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Assume you have basic knowledge of how a contract bridge game plays. There are 40 total HCPs in a set of cards. 4 for each Ace, 3 for each King, 2 for each Queen and 1 for each Jack.

What is the minimal HCP required for a partnership (you and your partner, e.g., N and S) to be able to

  1. Possibly claim all 13 tricks for an NT game (7NT +0)?
  2. Possibly claim all 13 tricks for a game with a given trump suit (e.g. 7♠ +0)?
  3. Secure a 7NT contract? (Guarantee a +0 result on NT Grand Slam)
  4. Secure a level 7 contract with a given trump suit (e.g. 7♠)?

To clarify, the first two cases assume you have stupid opponents - they play the best card for you, while the last two cases assume you have elite opponents - they play the best card against you (think it as double dummy). That's why the wording is different: "to possibly claim" vs. "to secure".

In all 4 cases, you're the declarer. The leading trick starts with the player on your left, and your partner is the dummy.

Note: You do not assume how cards are distributed - you control how cards are distributed, across all 4 players.

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  • $\begingroup$ Are we also to suppose an optimal opponents' card distribution for 1 and 2, and the worst possible one for 3 and 4? That's how I would interpret the "possibly claim" and "secure"; without knowing which opponent has which card, you won't have secured anything unless you can beat even the worst possible case. $\endgroup$ – Bass Apr 4 at 15:50
  • $\begingroup$ @Bass No. You distribute the card as you like, and in 1 and 2, opponents play as you like. Arranging the cards is exactly what this puzzle is about. $\endgroup$ – iBug Apr 4 at 16:36
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  1. (Assisted 7NT): Given optimal card distribution, just have the opponents discard the high cards:

0 HCP is enough. Example: We have ♠9876 ♦T98765432, the starting opponent has the rest of the spades, four of which get played first, while the other opponent dumps the ♦AKQJ.

  1. (Assisted 7♠): The ♠A can eat ♠K and ♠Q at most, so we must also have the ♠J, for a total or

5 HCP.

  1. (Guaranteed 7NT): First answer, which assumed that I cannot arrange the opponents' cards.

My best is 19 HCP: We have ♠AQJT98765 and the rest of the aces, dummy has ♠432, guaranteeing that the ♠K falls to the ace. Since we should suppose the worst case, trying to finesse seems to always result in the wrong opponent having the card we are trying to cut, so it seemed difficult to improve this any further.

  1. (Guaranteed 7NT): Edited in after OP clarified that I get to arrange the opponents' cards too, and that I know which cards they have while playing:

11 HCP seems to be enough: we get ♠AQJT98765432 ♦A, and the starting opponent has ♠K and the rest of the diamonds.

  1. (Guaranteed 7♠): Giving up high trump cards would require having even bigger cards in other suits, so the best seems to be

10 HCP, have all the trumps.

And if you are really clever, you can figure out a way to do the same with only half of those HCPs. Wow.

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  • $\begingroup$ You know how the cards are distributed so finesses can win whenever possible (e.g. left has KJ and dummy has AQ, you can finesse the left opponent with 100% confidence). $\endgroup$ – iBug Apr 4 at 16:42
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    $\begingroup$ For 3, you don't need QS because you can make both K and Q fall down with one Ace (from two sides), if K and Q are the only spades that L and R have. $\endgroup$ – iBug Apr 4 at 16:44
  • $\begingroup$ Ok, if I can arrange the opponents' cards too, the I can go a lot lower for 3, hang on a minute.. $\endgroup$ – Bass Apr 4 at 16:51
  • $\begingroup$ Forcible 7S can be done with 5HCP. See my answer for more details. $\endgroup$ – supercat Apr 4 at 22:18
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7S can be forcibly won with five HCP:

.                   ♠8765432/-/♦765432/-
♠K/♥AK/♦AKQJ/♣AKQJT9                         ♠Q/♥QJ/♦T98/♣8765432
.                   ♠AJT9/♥T98765432/-/-

If West leads a spade, South takes the ace. If West leads a heart, North ruffs and leads a spade to the ace. Otherwise South ruffs low and leads the ace of spades.

South can then ruff two hearts (or three, if desired), using spades or diamond ruffs to return to hand, and run the hearts until North has nothing but spades.

A slightly more detailed explanation, for those not familiar with bridge:

On the first trick, if West leads a spade (the king), North will have to play a spade 2-8 (all equivalent), East will be required to play the queen, and South can play the ace and take the trick. Neither East nor West will have any spades left.

If West leads anything else, it will be in a suit that either North or South doesn't have. A player without any card in a suit that's led may then play any card that is in their hand, including a spade (if one is possessed). The highest spade played on a trick, if any, will take the trick. Thus, either North or South will be able to play a spade on the first trick and then lead a spade, forcing East and West to play the king and queen while South takes the ace, again leaving East and West without any spades.

