10 per head and spend £120 on your own birthday dinner.
You pay your own way at an average of £20 each and spend £240 in total.
Think about it.

OTOH, each paying their own way allows people on a lower budget to have a better place for a meal.
Whereas one paying for all allows those with a higher budget to be at a better place.

The closest I've ever been is a friends birthday when they paid the bar bill and we paid for our own food.

Never heard of birthday girl laying g for everyone.

8 years ago, though. She really needs to let it go

I would expect to pay unless the birthday girl had said specifically it was her treat.