Two missed chances of throwing a 10 away

Discussion of hands from past BSkA tournaments (or other events) and how they were bid and played. Skat problems. General discussion of strategy and tactics.

Two missed chances of throwing a 10 away

Postby Gustavo » Thu Apr 06, 2017 12:50 am

I was HH in the hand below. By trick 5 I had the remaining 2 trumps. After trying to fish for the Ace of Spaces on trick 6 I obviously knew it was with FH and my Spades 10 was lost. By trick On trick 8 I trumped the Diamond 7 and immediately realised my mistake: I should have thrown the Spades 10 away in order to capture the last two tricks and the Ace of Spades with it.

Some might say that my mistake was way earlier, back on trick 4. Here I could have played Spades 10 instead of the D. If FH had the Ace of Spades (which he did), he would have cut my Spades 10, in which case I would have two guaranteed tricks in Spades with KD for later; otherwise he would have played a looser and I would have got that trick with my 10.

The question is: when holding 10-K-D, do you usually play the 10 to make the Ace fall and have the lead on the suit, or do you try to fish for it with K or D? What if the situation was 10-K-9, would you still play the 10?

Hands like this are the reason why this game is so fascinating!

Image
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Re: Two missed chances of throwing a 10 away

Postby John McLeod » Thu Apr 06, 2017 2:04 pm

I think you need to do a calculation, which on this hand is relatively easy. After trick 3 you know that the trumps are 3-2 so you will lose exactly three tricks: one in trumps (15 points), one in diamonds (quite likely 21 points) and one in spades. By playing the 10 of spades on trick 4 you limit the number of points you can lose in this trick to 21 and thereby guarantee to win the game if they beat the 10 with their ace, so this is what you ought to do. If the SA does not appear and you win the trick with your S10 you are still safe. You will lose a spade to the ace later but the third player has nothing better than a king left to smear for 19 points - all the aces and tens are already accounted for.

Your general question depends on a similar sort of assessment. From 10-K-Q you lead the 10 first if you can afford to lose 21 points. If you can't afford that you start with the king and hope that the suit breaks 2-2 or that the ace is singleton or is played prematurely. With 10-K-9 the decision is more difficult because if the suit breaks 3-1 you are liable to lose 2 tricks. I think in many cases I would try to avoid leading that suit and hope that the opponents open it for me. If my RHO led a small one through me I would probably put in the 9: if the third player takes with the queen at least I have got rid of that loser quite cheaply and moved the lead to my left. If my RHO led the queen I would probably cover it with the king. But it does all depend on the particular hand and how many points you need or can afford to lose.
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Re: Two missed chances of throwing a 10 away

Postby Gustavo » Thu Apr 06, 2017 11:14 pm

Thank you for your reply John. I'm not entirely sure I follow though:

John McLeod wrote:I think you need to do a calculation, which on this hand is relatively easy. After trick 3 you know that the trumps are 3-2...


Do you mean 3-1 instead of 3-2? After trick 3 there are 4 trumps left: I have SJ-CD-C8 and MH has C10.

John McLeod wrote: ... so you will lose exactly three tricks: one in trumps (15 points), one in diamonds (quite likely 21 points) and one in spades. By playing the 10 of spades on trick 4 you limit the number of points you can lose in this trick to 21 and thereby guarantee to win the game if they beat the 10 with their ace, so this is what you ought to do. If the SA does not appear and you win the trick with your S10 you are still safe. You will lose a spade to the ace later but the third player has nothing better than a king left to smear for 19 points - all the aces and tens are already accounted for.

Your general question depends on a similar sort of assessment. From 10-K-Q you lead the 10 first if you can afford to lose 21 points. If you can't afford that you start with the king and hope that the suit breaks 2-2 or that the ace is singleton or is played prematurely. With 10-K-9 the decision is more difficult because if the suit breaks 3-1 you are liable to lose 2 tricks. I think in many cases I would try to avoid leading that suit and hope that the opponents open it for me. If my RHO led a small one through me I would probably put in the 9: if the third player takes with the queen at least I have got rid of that loser quite cheaply and moved the lead to my left. If my RHO led the queen I would probably cover it with the king. But it does all depend on the particular hand and how many points you need or can afford to lose.


I guess I really need to work on my counting skills at this point. I must confess I don't count points while playing, and this would probably take my playing to the next level. Counting all cards and keeping track of void suits for each opponent already seems plenty though!
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Re: Two missed chances of throwing a 10 away

Postby John McLeod » Fri Apr 07, 2017 10:15 am

John McLeod wrote:
I think you need to do a calculation, which on this hand is relatively easy. After trick 3 you know that the trumps are 3-2...

Do you mean 3-1 instead of 3-2? After trick 3 there are 4 trumps left: I have SJ-CD-C8 and MH has C10.


I was referring to the original distribution of trumps between the two opponents. When you have 6 trumps the enemy trumps can be 3-2 (most likely), 4-1 or 5-0 (rare).
When you have three cards of a side suit the opposing cards can be split 3-1 (most likely), 2-2 or 4-0 (least likely).

I find it helpful to continue to think in terms of these original distributions even after some of the cards have been played.
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Re: Two missed chances of throwing a 10 away

Postby Mike Tobias » Fri Apr 07, 2017 10:40 am

In that game, I'd discard the D7 on the first trick, losing only 11pts and keeping the lead where I want it. Then when I get in, I'd lead the top two trumps. Although normally you want to win the third trump trick, in that instance, a 3-2 split either lets you capture HJ and therefore the rest, or concedes a maximum 13pt trick.
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Re: Two missed chances of throwing a 10 away

Postby Gustavo » Sat Apr 08, 2017 11:01 am

John McLeod wrote:I was referring to the original distribution of trumps between the two opponents. When you have 6 trumps the enemy trumps can be 3-2 (most likely), 4-1 or 5-0 (rare).
When you have three cards of a side suit the opposing cards can be split 3-1 (most likely), 2-2 or 4-0 (least likely).

I find it helpful to continue to think in terms of these original distributions even after some of the cards have been played.


Right, that's all clear now, thanks!

Since you've mentioned the 3-1 being more common than 2-2, would you mind sharing the reason for that? I was aware of the fact, but the reasoning behind it doesn't seem quite intuitive to me.

Mike Tobias wrote:In that game, I'd discard the D7 on the first trick, losing only 11pts and keeping the lead where I want it. Then when I get in, I'd lead the top two trumps. Although normally you want to win the third trump trick, in that instance, a 3-2 split either lets you capture HJ and therefore the rest, or concedes a maximum 13pt trick.


Interesting strategy for the first trick, and it makes a lot of sense. That would be 34 pts lost for certain, and you could still afford losing another 16. It would probably require some counting, although in the case any Ace smeared on that 13-point trick would only make my hand stronger (with the exception of Ace of Diamonds). Will keep these principles in mind.
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Re: Two missed chances of throwing a 10 away

Postby Mike Tobias » Sat Apr 08, 2017 1:17 pm

Gustavo wrote:Since you've mentioned the 3-1 being more common than 2-2, would you mind sharing the reason for that? I was aware of the fact, but the reasoning behind it doesn't seem quite intuitive to me.

Ignoring for the example, the chance of a 4-0 split, there are three possibilities:
Defender A 3 Defender B 1
Defender A 2 Defender B 2
Defender A 1 Defender B 3.

The most likely of these, as your intuition tells you, is the middle one, but either of the others gives a 3-1 split, and between them, they have a greater chance than the 2-2.

The difference is greater between 4-2/2-4 and 3-3, so when you have 5 trumps, you should expect the defenders' to be split 4-2
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