The Schnapsen Log
No Clear Winner (order of play)
Martin Tompa
We are looking at hands in which Katharina holds 2 aces, both in nontrump suits, she holds at least 2 additional cards in the combined suits of her aces, and her three “small” cards sum to at least 9 trick points (because we considered the case where they sum to at most 8 in an earlier case). Let’s call such hands “two-ace hands” for short.
You declare the marriage and your ♦Q wins the first trick, so that your hand now looks like this:
♠ T
♥ T
♣ T
♦ K
Only one of those tens is a winner, but which one? It seems as though you have absolutely no information to distinguish among them, but that is not the case. The information that you have is the card that Katharina discarded on your ♦Q, and it turns out to be informative!
I mentioned that, if you play your tens in an arbitrary order, there are 40 two-ace hands Katharina could hold that will cause you to lose. In 5 of those 40 hands, Katharina has ♦J and must play it on your first trick; that play of hers is absolutely uninformative, and all you can do is play your tens in some arbitrary order.
That leaves 35 two-ace hands in which Katharina will discard a nontrump card on your ♦Q. If you do the counting, you discover that, in 23 of those 35 hands, she is forced to discard a singleton in order to have any hope of preventing you from winning. This means that the odds are about 2 to 1 that the discard you see was a singleton and that the ten in that suit is a winner.
Based on this observation, let’s outline a better strategy for choosing the order in which to play your tens. If Katharina plays ♦J on your ♦Q, you play your tens in the order ♠T, ♥T, ♣T. But if Katharina discards a nontrump card on your ♦Q, you play the ten of that suit first, followed by your remaining two tens in the order ♠T, ♥T, ♣T. We can now enumerate all the two-ace hands Katharina can hold that defeat this strategy. Here are the ones containing ♦J:
- ♠AKQ, ♥A, ♦J
- ♠AK, ♥AK, ♦J
- ♠AK, ♥AQ, ♦J
- ♠AQ, ♥AK, ♦J
Here are the ones not containing ♦J:
- ♠AKQJ, ♥A
- ♠AKQ, ♥AK
- ♠AKQ, ♥AQ
- ♠AKJ, ♥AK
- ♠AKQ, ♥AJ
- ♠AKJ, ♥AQ
- ♠AQJ, ♥AK
In the 7 hands in this last list, Katharina’s best chance of winning is to discard a spade on your ♦Q; hence, you will play ♠T followed by ♥T and lose. Each of these 7 hands can be repeated with the long suit changed to either hearts or clubs and the short suit changed to spades, and the result is still a loser for you. That is, the total number of two-ace hands Katharina can hold that defeat your improved strategy is 4 + 3⋅7 = 25, as claimed.
Though I find this improved strategy fascinating and different from anything we have encountered so far in The Schnapsen Log, it hardly makes a difference to your expected gain: it decreases the total number of losing hands from 467 to 452, increasing your expected gain from 1.60 to 1.65 game points.
© 2014 Martin Tompa. All rights reserved.
