The Schnapsen Log

Almost Inseparable (solution)

Martin Tompa

The first thing Peter considers, having been well tutored by Hans, is closing the stock. If that succeeds, it guarantees Peter 2 game points.

The main concern is trump control. But Peter’s ♣Q and Hans’s void in that suit is key to the solution. If Peter can force Hans to trump by leading ♣Q, he can regain the lead with his small trump, pull Hans’s last trump with his ♥A, and cash his big spades for enough trick points.

Does he need to worry about Hans’s possible marriage? Not with the line of play we’ve just discussed. Hans would have only 39 trick points after trumping Peter’s ♣Q with ♥T, and would be on lead from a position something like this:

Hans: (39 points)
♠ J
♥ K
♣ —
♦ KQ

Peter: (21 points)
♠ AT
♥ AQ
♣ —
♦ —

Now Hans’s marriage brings his trick point total to only 59, after which he will take no more tricks. Peter trumps the diamond with ♥Q, pulls Hans’s last trump with ♥A, and then cashes his spades for a total of at least 69 trick points and a gain of 2 game points.

If Peter wants to gamble and cash ♠A or ♠T before leading ♣Q, all will be well provided Hans holds ♠J, which happens with probability 5/6. But with probability 1/6 that card is still in the stock, in which case the position looks like this:

Hans: (26 points)
♠ —
♥ TK
♣ —
♦ TKQ

Peter: (21 points)
♠ AT
♥ AQ
♣ Q
♦ —

If Peter tries to cash a spade immediately, Hans trumps with ♥T for 46 or 47 trick points, and then wins the deal by declaring his marriage. So you see that forcing with ♣Q immediately is a safety play, protecting against this 1/6 probability of failure.

“Well played, Peter,” Hans congratulates his brother. “Closing the stock and forcing me with your last club was a nice play.”

“I’m going to miss this very much, Hans,” his younger brother says.

“As will I, Peter.”

© 2013 Martin Tompa. All rights reserved.

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