The Schnapsen Log

New Arrival (solution)

Martin Tompa

With only 17 trick points, there is no reason to think too long about closing the stock. Even if Tibor could capture all the diamonds and the ♠A was miraculously still face-down in the stock, he would only get to 57 trick points.

Tibor truly does not have an appealing card to lead. He doesn’t want to open up either the spade or diamond suit, because of the possibilities of endplays in those suits. Both are suits that he would like to see Peter open. Suppose, for instance, that Tibor leads ♠T at trick 5. Peter will likely win with his ♠A, resulting in this position:

Peter: (41 points)
♠ KJ
♥ —
♣ A
♦ TQ

Tibor: (17 points)
♠ Q
♥ —
♣ TK
♦ AK

From this position, Peter has a nice elimination play. He cashes his winners ♣A and ♠K, getting down to this position:

Peter: (63 points)
♠ J
♥ —
♣ —
♦ TQ

Tibor: (17 points)
♠ —
♥ —
♣ T
♦ AK

Now Peter’s ♠J lead endplays Tibor, who must open up the diamond suit and lose 1 game point.

If, at trick 5, Tibor instead opens up the diamond suit by leading ♦K, he is likely to be endplayed in spades. Here is the position after Peter wins trick 5 with ♦T:

Peter: (34 points)
♠ AKJ
♥ —
♣ A
♦ Q

Tibor: (17 points)
♠ TQ
♥ —
♣ TK
♦ A

Peter again has an elimination play. He cashes his ♣A and exits with ♦Q, leaving Tibor on lead from this position:

Peter: (49 points)
♠ AKJ
♥ —
♣ —
♦ —

Tibor: (31 points)
♠ TQ
♥ —
♣ T
♦ —

Tibor is endplayed. He can cash ♣T in order to cross the 33-point threshold, but then must yield both spades and 1 game point to Peter.

Leading ♦A at trick 5 is the worst of all possible leads for Tibor: Peter will likely trump with ♣A and then cash ♦T and ♠A, winning 2 game points and the entire game.

By the process of elimination, Tibor is left with only one possible lead, ♣K. This seems like a waste of his valuable trump, but its advantage to Tibor is that it leaves the spade and diamond suits intact for possible endplays later. Let’s see what happens.

If Peter wins this trick with ♣A, he will be on lead from this position:

Peter: (35 points)
♠ AKJ
♥ —
♣ —
♦ TQ

Tibor: (17 points)
♠ TQ
♥ —
♣ T
♦ AK

From this position, neither player can get to 66 trick points, and the game point will go to whomever wins the last trick. Tibor has the advantage of having the only trump. The best Peter can do is lead a diamond loser whenever he is on lead; Tibor will respond by leading a spade loser whenever he is on lead, and will win the last trick with his trump. That is 1 game point for Tibor if Peter wins the ♣K lead at trick 5, so let’s go on to see what happens if Peter ducks the ♣K. The outcome will depend on which card Tibor draws from the stock.

If ♣A is in the stock, Peter’s best discard is ♠J, leaving Tibor on lead from this position:

Peter: (20 points)
♠ AK
♥ —
♣ T
♦ TQ

Tibor: (23 points)
♠ TQ
♥ —
♣ A
♦ AK

Now Tibor has his own elimination play, which is his reward for leaving the spades and diamonds intact. He cashes ♣A and ♦A, leaving him on lead from this position:

Peter: (20 points)
♠ AK
♥ —
♣ —
♦ T

Tibor: (58 points)
♠ TQ
♥ —
♣ —
♦ K

Now he exits with his ♦K, forcing Peter to open up the spade suit and give Tibor 1 game point.

If either ♠A or ♠K is the last face-down card in the stock, Peter’s best discard is again ♠J. In either case, Tibor has an elimination play in which he will force Peter to open up the diamond suit. For instance, suppose Tibor draws ♠A from the stock. He will be on lead from this position:

Peter: (20 points)
♠ K
♥ —
♣ AT
♦ TQ

Tibor: (23 points)
♠ ATQ
♥ —
♣ —
♦ AK

Tibor cashes ♠A and exits with ♠Q, putting Peter on lead from this position:

Peter: (34 points)
♠ —
♥ —
♣ T
♦ TQ

Tibor: (38 points)
♠ T
♥ —
♣ —
♦ AK

Peter is endplayed. He must open up the diamonds either now or after cashing his last trump, but in either case Tibor gets to 66 trick points exactly with the four diamonds.

We have now seen three cards that Tibor can draw from the stock, ♣A and ♠AK, with any of which he will gain 1 game point by an elimination play. The other three cards that Tibor can draw from the stock, ♦TQ and ♠J, will turn out to be losing draws for him. There is nothing very interesting to say about the ♠J draw: the outcome comes down to the last trick, which Peter will win due to the two trumps ♣AT that he holds.

If ♦Q is the last face-down card in the stock, Peter is best off discarding his ♦T on Tibor’s ♣K lead at trick 5, in order to void himself in diamonds. Tibor is then on lead from this position:

Peter: (20 points)
♠ AKJ
♥ —
♣ AT
♦ —

Tibor: (31 points)
♠ TQ
♥ —
♣ —
♦ AKQ

Tibor can declare the marriage for 51 trick points, but after winning that trick Peter can immediately lead ♠J to give up one last trick to Tibor’s ♠T. Tibor will reach only 63 trick points and Peter wins 1 game point.

The last possibility for the face-down card in the stock is ♦T, which has some interest. Peter’s best discard on Tibor’s ♣K at trick 5 is ♦Q, leaving Tibor on lead from this position:

Peter: (20 points)
♠ AKJ
♥ —
♣ AT
♦ —

Tibor: (24 points)
♠ TQ
♥ —
♣ —
♦ ATK

Tibor exits by leading ♦K, putting Peter on lead from this position:

Peter: (35 points)
♠ AKJ
♥ —
♣ T
♦ —

Tibor: (24 points)
♠ TQ
♥ —
♣ —
♦ AT

If Peter now continues with his ♣T, this lead squeezes Tibor. He cannot discard ♠Q, as that would relinquish all the remaining tricks. So he must discard ♦T, bringing Peter’s trick point total to 55. After that, Peter cashes ♠A and wins 2 game points.

To summarize this lengthy analysis, then, Tibor’s best lead at trick 5 is ♣K and Peter’s best response is to duck. After that, there are three cards Tibor can draw from the stock (♣A and ♠AK) that will allow him to win 1 game point, two cards (♠J and ♦Q) that will cause him to lose 1 game point, and one card (♦T) that will cause him to lose 2 game points. Therefore, his expected gain is ½(+1) + ⅓(−1) + ⅙(−2) = −1/6 game points. This means that, in the long run, he will nearly break even with this play, which is the best he can hope for in this deal.

© 2014 Martin Tompa. All rights reserved.

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