♠ A 7 2
♥ J 10 5 3 2
♦ A 6 3
♣ J 5
♠ Q J 9 ♠ 4
♥ K 6 ♥ Q 8 7 4
♦ J 10 8 ♦ Q 9 7 4
♣ K 9 8 6 3 ♣ Q 10 7 4
♠ K 10 8 6 5 4
♥ A 9
♦ K 5 2
♣ A 2
South opened 1♠, North raised to 3♠ and South bid 4♠.
Do you agree with North's limit raise? I do. If he bid 4♠, North-South could miss a slam on some layouts. The North hand has eight losers, the way I count them, and that is defined as a LR. There are two losers in spades, three in hearts, two in diamonds and two in clubs. That's nine losers, but I make adjustments. You have two aces, but no queens. I subtract one-half a loser for each ace and add one-half a loser for each queen. Supposedly, this version of losing trick count comes from an old book by Milton Work. I don't use it as a bible, but more as an additional tool when I'm on the fence about what to do.
West led the ♦J, and it looks like you have a loser in each suit because spades split poorly. How would you proceed?
You should win the diamond in hand, and cash the ♠K. Your next step should be to abandon trumps and lead the ♥A and another. West will win and continue with the ♦10. You should win in dummy and advance the ♥J. Suppose East covers -- over to you.
If you ruff, West will overruff, cross to East's ♦Q and another round of hearts will promote a second trump trick for the defense.
Instead, you should discard a diamond. All the defense can do is make one more trick with the queen of trumps -- the established ♥10 is available for a club discard.
Saving both of North's aces is key to making the contract. Simple when you think about it.