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Losers - counting example

#1 User is offline   Lesh18 

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Posted 2013-October-09, 12:16

Counting losers


Let's consider this example in counting losers. According to wikipedia, this hand has 1+1+2+2=7 losers. I don't get that. I see 3+3+2+2=10 losers.

So how many losers are in this hand? I guess it would be different for a NT and Trump contract. I see 10 losers if it played a NT contract. How many losers would there be if it played spade as trump?

Also, wikipedia refers to something as Basic Method when counting losers. Do you know any useful sources that explain counting losers? (The Learn to play bridge software does not really cover counting losers well, I believe it is straight forward but I'm still struggling with it)

Thanks
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#2 User is offline   Endymion77 

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Posted 2013-October-09, 12:20

If spades or hearts are trump and there's a good fit, the hand has 7 losers.
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#3 User is offline   Lord Molyb 

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Posted 2013-October-09, 12:25

View PostLesh18, on 2013-October-09, 12:16, said:

Counting losers


Let's consider this example in counting losers. According to wikipedia, this hand has 1+1+2+2=7 losers. I don't get that. I see 3+3+2+2=10 losers.

So how many losers are in this hand? I guess it would be different for a NT and Trump contract. I see 10 losers if it played a NT contract. How many losers would there be if it played spade as trump?

Also, wikipedia refers to something as Basic Method when counting losers. Do you know any useful sources that explain counting losers? (The Learn to play bridge software does not really cover counting losers well, I believe it is straight forward but I'm still struggling with it)

Thanks

How likely is it the 4th/5th spade and the 4th heart are losers when you have a fit and are playing in a suit contract?
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#4 User is offline   ArtK78 

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Posted 2013-October-09, 12:29

Using loser count (actually, I learned Modified Losing Trick Count, which amounts to the same thing in most instances), you count up to 3 losers in each suit, but no more than the number of cards in the suit. You subtract one loser for each A, K and Q held, but a singleton K or a doubleton Q does not reduce the loser count.

So, in the example hand, there are 7 losers - 1 in spades, 2 in hearts, 2 in diamonds and 2 in clubs.

Please keep in mind that losing trick count is a hand evaluation method, not a count of how many losers you will have in the suit.

As I mentioned, I learned Modified Losing Trick Count when I studied (and played) the Romex System in its original form in the mid 1970s. I assume that Romex still relies upon MLTC in evaluating hands for suit play, but I have not read any of Rosenkranz's later works.
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#5 User is offline   CSGibson 

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Posted 2013-October-09, 12:32

loser counting single dummy is different than when you have a full hand. Generally, you start with a maximum of 3 losers in any suit, if you have a 4+ card suit, the 4th and longer cards are not counted as losers. In this example hand, you have obviously counted 3 spade losers, but in a spade contract, do you really think those extra small trumps are negative features to your hand, or assets? How about in a NT contract? Those are tricks waiting to happen!

Anyway, the most losers you can have in a suit is 3, and that is only if you have 3+ cards in a suit. In general, you subtract one loser for each A, K, or Q in a 3+ card suit, each A or K in a 2 card suit, and each A in a 1 card suit for a very simple evaluation method.

If you want to refine it, you might note that aces are more likely to be winners than Q's, and that honor combinations often take more tricks than their loser count - for example, AQ tight is 1 loser in a simplified loser counting system, but 1/2 a loser makes more sense because the K will be onside 1/2 the time when the opponents have it. Similarly, the simple evaluation says that AJT is 2 losers, but 75% of the time when missing both honors, one will be onside. Heck, when they are both offsides, they sometimes lead it.

Anyway, using a very simple loser counting system, I would say that your example hand has 1+2+2+2 for 7 losers. Specifically, we are missing the Q of spades, the KQ of hearts, and the AK of both minors.

When you actually see dummy, then you should evaluate losers again, this time using the combined assets of the hand. Fluffy gave a good example of how to do so in one of your previous threads.
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#6 User is offline   ArtK78 

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Posted 2013-October-09, 12:56

Actually, in a bidding system using MLTC, you evaluate your loser count during the bidding. For example, suppose you open the bidding on this hand 1. Partner responds 2, which, in a system using MLTC, shows 1 1/2 - 2 potential cover cards (cards that cover losers. such as Aces, Kings and Queens, and, after a suit fit is found, shortnesses). Since you have 7 losers, that would lead to a combined 5 - 5 1/2 losers between the two hands, so you don't try for game. If partner made a limit raise, showing 2 1/2 - 3 cover cards, you would still pass on this hand.

Where MLTC and cover cards get interesting is in more involved auctions. You upgrade cards that are known to cover losers (Aces, Kings and Queens in suits that partner has shown length, Aces and possibly Kings in suits where partner is short) to full cover card status and downgrade cards that are known to be of limited or no value (Queens opposite shortnesses, Kings opposite singletons) to no cover card status. You also upgrade shortnesses which appear from the bidding to be working. If there is a good fit, a singleton can be worth up to two cover cards if it is opposite Axx(xx). A doubleton can be worth a cover card if not opposite a Q. With experience, you can become quite accurate in evaluating how well the hands fit and what the potential for the two hands can be.

