Losing Trick Count: Bridge Bidding Convention

Ron Klinger's Losing Trick Count

The aim of the Losing Trick Count (LTC) is to provide a more accurate tool for measuring the trick-taking potential of the partnership hands. We are all used to counting points and making point adjustments. For suitable situations, LTC claims to measure your trick potential better than point count.

Warning:

The LTC operates only with a trump fit (at least 8 trumps) OR when one partner has a self-sufficient suit (more later). The LTC usually comes into play only after a trump fit has come to light. You can use it to measure the potential of your own hand especially with a long, strong suit. The LTC is not suitable for no-trumps or for misfit hands. Use the point count for those.

Here is a basic example. Suppose you pick up: K86532 4 72 A943

Partner opens 1, you bid 1, partner 4. What do you do?

You have 7 HCP, about 11 or so counting distribution. Partner has about 19-20. Doesn't feel like a slam. I know that many would pass 4.

The two hands might be like this:

AQ74 J8 AK943 K5

K86532 4 72 A943

When the hand is over and you make 12 tricks, you ask partner: 'Could we have bid slam?' 'Nah, no way. No one will bid it.'

6 is a good bet, a piece of cake if trumps break 2-1. If trumps are 3-0, you will need to set up dummy's diamonds.

When you check the scores, you see one or two pairs have bid 6. 'Probably didn't know what they were doing,' you tell each other. Chances are they did know what they were doing. With the LTC you will find slams such as these child's play. We'll see more of these two hands later.

The LTC Formula

There are three steps in estimating your trick potential:

Count Your Losers.
Estimate Partner's Losers.
Add these together and deduct the total from 24.

The answer is the number of tricks your partnership will probably take, assuming normal breaks and half your finesses working.

Warning:

Remember, you need at least an 8-card trump fit or a self-sufficient suit for the LTC to operate. The answer for your trick potential is estimated to be at least 80 per cent effective. Don't expect the LTC to be accurate if trumps break 5-0 or if every finesse fails. Tough.

How To Count Your Own Losers

Each suit counts 3 losers at most.

For a 3-card or longer suit, count a loser for missing the ace, king or queen.

AKQ = 0 losers. So is AKQ4 or AKQ42 or AKQ432

AQ3 is 1 loser. The K is missing. 1 loser also for AQ32 or AQ532 or AQ7432.

A85 is 2 losers. The K and Q are missing. 2 losers also for A8532 or K43 or K863 or K87432.

Jxx or worse = 3 losers. So is Jxxx or longer.

Doubletons:

A-K = no loser
A-x, K-x = 1 loser
A-Q = 1/2 a loser
KQ is 1 loser (A missing)
Q-x or worse = 2 losers (A and K missing)

Singletons:

All singletons count as 1 loser except for ace singleton which is no loser.

VOID = 0 losers. Since you are playing in your trump fit, any losers in this suit in partner's hand can be ruffed.

How many losers do each of these combinations contain?

a. AK532, b. AJ73, c. 986432, d. K, e. KQ2, f. QJ53

Why not jot down your answers?

Remember there was mention at the beginning that paper and pen might come in handy.

AK532 = 1 loser, Q is missing
AJ73 = 2 losers, K + Q are missing
986432 = 3 losers, A + K + Q are missing
K = 1 loser, A is missing
KQ2 = 1 loser, A is missing
QJ53 = 2 losers, A + K are missing

The last example (QJ53) highlights a problem. Just how valuable are combinations headed by the queen?

A32 = 2 losers and obviously Q32 is not as good as A32.

On the other hand, we would all prefer to hold Q32 than 432. At least Q32 has some potential to take a trick or to build up winners in partner's hand

My suggestion is to count the queen as full value (1 trick) if the suit contains a second honor as well. Thus AQx or KQx count as 1 loser each. QJx or Q10x or longer count as 2 losers.

