The Schnapsen Log

Revised: June 1, 2014

No Clear Winner (solution)

Martin Tompa

At one extreme, if Katharina was dealt all four aces, you have only one winning card in your hand, namely one of your trumps. At the other extreme, if Katharina was dealt none of the aces, your cards are all winners and you want to close the stock immediately. How dangerous is it to close the stock now? How many aces do you expect her to hold?

Katharina is holding 5 of the 14 cards you cannot see, so you expect her to have about 5/14 of the 4 aces. That comes to about 1.5 aces in her average hand. This makes closing the stock appear promising: you expect to win at least 3 tricks, plus the 40 point bonus for the royal marriage. This is probably as much analysis as a good player could do at the Schnapsen table.

But, with some work, we can calculate the precise probability of winning if you close the stock. It is a good exercise in counting, and a fascinating bit of strategy will come up in the process.

Your cards:
♠ T
♥ T
♣ T
♦ KQ

Trump: ♦T

We can enumerate the hands Katharina could hold that will cause you to lose.

1. She could hold all 4 aces. There are 10 possible hands containing all the aces, because she needs one more card to complete her starting hand and there are 10 cards other than the aces and the 6 cards you can see.
2. She could hold any 3 aces. In that case, the best you can do is to win ♦K and one of your tens, and the most she will contribute to those tricks is two kings, for a total of 40 + 10 + 4 + 4 + 4 = 62 trick points, which is not enough. How many of her possible starting hands contain exactly 3 aces? There are 4 possible combinations of 3 aces that she could hold, one for each ace she might be missing. In how many ways can those 3 aces be filled out with 2 more cards to create her starting hand? There are again 10 cards to choose from, so it is the number of ways to choose a pair of cards from those 10. There are 10⋅9 / 2 = 45 such pairs, which we multiply by the 4 possible ways she could have 3 aces, for a total of 4⋅45 = 180 hands that contain exactly 3 aces.
3. She could hold exactly 2 aces, both in nontrump suits, and 3 “small” additional cards. In that case, the best you can do is to win tricks with one of your tens plus both of your trumps, for a total of 40 + 10 + 4 + 3 = 57 trick points, plus whatever Katharina discards on your three tricks. You will fail to reach 66 trick points if she has 3 jacks, or 2 jacks and a queen or king, or a jack and 2 queens. There are 4 ways in which she could have 3 jacks, 6⋅3 = 18 ways in which she could have 2 jacks and a queen, similarly 18 ways in which she could have 2 jacks and a king, and 4⋅3 = 12 ways in which she could have a jack and 2 queens, for a total of 4 + 18 + 18 + 12 = 52 ways in which her three discards add to at most 8 trick points. We multiply this by the 3 ways she could have 2 nontrump aces for a total of 3⋅52 = 156 hands containing exactly 2 nontrump aces and 3 other cards totaling at most 8 trick points.
4. There are some additional hands in which she holds 2 aces, both in nontrump suits, that will cause you to lose. These are hands where she holds at least 2 additional cards in the combined suits of her aces, and you play your tens in the “wrong” order. Once you declare the marriage and your ♦Q wins the first trick, you will play your 3 tens in some order. Let’s assume for the moment that the order is ♠T, ♥T, ♣T. Suppose the original hands were something like the following:

   Katharina:
   ♠ AQ
   ♥ AJ
   ♣ K
   ♦ —

   You:
   ♠ T
   ♥ T
   ♣ T
   ♦ KQ

   She discards ♣K on your ♦Q, wins ♠A, and forces out your last trump by leading ♠Q. You now have 54 trick points. When you next lead ♥T, she will win both remaining tricks. Had you somehow known to play your ♣T earlier, you would have won.

   If you play your tens in an arbitrary order, it turns out that there are 40 of these additional hands Katharina could hold that will cause you to lose. But if you are careful about the order in which you play your tens, you can decrease this number to 25. This is the “fascinating bit of strategy” I mentioned earlier.

5. Finally, there are certain hands she could hold containing ♦AJ and one other ace that would cause you to lose. (I thank Florian Wisser, the author of Doktor Schnaps, for pointing out these additional cases that I had missed in my original analysis.) One example of such a hand is the following:

   Katharina:
   ♠ AQJ
   ♥ —
   ♣ —
   ♦ AJ

   You:
   ♠ T
   ♥ T
   ♣ T
   ♦ KQ

   You declare your marriage and lead ♦Q, won by Katharina’s ♦A. She cashes ♠A and forces out your last trump by leading ♠J. Now she cannot be stopped from winning the last two tricks.

   A slightly more complicated example is the following:

   Katharina:
   ♠ A
   ♥ QJ
   ♣ —
   ♦ AJ

   You:
   ♠ T
   ♥ T
   ♣ T
   ♦ KQ

   You declare your marriage and lead ♦Q, won by Katharina’s ♦A. She leads ♥J, putting you back on lead from this position:

   Katharina: (14 points)
   ♠ A
   ♥ Q
   ♣ —
   ♦ J

   You: (52 points)
   ♠ T
   ♥ —
   ♣ T
   ♦ K

   Your best play, not knowing that she holds ♦J, is to try one of your remaining tens, retaining your trump to regain the lead so that you can try the other ten. But whichever ten you lead, Katharina wins, forces out your last trump by leading ♥Q, and wins the last trick.

   Any hand in which she hold ♦AJ, another ace, and any additional two cards, except for the case when the additional two cards are each singleton (that is, each alone in its suit), plays similarly. There are 3⋅3 = 9 hands in which the two additional cards are in the same suit as the nontrump ace, 3⋅3⋅2⋅3 = 54 hands in which one of the two additional cards is in the same suit as the nontrump ace, and 3⋅2⋅3 = 18 hands in which the two additional cards are together in their own suit, for a total of 9 + 54 + 18 = 81 hands.

In total, then, there are 10 + 180 + 156 + 25 + 81 = 452 starting hands Katharina could hold that would defeat your play.

How many starting hands are possible for Katharina in total? This is the number of combinations of 5 cards chosen from the 14 you cannot see, which is

(14⋅13⋅12⋅11⋅10) / (5⋅4⋅3⋅2⋅1) = 2002.

Therefore, the probability that you will fail to reach 66 trick points is 452/2002 = 0.226 and the probability you will succeed is 1 − 0.226 = 0.774. That is a good probability of success.

Armed with these probabilities, we can now compute the expected number of game points you will gain if you close the stock and declare the marriage: it is 0.774(+3) + 0.226(−3) = 1.65. That seems a good result. What is surprising is that you can achieve such a good result by closing the stock, even though you are missing all the aces!

If you instead leave the stock open and declare the royal marriage, there is some chance that you will lose the first trick and therefore limit your gain to 2 game points or less. Even if you win the first trick, there is a good chance that you will draw a small nontrump card from the stock, and then what will you do? In addition, leaving the stock open just increases the chance that Katharina picks up another ace. The situation with the stock left open is unclear to me, but I imagine your expected gain decreases slightly by leaving the stock open.

© 2014 Martin Tompa. All rights reserved.

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