The Schnapsen Log
Revised: December 31, 2014
A Vulnerable Hand (solution)
Martin Tompa
I hate to see a 4- or 5-card nontrump suit in my hand, even when it contains the ace and the marriage. The only ray of hope in today’s hand is that one small trump ♠Q: a long suit and no trump are the ingredients for taking no tricks at all, particularly against an opponent as aggressive as Uncle Tibor, who may well trump your first heart lead. In any case, I see few winners and a lot of expensive losers in your hand, and I foresee Tibor gaining some game points in this deal.
Did it occur to you to consider closing the stock at trick 1? Or is that too wild a notion, missing so many trumps? If Tibor has 2 or more trumps, you won’t get much further than declaring your marriage, and you certainly won’t get to 66 trick points. At the other extreme, when Tibor has no trumps things look much rosier: you will take all the tricks and, with your declared marriage, only need Tibor to contribute 15 trick points from his hand in order to reach 66. He is nearly certain to have more than that.
How many trumps do you expect Tibor to have been dealt? There are only 3 trumps you cannot see, and nearly twice as many face-down cards in the stock as in Tibor’s hand. That means that, on average, you expect Tibor to hold 1 trump and the other 2 to be in the stock. This in turn means that it is most important to consider what will happen if you close the stock and Tibor holds exactly 1 trump. You will begin by declaring the marriage and leading ♥K. (Somewhat counterintuitively, it will turn out to be slightly better to lead ♥K than ♥Q.) Tibor will trump either ♥K (if he doesn’t have ♥J) or ♥Q (if he does). Either way, your ♠Q is then the key to regaining the lead the trick after Tibor trumps, and you will take 4 tricks in all. You need Tibor to contribute 18 or 19 trick points to those 4 tricks since, with the marriage, you will contribute 47 or 48, depending on which heart he trumps. If he has at least one nontrump ten or ace, 18 or 19 trick points seems likely.
This is as far as I think any player would get analyzing the situation at the table in the heat of battle. We have deduced that you need Tibor to have 0 or 1 trump and a reasonable number of trick points. You might estimate from this that your expected game point gain is positive if you close the stock, and decide that is better than the alternative of leaving the stock open and having a negative expected gain.
Since we have the leisure now of being away from the heat of battle, with a calculator close at hand, let’s continue the closed-stock analysis more precisely. There are some interesting and useful techniques that will arise.
As outlined above, there are only three types of hands Tibor can hold that will allow you to reach 66 trick points after closing the stock:
- There are hands with no trumps and at least 15 trick points summed over all 5 cards held. Let x be the number of such hands.
- There are hands with 1 trump, ♥J, and at least 16 trick points summed over the other 3 cards held. Let y be the number of such hands.
- There are hands with 1 trump, without the ♥J, and with at least 19 trick points summed over the other 4 cards held. Let z be the number of such hands.
The total number of hands Tibor can hold that will give you a win, then, is x + y + z. How do we compute the probability of winning from this? That probability is x + y + z divided by the total number of possible hands Tibor could hold. To calculate this denominator, we need a useful bit of combinatorial mathematics.
There are 14 unseen cards, of which Tibor holds 5. The denominator we want to derive is the number of different combinations of 5 cards chosen from the 14 unseen ones. It will be useful later if we generalize this, and derive a formula for the number of different combinations of r cards chosen from n unseen cards. We will denote this number of combinations by the expression C(n, r), sometimes called a “binomial coefficient”.
We will derive the formula for C(n, r) by adding cards one at a time to your opponent’s hand. There are n possible unseen cards that could be the leftmost card in your opponent’s hand. For each of these possibilities, there are n-1 cards that could be the second leftmost. Continuing in this way, there are n(n-1)(n-2) … (n-r+1) different ways to place r cards in your opponent’s hand from the n unseen cards.
But this formula counts each combination of cards your opponent could hold multiple times, once for each of the ways your opponent could reorganize the r cards from left to right. In how many ways is each particular combination of r cards being counted? In other words, in how many ways can your opponent organize a particular hand of r cards from left to right? There are r cards your opponent could put in the leftmost position. For each of these, there are r-1 cards that could be put second leftmost. Continuing in this way, there are r(r-1)(r-2) … 3⋅2⋅1 different ways of organizing a particular combination of r cards. Since this is the number of ways each combination is being counted, the final formula is
An example should make this formula more concrete. In our desired denominator, there are 14 unseen cards and Tibor holds 5 of them. The total number of possible combinations of cards Tibor could hold is
Therefore, the probability that you win after closing the stock is (x + y + z) / 2002. What remains now is to calculate each of the three values in the numerator.
© 2014 Martin Tompa. All rights reserved.
