I'm Caucasian and my fiancé is Chinese - I met her on a trip to Tahiti though as chance would have it, it turned out that we actually grew up in adjacent towns in Ohio.

We decided to have a very intimate wedding and only invite our respective immediate family members. We had yet to ever introduce one another to our respective families. We had no extended family and all our grandparents had already passed on, so with each of us having just one sibling, the total number of invites sent out was 4 (one to each set of parents, and one to both my and my fianceé's siblings).

Everyone RSVPed. My sibling said they'd be bringing another guest, as did my fiancée's.

However on the day of the wedding only 6 people turned up and yet everyone we were expecting was at the affair!


Hint 1: The siblings did not bring one another as their +1.

Hint 2: Both my and my fianceé parents are happily married couples and there's no unconventional hanky panky going on between any members of my or my fianceé's family.

Hint 3 (huge hint - possible spoiler... not needed to solve)

Both our siblings are half Chinese, half Caucasian females.


3 Answers 3


Your parents have separated. So have your fiancée's. One of your parents has subsequently started a relationship with one of hers, and they had a child - your half-sibling, and also your fiancée's half-sibling. You have one sibling each, it just happens that those siblings are the same person. (Also, you may be your fiancée's step-brother)

Guest numbers:

Your parents (2), your fiancée's parents (2), your shared sibling (1) and their date (1) for a total of 6

  • $\begingroup$ very nicely done! $\endgroup$ Nov 2, 2015 at 9:52
  • $\begingroup$ The precise solution I was looking for (and it's based on a hypothetical reality in my actual family - sans any awkwardness) is that my sibling's parents split up (Caucasian and Chinese). They each then started new relationships - the Caucasian with another Caucasian and the Chinese with another Chinese. Each new couple had a child (me and my fiancé). The initial child is both our sibling and yet my fiancé and I have no genetic ties of any kind. I'll gladly take pointers on how to better reword the puzzle to avoid the fracas in the answer arrived at by CodeNewbie $\endgroup$ Nov 2, 2015 at 9:58
  • $\begingroup$ This puzzle's definitely easier to solve if you already have a mess of step- and half- siblings. Though I'm the older one in my family, which probably affects how I approached the puzzle (no younger ones in a position to get married - I'm not sure they've actually met!) $\endgroup$ Nov 2, 2015 at 10:38
  • 5
    $\begingroup$ Based on the question, neither of you have ever met your sibling, which seems unlikely if you're going to invite them to such a small wedding. You also stated "both" siblings, which means there are two separate people involved. That could be reworded as "My sibling said they'd be bringing another guest, as did my fiancée's". Another scenario is that your sibling takes one of your fiancées parents and your fiancées sibling takes one of your parents. $\endgroup$
    – thelem
    Nov 2, 2015 at 11:25
  • 4
    $\begingroup$ " I had never met her family members and she had never met mine" this statement makes the answer completely wrong. $\endgroup$
    – AeJey
    Nov 2, 2015 at 13:27


Each sibling was thinking of bringing the other

  • $\begingroup$ That works out logically, but not the answer I was looking for - I'll edit the question to remove that possibility. $\endgroup$ Nov 2, 2015 at 8:34

An oddball answer, but fits the solution well.

There is one (awkward!) coincidence that could have taken place. The parents of the bride and groom must be divorced, and the siblings are dating each other's father. So the siblings end up there with their dates, who also happens to be the fathers of the bride and groom. And the mothers are there too, bringing up the total to 6.

  • $\begingroup$ hahah was thinking the same lol $\endgroup$
    – Mekalikot
    Nov 2, 2015 at 9:23
  • $\begingroup$ holy jesus. That also works... and now I'm stumped as to how to reword the puzzle again so as to prevent this bizarre mishmash :) I'll think of something. +1 for creativity, but the answer is actually dramatically less convoluted! $\endgroup$ Nov 2, 2015 at 9:51

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