<?xml version="1.0" encoding="UTF-8" standalone="yes"?><oembed><version><![CDATA[1.0]]></version><provider_name><![CDATA[Chaos at the Sky]]></provider_name><provider_url><![CDATA[https://chaosatthesky.wordpress.com]]></provider_url><author_name><![CDATA[chaotic_iak]]></author_name><author_url><![CDATA[https://chaosatthesky.wordpress.com/author/chaoticiak/]]></author_url><title><![CDATA[Grand Slam with 26&nbsp;HCP]]></title><type><![CDATA[link]]></type><html><![CDATA[<p>N KT643S 4H AQ753D 42C<br />
E 9752S J5H 86D K9653C<br />
S AQJ8S AKQT876H J8C<br />
W 932H KJT942D AQT7C<br />
S1H W2D Nnb Enb S4H Wnb Nnb Enb<br />
South plays for 4 Hearts.</p>
<p>Definite win? Of course. Grand slam? It&#8217;s quite some luck.</p>
<p><!--more--></p>
<p>We hold great 8 hearts, distributed 9 spades, bad 4 clubs, and non-fit 5 diamonds. The last thing we want is to have East lacking spades and West leads spades, for a trick to them. Luckily, it&#8217;s West who is void instead, and so leads Jack of Diamonds.</p>
<p>Finesse time. Queen of Diamonds, to be trumped if East plays King of Diamonds, then a clearing of trumps, then entering North&#8217;s hand via spades. Otherwise, South has one chance to discard a club. So&#8230;East plays 6 of Diamonds.</p>
<p>Seeing that North has 5 diamonds and South none, it&#8217;s natural to assume the others distributed in at least 6-2, better if 5-3 or 4-4. So the Ace scores, and both South&#8217;s clubs are gone. Yay.</p>
<p>Now lacking in diamonds, North gives control and also stripsthe defenders from their trumps, which happen to be distributed 3-2. 4-1 also wins if the Jack is in the 1. So three trump tricks and all trumps are gone, and the rest is trivial.</p>
<p>I must try to be more productive here 😦</p>
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