'The most interesting questions of visitors from the Dutch sister site (Bridgevaria.nl)'. 

Dear Ed and Peter.

We play Rubensohl if an opponent overcalls 2♣ or 2♦ over our 1NT opening. If partner has one or two major four-card suits, she has to reply 3♣.
Our problem is that we play 2NT and up as transfer bids (2NT transfers to clubs, 3♣ transfers to diamonds). Or should we only do so without intervention?

Thank you in advance.

Best regards.
Els

After Ed & Peter's replies Els (and her partner Mimi) asked for more Rubensohl-examples. In response to that Peter has extended his answer, resulting in his final reply below. 

Reply Ed Hoogenkamp (South)

Dear Els, 

I have the feeling you have mixed up Rubensohl and other agreements. In Rubensohl you cannot use 3♣ to enquire after major four-card suits, since it is already in use (as a transfer, as you observe yourself). In a nutshell Rubensohl looks like this:

1NT - (2♦) - ??

double : for penalties

2♥/♠ : to play

2NT: transfer to 3♣

3♣ : transfer to 3♦ (= the opponent's suit) -> enquires about a four-card major suit

3♦ : transfer to hearts (invitational strength or better)

3♥ : transfer to spades (invitational strength or better)

3♠ : transfer to 3NT -> no four-card major suit, no guard in the opponents' suit (often
        called HELP) 

3NT : to play (showing a guard in the opponents' suit therefore)

The same principles apply if the overcall is in another suit. The transfer to the opponent's suit always enquires about a four-card major suit (Stayman-like), 3♠ is always HELP.
Perhaps Peter has something to add about Rübensøhl.

Un saludo desde Barcelona

Dear Els,

As Ed says: you are mixing up a few things.
1. You write that if an opponent overcalls 2♣ or 2♦ over your 1NT opening, partner has to reply 3♣ if she has one or two major four-card suits. This 3♣ bid is a replacement for Stayman, it is not Rubensohl.
You either play this Stayman-replacement (a simple but limited solution* for the problems created by the overcall) or Rubensohl (more complicated but a better solution); you cannot play both!
2. Reading your question I sense that after an overcall in a minor you play the Stayman-replacement and after an overcall in a major you play Rubensohl. I advise against this: play either full-blown Rubensohl (meaning: over a 2♣/♦ as well) or don't play Rubensohl at all.

Furthermore I have some additions to make to the transfer to the opponents' suit in Rubensohl.
In the first place it is game forcing (Ed doesn't mention this at the 3♣ bid in his scheme).
In the second place: by way of such a transfer to their suit you enquire not only about a major four-card suit (so it is a kind of Stayman; that is why you don't need your 3♣ Stayman-replacement) but to cater for the possibility that no major 4-4 suit is present, the 1NT opener must also be able to show whether he has a guard in the opponents' suit.
It goes like this:

1. If the overcall is in a minor and the 1NT opener does not have a guard in that suit, he begins by 'obeying' the transfer, whether he has a four-card major or not. His partner continues by bidding his own major suit, after which the fit (if there is one) is always found (as both majors are unbid so far). An example:

South will now bid his major four-card suit (with both majors he will bid 3♥) and the rest of the bidding is natural:

If North now bids:
- 3♠, he shows four spades and denies a heart fit
- 3NT, he denies a four-card major; remember: he has already denied a guard in diamonds, so South will not pass 3NT if he hasn't got a diamond guard either!
- 4♣/♦, he shows a control and a maximum hand and establishes hearts as trumps (he has at least four)
- 4♥, he shows four hearts and a minimum hand

If the 1NT opener does have a guard in the opponents' suit, he bids his major four-card suit directly over his partner's transfer to the opponents' suit. Having a guard but no major four-card suit, he bids 3NT. In diagrams:

2. If the overcall is in a major suit and the partner of the 1NT opener transfers to the opponents' suit, he is directly known to have a four-card suit in the other major. Since, as mentioned, the transfer to the opponents' suit is game forcing, the Principle of Fast Arrival applies. An example:

To conclude: all transfers in Rubensohl to a suit of our own show at least invitational strength and a five-card suit. In the case of 2NT/3♣ (transfers to 3♣/♦) that may be 'slightly' invitational.
I repeat: this does not apply to the transfer to 'their' suit (the 'Stayman-like' bid), since that transfer is game forcing.

Hilsen fra Orkanger

* If you think Rubensohl is too much for you: here is the simpler way of dealing with 'their' intervention over 'our' 1NT opening, which includes the 3♣ Stayman-replacement. This is the method I mentioned at the start of my answer.