The Schnapsen Log
More on Race to the End
Martin Tompa
Florian Wisser and I have been having some intriguing conversations about a column I posted recently entitled Race to the End. (Faithful readers will recognize Florian as the author of the formidable Doktor Schnaps program.) What emerged from those conversations is that there is much more to the deal I presented in that column than I had realized. Here was the position I provided:
Unseen cards:
♠ K
♥ K
♣ AKJ
♦ TYour cards:
♠ A
♥ A
♣ TQ
♦ JTrump: ♥J
Stock: 1 face-down card
Game points: Katharina 1, You 3
Trick points: Katharina 21, You 25
On lead: You
In my analysis, I concluded that you should close the stock, play ♥A to pull Katharina’s last trump, play ♠A to pull her last safe exit card, and throw her in by leading ♦J. She would then be forced to open up the club suit, and the trick you would win with your ♣T would bring you to at least 67 trick points. I admitted that this line of play would fail if ♦T was still in the stock, since in that case your throw-in would fail, but that this still gave you a 5/6 probability of winning 2 game points.
What I missed was the possibility of a clever play by Katharina, if either ♥K or ♠K was still in the stock. In either of those cases, Katharina might make an unblocking play by discarding her ♦T! If she does this, you will have 61 trick points, but your throw-in will be foiled. You can cash ♦J, collecting her ♣J and 4 more trick points, but then you are the one who will have to open up the clubs and Katharina will take the remaining two tricks. If Katharina makes this insightful unblocking play, you have only a 3/6 probability of winning the deal, not 5/6, because you will fail if any of ♦T, ♥K, or ♠K is in the stock.
Florian pointed out a clever alternative play that has a better probability of success. You can close the stock, forgo pulling trumps, and lead ♠A, with the intention of next leading ♦J. This line of play will fail if ♠K is in the stock: Katharina will trump your ♠A, cash ♦T and ♣A, and you will end up with only 52 trick points. But the beauty of Florian’s play is that the ♦J will succeed in throwing Katharina in if either ♥K or ♦T is in the stock. In the latter case, you have cleverly left her a trump with which she can be thrown in.
But if, instead, one of those clubs is in the stock, your throw-in will fail. Katharina will win ♦T, but will still have ♥K as an exit card, whose lead will bring you to 55 trick points. Now the outcome will depend on exactly which club is in the stock. If it is ♣J, you will fail, because you are left with two losing clubs. But if it is either ♣K or ♣A, leading your ♣Q next means that you will win the last trick with ♣T, arriving once again at 67 trick points.
The conclusion is that Florian’s line of play only fails if either ♠K or ♣J is in the stock. This gives you a 4/6 probability of winning the deal, better than the 3/6 probability we saw above if you adopt my line of play and Katharina makes the unblocking play when possible.
Having said all this, I would like to go back to my suggested line of play (close the stock and lead ♥A, ♠A, and ♦J) and ask the question, “From Katharina’s point of view, if she is missing either ♠K or ♥K, is discarding ♦T the right play for her to make?” Because remember, she does not know that that missing card is in the stock. From her point of view, it may instead be in your hand.
Let’s suppose first that it is ♠K that is in the stock. In order to analyze the situation from her point of view, we have to stand up, walk around to the other side of the table, and sit in her seat. Here is what the position looks like to her:
Unseen cards:
♠ AK
♥ A
♣ TQ
♦ JKatharina’s cards:
♠ —
♥ K
♣ AKJ
♦ TTrump: ♥J
Stock: 1 face-down card
Game points: Katharina 1, You 3
Trick points: Katharina 21, You 25
On lead: You
You close the stock and lead ♥A, on which she plays ♥K, bringing you to 40 trick points. Next you lead ♠A and Katharina has arrived at the critical juncture. Is her best play on this trick to discard ♦T? If you do not hold ♠K, we know that it is not only her best play, but the only play she can make to stop you from winning the deal. But if you do hold ♠K, it is a mistake for her to discard ♦T on your ♠A. That trick will bring you to 61 trick points, and cashing ♠K will then give you enough. If she instead discards ♣KJ on your two spades, you will reach only 61 trick points in total, and she will have to win the last two tricks with ♣A and ♦T whatever your remaining cards are.
So, from Katharina’s point of view, is it more likely that you hold ♠K or that you do not? From a purely probabilistic point of view, with no other information, there is only a 1/4 probability that ♠K is the one card in the stock out of the four Katharina has not seen up to this point.
But this point of view is foolish, because Katharina has other information she should use. In particular, she knows that your specific cards warranted closing the stock. Would you have elected to close the stock holding only the three winners ♥A and ♠AK, knowing that all she has to do to defeat you is to discard three low cards on these winners? This is a difficult question to answer. If Katharina’s answer to this question is no, then she will conclude that ♠K must be in the stock and her correct play is to unblock her ♦T on your ♠A. But if that is how she is certain to play, then her answer to the question should be yes, you would close the stock with only those three winners, since Katharina is certain to give you the gift of ♦T. The logical conclusion of this is that Katharina cannot be certain you wouldn’t close the stock holding ♠K. She is truly in a dilemma.
The situation is a little harder for Katharina if it is ♥K in the stock rather than ♠K, because in that case she has to make her unblocking decision one trick earlier.
This deal turned out to be much more involved and complex than I had originally realized. Many thanks to Florian for pointing this out and pursuing this interesting analysis with me. It’s much more fun doing it together with a reader!
© 2012 Martin Tompa. All rights reserved.