If South plays a suit that North doesn't have but the opponents do (hearts), North will be able to play a spade which will prevail over any heart that East or West plays. The first time South takes control, it will be after spades have been led once, and South will have played at most of his spades prior to that, so South will have at least two high spades left. Thus, the first two times North trumps a heart, North will be able to lead a spade that South can capture to regain the lead.

The first two times hearts are led, the opponents will each have to play one. After that, nobody except South will have any hearts; if South leads hearts and North discards diamonds until there are none left, the hearts will take tricks. At that point, North will have nothing but trump cards, so on every remaining trick, no matter what North and South do, either North will lead a trump which is captured by a higher trump in South's hand, South will lead a heart which gets captured by a trump in North's hand, or North will lead a spade upon which South discards a heart.

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  • $\begingroup$ Nice! Since this isn't a Bridge forum, most of the people here will probably have a hard time reading this particular notation. (Also, I kind of get how S can "ruff" hearts although his hand is full of hearts, but as a non-native speaker, it took a little time to puzzle that one out.) $\endgroup$ – Bass Apr 4 at 22:38
  • $\begingroup$ Wow! Real genius! $\endgroup$ – iBug Apr 5 at 0:18
  • $\begingroup$ @Bass: Is that clearer. $\endgroup$ – supercat Apr 5 at 16:12
  • $\begingroup$ Great! I added some suit symbols to make it even easier to read. $\endgroup$ – Bass Apr 6 at 12:19
  • $\begingroup$ My friend just told me that 7NT can be forcibly won with only 6 HCP. Do you have any idea how it's possible? $\endgroup$ – iBug Sep 7 at 11:09
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I think for 1, the answer can be as low as

0

A possible distribution of cards, and a draft of play:

me - opp on left - dummy - opp on right
10$\heartsuit$ - 6$\heartsuit$ - 2$\heartsuit$ - A$\diamondsuit$
9$\heartsuit$ - 5$\heartsuit$ - 6$\diamondsuit$ - K$\diamondsuit$
8$\heartsuit$ - 4$\heartsuit$ - 5$\diamondsuit$ - Q$\diamondsuit$
7$\heartsuit$ - 3$\heartsuit$ - 4$\diamondsuit$ - J$\diamondsuit$
10$\diamondsuit$ - A$\heartsuit$ - 2$\diamondsuit$ - A$\clubsuit$
9$\diamondsuit$ - K$\heartsuit$ - 3$\diamondsuit$ - K$\clubsuit$
8$\diamondsuit$ - Q$\heartsuit$ - 8$\spadesuit$ - Q$\clubsuit$
7$\diamondsuit$ - J$\heartsuit$ - 7$\spadesuit$ - J$\clubsuit$
10$\clubsuit$ - 6$\clubsuit$ - 2$\clubsuit$ - A$\spadesuit$
9$\clubsuit$ - 5$\clubsuit$ - 5$\spadesuit$ - K$\spadesuit$
8$\clubsuit$ - 4$\clubsuit$ - 4$\spadesuit$ - Q$\spadesuit$
7$\clubsuit$ - 3$\clubsuit$ - 3$\spadesuit$ - J$\spadesuit$
10$\spadesuit$ - 6$\spadesuit$ - 2$\spadesuit$ - 9$\spadesuit$

The key idea is:

Let one of the opponents (in this case, the on on your right) be void in one suit (hearts in our case), so he can start discarding high cards already in the first trick. Once He gets rid all the HCP-worth cards in one suit, a change of suits (both on his, and our side) can take place. Meanwhile the other opp has to be long in the last suit, so he can foolishly hold back the highest cards of that suit to the point, where we get void in that suit.

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Assuming you use the 4-3-2-1 scoring for A/K/Q/J with no points for other cards and no adjustments for suits:

You can win a 7S having AJ1098765432S 109D, and your partner having 0 pt hand with the seven lowest diamonds, and your opponents splitting the four high diamonds at two card a piece, and the two remaining spades at one a piece. First trick, you take the KQ S with your AS. Second and third tricks you take AKQJ D with any two of your S. You are then guaranteed the next 10 tricks by virtue of having all the trump cards and highest remaining of a suit. Total of 5 points.

You can win 7NT with an identical hand, and identical pattern of play.

I'm not an experienced enough bridge player to figure out how to guarantee a seven trick hand with the lowest HCP total. It obviously can be done for 7S with all thirteen spades (10 points).

What I originally thought was possible for guaranteeing 7NT was impossible, but it is definitely doable with your partner having all four aces, and you have the 2 of all four suits and KQJ1098765 of any one suit, for a total of 22 points

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  • $\begingroup$ You gave the right answer for 7S (7 with trump), but as answered by elias, 7NT is doable without a single HCP. $\endgroup$ – iBug Apr 4 at 16:39

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