The Schnapsen Log

All the Aces (solution)

Martin Tompa

If you lead one of those nontrump aces, Hans is sure to trump it. If he can trump it with ♣T, that’s the end of this deal and your beautiful hand of aces. It doesn’t seem as though leading ♦Q is a good idea either, as it is likely to just give Hans more trick points without accomplishing much.

Hopefully it has crossed your mind to close the stock, so that you can cash those aces. If you succeed in cashing all four aces, and if Hans drops just one ten on some ace, that will bring you to 71 trick points. On the other hand, if he happens to have all four kings, your four aces will bring you only to 65 trick points. Another problem is that he might trump one of your aces. Perhaps this can be avoided if you first cash ♣A?

We’re not done until we come up with some order in which to play your cards. Let’s assume that you close the stock, cash ♣A, and then play the other three aces in an arbitrary order. Under these assumptions, let’s count the possible hands Hans could hold that allow you to win.

1. You would win if he has at least one card in each suit, but doesn’t have all four kings. Here’s an example of this situation:

   Hans: (50 points)
   ♠ K
   ♥ T
   ♣ TK
   ♦ K

   You: (5 points)
   ♠ A
   ♥ A
   ♣ A
   ♦ AQ

   There are 16 possible hands Hans could hold that satisfy these conditions. (It’s a good exercise to figure out why there are 16.)
2. You would win if he has all four kings, but has neither ♣T nor ♦T. Here’s an example of this situation:

   Hans: (50 points)
   ♠ K
   ♥ TK
   ♣ K
   ♦ K

   You: (5 points)
   ♠ A
   ♥ A
   ♣ A
   ♦ AQ

   In this case, as we saw, your four aces would only get you to 65 trick points, but your ♦Q would take the last trick. There are 2 hands satisfying these conditions, since his fifth card would have to be either ♠T or ♥T.
3. You would win if he has a void (0 cards) in some suit, but doesn’t have ♣T. Here’s an example:

   Hans: (50 points)
   ♠ TK
   ♥ —
   ♣ K
   ♦ TK

   You: (5 points)
   ♠ A
   ♥ A
   ♣ A
   ♦ AQ

   In this case, your ♣A will pull his only trump, and your other three aces will pick up one of his tens, because he must have TK in each of two suits. There are 3 hands satisfying these conditions, corresponding to the 3 suits in which he could be void. (He cannot be void in trumps.)

In total, then, you would win if Hans holds any of 16+2+3 = 21 possible hands.

How many possible hands are there for Hans to hold in all? Since you know one of his cards, there are (7⋅6⋅5)/(3⋅2⋅1) = 35 such hands. (It is another good exercise to work out why this is the correct formula.) This means that your probability of winning with this line of play is 21/35 = 3/5. Not bad. But can you do better? This is the key point where a master of the game could make a subtle improvement.

We’ve done as well as we can when Hans has no void suits. But there are hands he could hold with ♣TK and a void in either ♥ or ♠ against which we can do better. Here is an example of such a hand:

Hans: (50 points)
♠ TK
♥ —
♣ TK
♦ K

You: (5 points)
♠ A
♥ A
♣ A
♦ AQ

With the play we’ve been analyzing, he will trump one of your aces with ♣T and reach 71 trick points. How can you avoid having him trump an ace? The key is your ♦Q, which you can afford to let him trump. If he has exactly one ♦, then by playing your cards in the order ♣A, ♦A, ♦Q, you will reach 66 trick points before Hans does. He will trump ♦Q with ♣T, but will then have to let you back in with one of your aces, whichever suit he has left in his hand, and will contribute a ten to your final trick.

Notice that there are 2⋅2 = 4 hands Hans could hold in which he has ♣TK, one diamond, and one void, because there are 2 possibilities for which diamond he holds and 2 possibilities for the void suit.

At this point you should go back and convince yourself that inserting ♦Q into the middle of your play like this will not hurt you in any of the 21 hands we analyzed earlier. If he has no void, you will take the same tricks you would have taken had you run your four aces in a row. If he has a void but doesn’t have ♣T, you will likewise take the same tricks you would have taken.

Leading ♦Q at the critical point is another sort of safety play: it insures you against the possibility that Hans holds one of these four additional hands. With this enhancement, you will win against any of 21+4 = 25 possible hands Hans could hold. You have improved your probability of winning to 25/35 = 5/7.

This means that the expected number of game points you will gain is ⁵⁄₇(+1) + ²⁄₇(−2) = 1/7. It may seem disappointing to you, after all this work, that your expected gain is only slightly better than 0. But remember what your alternatives to closing the stock are.

1. You could lead ♣A and hope to get a good draw from the stock. Hans is quite likely to void himself in some suit on your ♣A and likely will have gained trump control. You will now have 20 or 26 trick points, and if you close the stock you will probably be able to cash only two of your aces. If you don’t close the stock, you are likely to lose 2 game points when Hans trumps.

2. You could lead ♦Q and hope your draw from the stock is ♣T or some king (so that Hans’s ten in that suit, if he has it, is singleton). Hans is likely to win this trick and will have 57 or 63 trick points to your 5. If Hans now has ♣TK and either a void or all four kings, he will close the stock himself and you will lose 2 game points. Otherwise, he won’t close the stock and you may possibly win 1 game point. To complete this analysis thoroughly is complicated and involves many cases. But let’s approximate it. Let’s suppose Hans wins this trick with ♦K, which is the better case for you than if he wins with ♦T. After replenishing his hand from the stock, there are 6 hands Hans could have that contain ♣TK and a void: 4 possibilities if the void is ♦, and 1 possibility for each of the two other possible void suits ♥ and ♠. There are (6⋅5)/(2⋅1) = 15 possible hands Hans could hold in all. Therefore, the probability that Hans can close the stock and win is 6/15 = 2/5, and the expected number of game points you will gain is at most ⅖(−2) + ⅗(+1) = −1/5. And this is an optimistic analysis for you, since it’s more likely than not that Hans could win the trick with ♦T instead of ♦K, and we have simply given you the game point if Hans doesn’t close the stock.

This turned out to be a complicated analysis, one that would be impossible to do completely in the heat of battle. The main lesson to take away, though, is the safety play of forcing out Hans’s possible second trump with your ♦Q as early as possible.

© 2012 Martin Tompa. All rights reserved.

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