The Schnapsen Log
To Close or Not To Close (solution)
Martin Tompa
Concealed cards:
♠ AKQJ
♥ KJ
♣ TKQJ
♦ TKQJYour cards:
♠ T
♥ AT
♣ A
♦ ATrump: ♥Q
In the actual game, Jeff decided to close the stock. His opponent held on to ♠A until the last trick and Jeff fell short of 66 trick points.
In the heat of battle, I would have done the same. Among good Schnapsen players, it’s a sound adage that if you never lose points when you close the stock, you’re not closing nearly often enough. In this instance, the formula for the expected number of trick points your opponent has in hand is 5(120−56) / 14 ≈ 23. This means that, if ♠A is in the stock (probability 9/14 ≈ 2/3) you expect to reach 53+23 = 76 trick points, plenty to win. With this very heuristic analysis and the idea of a possible squeeze even if DrS does hold ♠A, I’m sure I would have closed the stock too. Let’s do a more careful analysis and see whether or not it’s a good play.
First, let’s start with the squeeze aspect. I don’t think there’s much possibility of your opponent discarding ♠A. We will see this by putting ourselves in your opponent’s seat, what I have previously called role reversal. I will use the symbol “x” to denote any small card — K, Q, or J — in the suit.
Let’s suppose, without loss of generality, that you play your cards in the order ♥A, ♥T, ♣A, ♦A, ♠T, because this order is as good as any. If your opponent holds ♠A and ♦Tx, the only way he or she will discard ♠A is if these are the last 3 cards in hand and your opponent decides to hold on to ♦Tx and discard ♠A on ♣A. That would be a poor decision, I think, relying on the slim chance that your last two cards happen to be ♦Ax. If, on the other hand, your opponent holds ♠A and ♣Tx, then the last 2 cards in hand would be ♠A and ♣T, faced with the decision of which to discard on ♦A. Here I think there are 2 reasons to favor discarding ♣T:
There are probably at least as many unseen spades as clubs, unless opponent’s other 2 cards were both spades.
If you held ♣Ax and ♦A, you would probably play ♦A before ♣A rather than the other way around, in order to give opponent a chance to mistakenly discard a club on ♦A. You wouldn’t play them in the order ♣A, ♦A, ♣x, which would be the only reason for opponent to hold on to ♣T rather than ♠A.
Ignoring the squeeze possibility simplifies the situation: we will do the remaining analysis assuming that, if your opponent holds ♠A, you cannot possibly win a trick with ♠T. There are two cases to consider, when ♠A is in the stock and when it’s not.
Case 1 (♠A is in stock): In this case you’re going to make 53 trick points from your hand plus whatever opponent contributes. You’ll only fail to reach 66 if opponent has either 4 jacks plus a queen or king (7 hands), or 3 jacks plus 2 queens (4⨯3 hands). So the probability of failure is 19 / C(13,5) ≈ 0.015, where C(13,5) is the number of possible combinations of 13 cards taken 5 at a time, which is (13⨯12⨯11⨯10⨯9) / (5⨯4⨯3⨯2⨯1). (Why 13, when there are 14 unseen cards? Because one of the cards is ♠A, which we are assuming in Case 1 cannot be in your opponent’s hand.)
Case 2 (♠A is in opponent’s hand): In this case you are going to make 43 trick points from your hand plus the four cards from opponent’s hand other than ♠A. You can only reach 66 if opponent has both missing tens (C(11,2) possible hands, the number of ways of choosing the last 2 cards of opponent’s hand from the 11 unseen cards other than ♠A and the two tens). Thus the probability of success is C(11,2) / C(13,4) ≈ 0.077. (Why C(13,4)? Because this is the total number of ways of choosing opponent’s 4 cards other than ♠A from the 13 unseen cards other than ♠A.)
Putting this all together, your probability of success is 9/14 (1 − 0.015) + 5/14 (0.077) ≈ 0.66. Thus, your expected gain is 0.66(+3) + 0.34(-3) ≈ 0.96 game points by closing. That’s not great. With this powerful hand, you surely hope to gain at least 1 game point, so leaving the stock open and leading ♣A would be better.
Where was my original, heat-of-battle intuition wrong? My approximation that you’ll succeed if ♠A is in the stock and fail if not was fine: the probability of succeeding if ♠A is in the stock is 0.985 and the probability of failing if ♠A is in opponent’s hand is 0.923. Where my intuition was off is that failing 5/14 of the time isn’t a great result: 5/14 is approximately 1/3, and if you fail 1/3 of the time your expected gain is exactly 1 game point. A good lesson to remember.
© 2015 Martin Tompa. All rights reserved.
