I will add the last hand now (which actually has more questionable play elements than bidding, but one of the bids does stand out for me).
The hand (I have checked it this time):
Again, only fifteen tables played this hand. East obviously opened 1
♦. At one table, South bid 5
♣ over that, and the Robot duly punished him for -1100 (although it does seem a bit unlucky that not a single table bid a very high percentage slam).
Seven Souths never got into the auction:
At every table with this auction, North led the
♠J, and the Robot chose to take diamond finesse (which seems a bit unusual to me with no opposition bidding), and had 13 tricks without the heart finesse or the fourth spade. The play (which looks slightly suboptimal to me) went win the first spade in hand, heart to the K, spade back to hand, and then take the diamond finesse.
At three tables (including mine) South threw in a 3
♣ bid:
Obviously, declarer had no trouble with the contract, winning the club lead, playing a diamond to the king, cashing two spades and two hearts, and then taking the diamond finesse for all 13 tricks.
At one table, South made a delayed 3
♣bid:
This worked out very well for him, as the play went:
♣5-
♣A-
♣2-
♣K (which makes me want to scream "NO!" at my monitor, but declarer still has 13 tricks available through trick 6),
♥ to A,
♠ to K,
♠ to A,
♥ to K,
♠ to Q,
♦ to A,
♦ to K, cash the long
♠, and conceded four club tricks to South.
At one table, this South received an undeserved reward for his bidding:
The play began:
♣7-
♣K-
♣5-
♣A (double UGH! - South deserved to have (on a different layout) declarer win the trick with the
♣9). The Robot then played a
♠ to the Q,
♥ to the K, and then a
♥ to the J, which promptly led to down 3. I know you will say "double dummy, declarer still has all 13 tricks, even after playing both club honors at trick one," when I ask "Why?", but still the logic to fix that doesn't appear particularly challenging. At all of the tables (including the next two), only this declarer took the heart finesse (taking it the "wrong" way, presumably placing the
♥Q in South for the 2
♣ bid).
Finally, we get to the "questionable" bid which occurred at two tables on this auction:
I will finish the play at those two tables before I ask the obvious question. Of course, 4
♥ can make 12 tricks (13 with the opening lead of either the
♥Q or the
♥4). At one table, declarer won the club lead with the A, played a
♥ to the A,
♥ back to the K, and a third
♥, North winning the Q. North played a second
♣, declarer winning and drawing the last trump. Cashing the spades now (completing the count on the South hand) becomes the obvious play, but declarer instead played a
♠ to the K, a
♦ back to the K, a
♦to the A, a
♠ back to the A (still 11 tricks available), and then... conceded a diamond (and the rest of the tricks). I won't bother to create a diagram of the three-card end position, but declarer had the good
♠Q (the trick that makes the contract), a small spade (which becomes an overtrick when spades break 3-3), and a small diamond. Play proceeded similarly at the other table with this auction, but this South had discarded too many clubs, and had to give declarer a spade at the end, which allowed the contract to make.
I don't like the 4
♥ bid, of course. Why play in a 4-3 (at best) fit, with the bidding suggesting bad breaks (although South only had seven clubs), when an eight-card (at least) diamond fit exists? I can't blame the Pass by East (but I would expect a fifth heart, at least, in the West hand for 4
♥). Looking just at the West hand, I also want East to declare this. If North has a singleton club, it will certainly get led, where (suppose East has xx of clubs, and South actually has eight of them) on opening lead, South might not lead the
♣A missing the
♣K.