# The Schnapsen Log

## Remembrance (solution)

#### Martin Tompa

Peter begins by thinking about closing the stock, knowing that this is the first thing to consider when on lead to the last trick before the stock is exhausted, as he is now. Closing the stock has the added bonus of freezing Hans’s trick points while still under 33. Peter adds up the points that he would accumulate by cashing his winners, ♥Q and ♠A. If Hans holds both missing queens, which is more likely than not, Peter’s trick point total will only reach 65, one point short.

Where can he find another trick? His ♠AK opposite the missing ♠TQ clearly suggests the possibility of an endplay. It would be an elimination play, since he needs to lead his ♥Q to eliminate it as a winner before throwing Hans in with either a club or diamond. Suppose Hans holds both ♣T and ♦Q and Peter chooses to exit by leading his ♣K. This would be the position when Hans gains the lead:

Unseen cards:

♠ TQ

♥ —

♣ —

♦ AT

Peter’s cards:

♠ AK

♥ —

♣ —

♦ K

Trick points:Hans 34, Peter 51

As long as the ♠T is not the last card in the stock, Hans is endplayed. He can cash two high diamonds (if he has them) and get to 63 trick points, but at some point must open up the spade suit, allowing Peter to capture his ♠T. On the other hand, if the ♠T is still face-down in the stock, then Peter ends up with the same 65 trick points we saw earlier and the elimination play fails.

Thus, the expected number of game points Peter wins by closing the stock is ⅚(+2) + ⅙(−2) = 4/3. This is good, but Peter doesn’t make a move yet. Before proceeding, he wants to consider if he could do better by not closing the stock.

How can Peter do better than 4/3 game points with the stock open? Only by keeping Hans from crossing the 33-point threshold with high probability. That means leading a card now that Hans cannot win. Since Peter probably wants to retain his trump to pull the ♥J Hans will draw, the obvious lead is ♠A. Hans will discard some queen on this lead and Peter will have 59 trick points. That means that any winner Peter draws from the stock (♠T, ♣T, or ♦A) will earn him 2 game points (after pulling Hans’s trump, if necessary), as will any marriage partner he draws from the stock (♠Q or ♦Q).

The only remaining card Peter can draw from the stock is ♦T. In that case, Hans will have discarded his losing ♦Q on Peter’s ♠A, which would leave Peter on lead from this position:

Hans:(20 points)

♠ TQ

♥ J

♣ T

♦ A

Peter:(59 points)

♠ K

♥ Q

♣ K

♦ TK

Peter’s best play is to lead ♦K, putting Hans on lead from this position:

Hans:(35 points)

♠ TQ

♥ J

♣ T

♦ —

Peter:(59 points)

♠ K

♥ Q

♣ K

♦ T

Hans is too good a player to blunder from this position. He will cash his ♠T and continue with another spade to force Peter to trump:

Hans:(49 points)

♠ —

♥ J

♣ T

♦ —

Peter:(65 points)

♠ —

♥ —

♣ K

♦ T

One trick point short of his goal, Peter must concede the last two tricks and 1 game point.

Thus, if Peter leaves the stock open, the expected number of game points he will gain is ⅚(+2) + ⅙(−1) = 3/2. This is slightly better than his expected gain of 4/3 game points if he closes the stock. The improvement can be attributed to the fact that, if Peter’s play fails, he loses only 1 game point with the stock open instead of 2. Having done all this calculation in his head in less than a minute, Peter finally leads ♠A.

© 2014 Martin Tompa. All rights reserved.