For weaker holdings, Q9x or worse, deduct half a trick. Count these normally as two and a half losers. If partner shows great strength in the suit, upgrade Qxx or longer to two losers.

How many losers does this bridge hand contain?

K86532 4 72 A943

Why not jot down your answers, suit by suit?

K86532 is 2 losers, 4 is 1 loser, 72 is 2 losers, A943 is 2 losers. Total losers = 7.

Now, how many losers in this bridge hand?

AQ74 J8 AK943 K5

AQ74 is 1 loser, J8 is 2 losers, AK943 is 1 loser, K5 is 1 loser. Total losers = 5.

AQ74, J8, AK943, K5

K86532 4 72 A943

This was the first example hand.

North 5 losers, South 7 losers.

Total 12. 24 minus 12 = 12 tricks as the likely outcome. Head for 6.

1	Pass	1	Pass
4	Pass	4NT	Pass
5	Pass	6

We would all like to reach 6 on these cards but note that North has nothing special for 4. No void, no singleton, only 17 HCP.

AQ74 J8 AK943 K5

It is easy to measure this hand as 5 losers when you see it.

How do you work that out during the bidding?

The average minimum opening hand has around 13-14 points. Likewise, the average minimum Opener has 7 losers.

How many losers here?

KQ74, A8, 872, K943

Only 12 HCP but the hand has 7 losers.

1 in spades, 1 in hearts, 3 in diamonds and 2 in clubs

Take a minimum opening as a 7-loser hand

J3 AQ632 K72 K95

13 HCP and 7 losers (1 in hearts, 2 in each other suit).

This is normal expectancy for minimum openings with nothing special in shape.

As strength increases, there are more tricks.

More points, fewer losers.

16-18 points with ordinary shape will usually have 6 losers.

As strength decreases, there figure to be fewer tricks.

Fewer points, more losers.

10-12 points, no special shape: expect 8 losers.

40 HCP in the pack. . . 13 tricks. . . so roughly 3 HCPs = 1 trick. So if 13-15 points = 7 losers, 3 extra points, 16-18, should be 6 losers. 19-21 points, expect 5.

3 fewer points, 10-12: expect 8 losers.

Similarly, 7-9 points is usually 9 losers.

7-9 = 9, 10-12 = 8, 13-15 = 7, 16-18 = 6, 19-21 = 5, and so on. If you know partner's points, you can estimate partner's losers.

Partner opens 1. . . Losers expected: 7 or fewer. Could be a minimum opening, could be stronger.

You respond 1 and partner rebids 2. This confirms a trump fit and only a minimum opening. Place partner with 7 losers.

Count your losers, add on 7, deduct the total from 24. The answer = the number of tricks you are likely to win.

1	Pass	1	Pass
2*

* minimum = 7 losers.

How many losers would Opener show if the bidding went:

1	Pass	1	Pass
3*

* one trick more and so 6 losers. 16-18 points = 6 losers

What if it went:

1	Pass	1	Pass
4*

* two tricks more than 2 and so 5 losers expectancy.

OK, back to the first hand.

Opener: AQ74 J8 AK943 K5

You open 1 and partner bids 1. What next?

Trump fit exists, so count losers. You have 5. 1-level response is 9 losers or better. Assume the worst, 9. So, 5 plus 9 = 14. 24 minus 14 = 10, so ten tricks expected, so insist on game.

With a singleton or void, you could splinter. As it is, bid 4.

Now, Responder's decision with K86532 4 72 A943

Partner opens 1, you bid 1, Opener rebids 4. What now?

Trump fit exists, so count losers. You have 7. Opener's jump to game opposite 6+ points should be 19+ points, so expect 5 losers.

7 plus 5 = 12. 24 minus 12 = 12, so twelve tricks expected. Explore slam.

With spades headed by the king, a singleton heart and clubs headed by the ace, Blackwood (or RKCB) is suitable. You can risk your rag doubleton diamond as partner opened 1 and has shown a strong hand.

