## Problem no. 37, 2009

### The problem

Giorgio and Benito are two young, promising bridge players. They are right now at the local café, drinking a beer and discussing bridge. Giorgio played a local duplicate yesterday, which Benito did'nt. "Here is something pretty", Giuorgio says and writs down an ending on his napkin.

 A 10 A 3 2 K 8 A J Q 9 4 2 7 J 10 8 J 6 10 7 4 Q J Q 9 5 4

"I was declarer in 4 and needed the rest of the tricks. I cashed the queen of diamonds and played another diamond, hoping for a low ruff in dummy. Then, I could cash my side suit winners and crossruff the rest."
"But West ruffed ahead of dummy with the four of hearts", Benito said.
"Exactly. And after overruffing in dummy, I had a choice of three plans. I could still play for a cross ruff, which succeeds if West hasn't the nine of hearts, or I could play for a squeeze."
"I can understand what you did", Benito laughed.
"Yes, but after overruffing I had a choice of squeezes. If East is long in spades, I shall cash the clubking and draw two rounds of trumps. East ins then squeezed between diamonds and spades." Girogio paused briefly. "But I thought West had more spades than his partner. Therefore I cashed the ace of spades instead before cashing my trumps. Om my last heart, there was a double squeeze. Westmust discard clubs, and when dummy parted with the spade ten, it was East's turn to feel the pressure."
"Very nice ...if this is what really happened" Benito replied. "I think you try to fool me!"

### Solution

Benito is right! There is nothing wrong with the analysis, and the ending is pretty, but ... Giorgio either doesn't remember all details, or has made it up.
The ending above cannot have appeared in real life. In the diagram are no voids, so no discards have been made. Each played trick consists of four cards of the same suit. Therefore, there should be thirteen, nine or five cards left in each suit – but that isn't the case. Six spades, seven hearts, eight diamonds and seven clubs is simply impossible.

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