The Schnapsen Log
Revised: December 31, 2014
A Vulnerable Hand (conclusion)
Martin Tompa
It will help in our calculations to be able to refer to a table of the 14 cards you cannot see:
Concealed cards:
♠ ATJ
♥ J
♣ ATKQJ
♦ ATKQJ
The number x of hands with no trump and at least 15 points
Since there are 11 concealed nontrump cards, the total number of possible hands Tibor could hold that have no trump is C(11, 5) = 462. Only a few of these contain fewer than 15 trick points, and it will be simplest to count those few and subtract them from 462. Looking at just the nontrump suits in the table of concealed cards above, the hands that contain fewer than 15 trick points are KKJJJ, KQQJJ, KQJJJ, and QQJJJ. There is only 1 way to construct the hand KKJJJ from the concealed nontrump cards, but 2⋅3 = 6 ways to construct the hand KQQJJ, 2⋅2 = 4 ways for KQJJJ, and 1 way for QQJJJ. Therefore, x = 462 - (1+6+4+1) = 450.
The number y of hands with 1 trump, ♥J, and at least 16 points in the other 3 cards
The number of ways to choose 3 cards from the club and diamond suits is C(10, 3) = 120. Of these combinations, the ones that contain fewer than 16 trick points are AJJ, TQJ, TJJ, and any 3 cards chosen from KKQQJJ. There are 2 ways to construct each of the combinations AJJ and TJJ, 2⋅2⋅2 = 8 ways to construct the combinations TQJ, and C(6, 3) = 20 ways to choose 3 cards from KKQQJJ. Therefore, there are 120 - (2+8+2+20) = 88 ways to choose 3 cards that contain 16 or more trick points from the club and diamond suits. Since there are 3 possible trumps Tibor could hold for each of these 88 combinations, y = 3⋅88 = 264.
The number z of hands with 1 trump, no ♥J, and at least 19 points in the other 4 cards
The number of ways to choose 4 cards from the club and diamond suits is C(10, 4) = 210. Of these combinations, the ones that contain fewer than 19 trick points are AQJJ, TKJJ, TQQJ, TQJJ and any 4 cards chosen from KKQQJJ. There are 2⋅2 = 4 ways to construct each of AQJJ, TKJJ, TQQJ, and TQJJ, and C(6, 4) = 15 ways to choose 4 cards from KKQQJJ. Therefore, there are 210 - (4+4+4+4+15) = 179 ways to choose 4 cards that contain 19 or more trick points from the club and diamond suits. Since there are 3 possible trumps Tibor could hold for each of these 179 combinations, z = 3⋅179 = 537.
The probability of success and the expected gain
Putting all this together, the probability that you win after closing the stock is p = (x + y + z) / 2002 = (450+264+537) / 2002 = 1251 / 2002 ≈ 0.625. Therefore, the expected number of game points you will gain by closing the stock is p(+3) + (1−p)(−3) ≈ 0.749. Our earlier guess that your expected gain would be positive turns out to be correct.
(If you lead ♥Q instead of ♥K when you declare the marriage, your expected gain is 0.695 game points instead of 0.749. The analysis is very similar: y turns out to be 54 less and z to be 36 more. I find this counterintuitive: Tibor is less likely to be holding ♥J than not to be holding it, so my intuition says I should lead the cheaper ♥Q for him to trump.)
After all this work, are you disappointed to find that you expect to gain less than 1 game point by closing the stock? I don’t think you should be disappointed: remember that, if you leave the stock open, the forecast is for you to lose 1 or 2 game points, because this hand does not play well with the stock open.
(I am once again very grateful to Florian Wisser, author of the devilish program Doktor Schnaps, who kindly ran the good Doktor’s engine on this problem for me to confirm the calculations that I did.)
© 2014 Martin Tompa. All rights reserved.