So, if you open a 5 loser hand (a maximum one-bid) and you find a good fit and partner shows that he has 4 working cover cards, you have the makings of a small slam.

Romex is the only bidding system that I have ever seen that relies on MLTC in its bidding structure. But MLTC can be used effectively within any bidding structure even if the bids are not based on losing trick count and cover cards.

You will find that a large number of players dismiss losing trick count as an accurate method of hand evaluation. I have found that it has its place and, on the right hand, can be quite valuable.
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#7 User is offline   Stephen Tu 

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Posted 2013-October-09, 16:25

Lesh, I think you are confusing counting losers in the play of the hand (where you get to see dummy first), with "losing trick count" which is a hand evaluation alternative to point count (4321 HCP, plus distributional adjustments, plus adjustments for working/non-working cards based on partner's and opponent's bidding), where you only get to see your own hand. I personally don't think "losing trick count" is worth learning; by the time you make enough adjustments to make it accurate, IMO you might as well just have used point count to begin with. When you get better eventually you start visualizing partner's possible hands based on the bidding, and play the hand out in your mind, to estimate how much you want to bid.

But for counting losers in the play of the hand, like I said in the other thread, get a good declarer play book like Root's; all of them cover counting losers in some detail. I don't think there are any web resources that can really give as comprehensive a lesson as the many declarer play books out there. More pages = more examples, more explanation, people don't usually publish such things on the web.
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#8 User is offline   blackshoe 

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Posted 2013-October-09, 17:38

For my money the best sources on counting losers are The Modern Losing Trick Count, by Ron Klinger and Chapter One (I think) of Bid to Win, Play for Pleasure, by George Rosenkranz, which is titled Hand Evaluation. Both are available, I believe, from Baron-Barclay or Amazon.
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#9 User is offline   SteveMoe 

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Posted 2013-October-09, 19:11

Lesh, Counting losers helps identify threats whether in a trump or a NT contract. In a trump contract we look to eliminate losers in one hand so we can ruff losers from the other. Usually in NT the loser count in a long suit must be less than or equal to the number of entries we have to the hand with that suit, so we can set up length tricks then get to them to cash them.

Take just one suit into consideration with NO high cards: if it splits xxxx opposite xxxx then as long as the remaining card split 3-2 you have one sure winner. xxxx opposite xxx gives you a sure winner only when the remaining cards split 3-3.

For xxx opposite xxxxx you will have 2 winners if the suit splits 3-2 (losing three top tricks). Likewise xx opposite xxxxxx might yield 3 tricks if the top cards behave evenly.

So we conclude that length per se offers tricks as long as we have the time to develop them.

For AKxxx you have 5 of the 13 cards in your hand. Each player rates to have 8/3 or 2 2/3 cards in the suit. You will likely have only 1 loser if they split evenly (and partner doesn't have the Q). Another way to think about it is if the A and K each take 3 cards then the first 2 tricks exhaust 8 of the 13 cards - you have 3 of the 5 left. If the rest split 2-0 unfavorably you will lose 2 tricks. If they split 1-1 you will lose only one trick.

Now consider AKQx AKQ AKQ AKQ. You would likely consider this a 1 loser hand. It might be. All depends on whether partner has LTE 2, 3, 4, or more {sp}s. Where we own 8 or more cards in the suit on this hand the chances improve that this hand has NO losers.

Losing trick count is a technique for counting losers by assigning winners as A, K, or Q. You cannot hold more losers than cards in any suit. A is always a winner. Singleton K and doubleton Q are winners only when partner has an honor in that suit. No way to know until bidding confirms it. So stiff K is 1 loser and Qx is 2 losers. Kx is 1 loser.

Losers are any A, K, or Q you do not hold. So for AKQx AKQ AKQ AKQ There are NO losers. For a companion hand, xxxx xxx xxx xxx there are 12 losers (the max possible). So in any 2 hands there are 24 possible losers. If we total the number of losers in both hands in our example, we get 0+12=12. 24-12=12 are the number of tricks we expect to take, assuming we have an 8-card trump fit. So if partner has xxxx xxx xxx xxx we stand to make slam (any 3-2 split). If partner holds xxx xxx xxx xxxx, then we are likely to make 12 tricks only 36% of the time (when split 3-3. Note a 3-3 break helps only if are trumps because we have no entry to partner's yarborough. I can't say how anyone would know which black suit to choose.

So, AKxxx Axxx Qx xx are all losers.

Finally if we hold Axx opposite xxx, we expect 2 losers. If we hold Axx opposite Kxx we expect 1 slow loser (slow because they have to knock out both the A and the K). If we hold Axx opposite Qxx we expect 1.5 losers - half the time the K will be well placed. Slow losers are important to identify in trump contracts because you want to pitch them on winners or other losers (loser on loser) to reduce the number of threats the defense holds against our contract.

Hope this helps...
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#10 User is offline   paua 

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Posted 2013-October-11, 06:20

http://www.mrbridge....?choice=bidding
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