[beowulf]

Yesterday in an ACBL robot tournament (#72) I made a Soloway Jump Shift. My robot partner then launched into RKCB and discovered that I had 0 or 3 "aces" (obviously, I had 3). But the robot tried to sign off just in case I had made a jump shift without any aces. The problem is that I didn't know how many aces my robot partner had. I bid 6NT and got a good board (83%). But 7NT would have been somewhat better (98%).

I held: ♠AQT ♥8 ♦AKQJ972 ♣J8. Robot held: ♠KJ85 ♥AJ652 ♦53 ♣A6

Complete auction: 1♥ p 3♦ p 4♣ p 4♦ p 4NT p 5♣ p 5♦ p 6NT p p p.

Is it possible to teach the robots this kind of deduction?

[Bbradley62]

Robot has fairly recently been instructed that when RKC responder has the greater of his two possible holdings, he will bid again. So, he knew what you had, but couldn't proceed over 6N. If you had bid 5♠, he would have known you had 3 keycards and might have asked for kings rather than signing off in 6, in which case you would have known he had both missing aces.

Edit: But it still looks like it's going to be hard for either of you to count to 13 tricks.

[beowulf]

Bbradley62, on 2014-June-23, 18:03, said:

Robot has fairly recently been instructed that when RKC responder has the greater of his two possible holdings, he will bid again. So, he knew what you had, but couldn't proceed over 6N. If you had bid 5♠, he would have known you had 3 keycards and might have asked for kings rather than signing off in 6, in which case you would have known he had both missing aces.

Edit: But it still looks like it's going to be hard for either of you to count to 13 tricks.

True enough about the thirteen tricks, though if he'd found out I had ♠K and ♦Q by bidding 5♥ it would have helped. But my main point is that a hand that has jump-shifted must have at least one ace and 99% will have three or more key-cards.