# The Schnapsen Log

## Peace Blooms (solution)

#### Martin Tompa

The first thing for Liesl to consider, when on lead to the last trick before the stock is exhausted, is closing the stock. If she does so, then even if the ♥T is in my hand, she will still need to win one of the tens in her hand to reach 66 trick points. Since there is only a small chance of her winning the ♦T (probability 1/6 that ♦A is still in the stock), her best chance is by winning ♠T. This suggests an elimination play to force me to open up the spade suit.

Liesl first cashes ♥A, to eliminate my ♥T as a safe exit card. Assuming that I indeed hold ♥T, she can then exit with either a club or diamond. Let’s assume she exits with ♦T, and that ♦A is not in the stock. (If it is in the stock, she reaches 67 trick points immediately.) This puts me on lead from this position:

Unseen cards:

♠ AKQ

♥ —

♣ Q

♦ —

Liesl’s cards:

♠ TJ

♥ —

♣ J

♦ —

Game points:Peter 2, Liesl 2

Trick points:Peter 27, Liesl 54

If I hold the ♣Q, the best I can do is cash it and ♠A, but then I have to let Liesl win the last trick with her ♠T for 67 or 68 trick points. If I do not hold the ♣Q, then all I can do is cash ♠A and declare the marriage, for a total of 60 trick points, before letting her win ♠T for 67 trick points. Either way, her elimination play succeeds and she wins 2 game points and the game.

The only possible face-down card in the stock we still need to consider is ♥T, which I have assumed until now was in my hand. If it is in the stock, then this is the position when Liesl closes the stock:

Peter:(6 points)

♠ AKQ

♥ —

♣ Q

♦ A

Liesl:(33 points)

♠ TJ

♥ A

♣ J

♦ T

When Liesl cashes ♥A, I will break up my marriage and discard
♠Q, bringing her to 47 trick points. Even though the elimination
play still works and she will eventually win her ♠T, she will only
reach 61 trick points, and she will lose 2 game points and the game.
(It is interesting to note here that I *must* break up my marriage: I
cannot afford to discard ♣Q, for that establishes her ♣J as a
winner. That winner plus the elimination play would give her exactly
66 trick points.)

Therefore, there is a probability of 1/6 (♥T remaining in the stock) that Liesl will lose the game and a probability of 5/6 (any of the other 5 unseen cards remaining in the stock) that Liesl will win the game. Since we are both so close to 0 game points, the usual expected game point analysis does not apply. But we can, instead, simply conclude that, if this situation were repeated a large number of times, she would expect to win 5/6 of the games, a good result.

Can she do better leaving the stock open? To answer this is somewhat complicated. Her best lead is possibly ♦T. If I were to win this trick, say with ♦A, I would be on lead from this position:

Peter:(27 points)

♠ AKQ

♥ T

♣ Q

♦ —

Liesl:(33 points)

♠ TJ

♥ AK

♣ J

♦ —

I cannot quite get to 66 trick points from this position, and Liesl will win 1 game point. (The only way I can stop her from winning her critical ♠T is to cash my ♣Q and exit with my ♥T. But then she leads ♠J and is guaranteed to win the last trick with her last trump.)

Alternatively, I might duck ♦T by discarding a spade, bringing her to 46 or 47 trick points. Now everything depends on what she draws from the stock. If she draws ♠A or ♥T, cashing that winner plus ♥A will give her 2 game points and the game. If she draws ♦A, ♠K, or ♠Q, I will win the last trick for 1 game point. Surprisingly, the most interesting card she can draw from the stock is the lowly ♣Q. This would be the position, with Liesl on lead:

Peter:(6 points)

♠ AK

♥ TK

♣ —

♦ A

Liesl:(46 points)

♠ TJ

♥ A

♣ QJ

♦ —

From here she forces me by leading ♣J, putting me on lead from this position:

Peter:(18 points)

♠ AK

♥ K

♣ —

♦ A

Liesl:(46 points)

♠ TJ

♥ A

♣ Q

♦ —

I am now endplayed by a counterforce play. I cannot force her back by leading ♦A, because that would give her enough points to win. I cannot lead ♥K, because that would pull the last trumps and she could cash her ♣Q for 68 trick points. And I cannot open up the spade suit, because cashing ♠A doesn’t even get me over the 33 point threshold. Whatever I do, Liesl wins 2 game points and the game.

To summarize so far, if Liesl leaves the stock open and leads ♦T, and if I duck this trick, she will win the game if any of ♠A, ♥T, or ♣Q is face-down in the stock, and she will lose 1 game point if any of the other three cards is in the stock. Liesl can now ask herself this question: would I duck her ♦T if I were missing any of ♠A, ♥T, or ♣Q? To answer this, she must do a role reversal and imagine herself in my position. In the most interesting of these three cases, if ♣Q were the last face-down card in the stock, here is what it would look like from my point of view:

Unseen cards:

♠ TJ

♥ A

♣ QJ

♦ T

Peter’s cards:

♠ AKQ

♥ T

♣ —

♦ A

Game points:Peter 2, Liesl 2

Trick points:Peter 6, Liesl 33

I would recognize that ducking her ♦T lead is no good for me, because I would see the whole counterforce play unfolding like clockwork, exactly as described earlier. So no, I would not duck. It would be even easier for her to see that I would not duck if either ♠A or ♥T were still in the stock.

So, in truth, I would win her ♦T lead if any of these three cards were in the stock, and I would lose only 1 game point. If any of the other three cards were in the stock, I would duck and I would win 1 game point. Therefore, Liesl’s expected gain by leaving the stock open and leading ♦T is ½(+1) + ½(−1) = 0 game points. Thus, she is better off closing the stock and expecting to win the entire game with probability 5/6.

That was a complicated analysis, and so it is high time that I close.

With many kisses,

your Peter.

© 2014 Martin Tompa. All rights reserved.