Jeff Reubens Play Problem

Jeff Rubens Vacant Places Problem: How Should The Play Go?

There is another thread with this hand which discusses the odds from a Vacant Places perspective.This thread is about the play of said hand.

In his book, Expert Bridge Simplified, Jeff Reubens posed the following "Gedanken Experiment"

North: 2 / T98 / 9876 / 98765

South: - / AQJ / AKQJT / AKQJT

South is playing in NT. Bidding not given!! This is a theoretical exercise to clarify the theory of Vacant Places.

As a play problem it also has some interest, because at trick 1 West leads the spade Ace and East shows out! West has 12 running spades, and one more card.

What are the chances that E/W can win trick 13? And how should East play to maximize those chances?

Conversely N/S would like to win trick 13 and how do they maximize their chances?

From South's perspective after trick 1, he knows that East's shape is either:

1. A) 3 Hands: 0=7=4=2
2. B) 4 Hands: 0=7=3=3
3. C) 6 Hands: 0=6=4=3 With the Heart King
4. D) 1 Hand: 0=6=4=3 Without the Heart King.
5. Total: 14 possible hands.

The play to the first 10 tricks is routine. West runs spades. East discards 5 hearts, 3 diamonds, and 2 clubs.

At that point South is still in the dark as to what is East's actual shape.

North comes down to the Heart Ten, the d-9 and the c-9.

South has come down to his three Aces. (East should know that South has all three otherwise his partner would have claimed at trick 2!)

East therefore knows which is West's 13th card, and must do his best not to let South figure this out.

• The situation is now one of:
• [#] West   North          East   South       # of Possible Hands
• [1] ssc     h-T d-9 c-9    Khd   h-A d-A c-A    3
• [2] ssd     h-T d-9 c-9    Khc   h-A d-A c-A    4
• [3] ssh     h-T d-9 c-9    Kdc   h-A d-A c-A    6
• [4] ssK     h-T d-9 c-9   hdc    h-A d-A c-A    1

So the crux of the problem is what should East discard at trick 11?. If North discards his heart, then South must guard the hearts and East must pitch so as to give N/S a guess as to which minor to keep at trick 13.Suppose he discards a minor suit card; then N/S will have a complete count of that minor and South will keep his heart Ace at trick 13, North will keep the correct minor suit Nine, and N/S will always win trick 13, either with the heart Ace, (if West is left with a heart see [3] or[ 4] above), or with North's 9 (if West is left with a minor see [1] or [2] ).

So East cannot discard a minor suit card.

So he must discard a heart, the King in [3] and a small one in [4]. What should he do in [1] and [2]?Say that he discards a small heart. Then South will "know" that situation [3] cannot arise. He will discard his h-A at trick 12, keeping a d-9 in North, and the c-A in South. This will win in [1] and [2] and lose only in [4]; E/W will win trick 13 only once.East's other choice is to discard the heart King at trick 11. This will eliminate [4] but leave open [1], [2], and [3]. Now South must keep his heart Ace to guard against [3] and North must guess as to which minor to keep. Since there are 4 hands where West could have a diamond and only 3 where West could have a club, North will keep the diamond 9. This leads to E/W winning 3 hands out of 13 in [1],[2],[3] and one hand in [4] for a total of 4 hands out of 14. N/S will win trick 13 ten times out of 14.

All of this is presuming that North discards his heart Ten at trick 11. Could N/S do better if North discards a minor? Say he discards a club on the assumption that West is more likely to have been dealt a diamond.East as before throws the heart King, and South can now afford the heart Ace, since North guards hearts. At trick 12, North will throw his other minor 9 and East will throw the small heart from [1] and [2] leaving South with a guess as to which minor to keep; as before he will keep the d-Ace, winning in [2] and losing in [1]. From [3] and [4] East and South do not matter as either West wins with the King, or loses with a small heart to North. So once again N/S will win a total of 10 times out of 14 and E/W 4 times. The situation is similar if East throws a small heart at trick 11 instead of the King. South will keep the diamond Ace at trick 12 regardless, since West is more likely to have a diamond than the h-K, and can no longer have a small heart.

So N/S should win trick #13 ten times out of 14.

There is one final wrinkle to all this: Above we stated that if East discards a small heart, South should discard his heart Ace and keep the club Ace. But since South "knows" that East should discard his heart King if he has it, and would only discard a small heart in [4], when East DOES discard a small heart at trick 11 is South entitled to discard his c-A and keep his h-A at trick 12 playing West to have the h-K? i.e that [4] applies?If he does so he is going against the odds; East might have "false carded" the small heart from [1] or [2] hoping to lead South into that error. So South's original strategy is unchanged: Keep the heart Ace at trick 13. Aside: Kind of neat to think that the "true' card is the big one (K), and the "false" card is the small one in this instance.

Notice that the above analysis has built in assumptions about the odds at trick 12; for example that [1] is three times as likely as [4], because 'a-priori' West being dealt a club is 3 times as likely as being dealt the heart King, and nothing about East's discards change that, unless he gives a complete count of the club suit, in which case we eliminate those hands from the hands still possible. This preserves the concept of 'shown' cards versus 'known' cards which is also a key concept in the theory of Vacant Places.