What would you do after:

1	Pass	1	Pass
4	Pass	?

AQ74 J8 AK943 K5

K86532 42 7 A943

Slam is now risky.

Some players might take their chances, ask with 4NT and bid 6, but against you and me, they always lead hearts, right?

With the danger in hearts, South should cue-bid 5 after 1 - 1 - 4.

North cue-bids 5 to show the Ace of diamonds.

South now bids 5 (Bypassing 5 says, 'I'm worried about hearts). North passes ('Me, too.')

OK, let's look at a few other actions and the losers expected.

How many losers for a takeout double?

How many losers for a 1-level response?

What about for a 2-level response?

Takeout double = about opening strength. Expect 7 losers or fewer
1-level response = 6 or more points. Expect 9 losers or fewer.
2-level response = 10 or more points. Expect 8 losers or fewer.

How about Opener's reverse? What would you expect for that?

Since Opener's reverse is around 16 points, you can expect 6 losers or fewer.

As Opener normally has at least a 5-4 pattern, you will find that Opener's reverse more often turns up with only 5 losers.

62 AK42 K2 AQJ43

Open 1. Over 1, rebid 2. The hand has only 5 losers. So, basic expectancy for Opener's reverse is 5 losers.

You are entitled to reverse with fewer than 16 HCP as long as the hand has no more than 5 losers.

Since a minimum opening is 7 losers, a 5-loser hand has a playing strength two tricks better than a minimum.

It is reasonable to reverse with such strength. It would be OK to open 1 and rebid 2 over 1 with 62 AQJ2 AK10942 3

What about Opener's jump shift, say:

1	Pass	1	Pass
3

How many losers?

Given that Opener's jump shift is forcing to game and Responder might have only 6 points, Opener's expectancy is 19+ points.

That would translate to 5 losers or better, but as Opener is normally at least 5-4, expect the jump shift to be a 4-loser hand.

A2 AQ742 87 AKQ4. Open 1, rebid 3 over 1. 4 losers.

What about a 1-level overcall by partner? You know your partner's habits better than I do, but if partner can be expected to have a decent 5-card or longer suit and around 8 HCP or better, the 1-level overcall figures to be 8 losers (minimum) up to 6 losers (maximum).

These are respectable overcalls of 1:

AKQ62 95 643 862 or KQ762 95 A43 862. Each has eight losers.

An overcall type hand with 5 losers is usually too strong for a simple overcall. With 5 losers, double first, bid your 5+ suit next.

Double then jump in a new suit shows an overcall hand with 4 losers.

Suppose Right hand Opponent (RHO) opens 1 and you hold:

AQJ762 KQ5 A4 62. What is your bidding plan?

You have a 5-loser hand. Double and bid again shows this.

Double and rebid 1 over 1 or 2 over 2.

What if partner replies 1NT to your double?

AQJ762 KQ5 A4 62

You doubled 1 and partner replied 1NT. What now?

The 1NT response to a takeout double usually has 6-9 points. That will generally produce two tricks. If so, partner's two winners will reduce your 5 losers to 3.

Partner 1NT response should contain 2-3 spades. Bid 4.

Why Deduct From 24?

Where does that figure of 24 come from?

Maximum losers in a suit? Three. So, maximum losers in 4 suits? 12.

So maximum losers in your hand = 12 and maximum in partner's = 12. Total 24. By deducting actual losers from the maximum possible what accounts for the difference? The tricks your side should win.

How Accurate is the LTC?

Suppose one of your suits is AQ5. What are the odds that this will be only one loser, or in other words, what are the chances that your AQ5 will translate into two tricks?

What do you think? 50 percent? 65? 75? 80? 85? 90?

If partner has rags only, the chance for the finesse is 50 per cent, but partner need not have rags. Partner could have the king.

The king can be in 3 spots. If the king is with partner, you have two tricks. If the king is on your right, you have two tricks.

That is already 2/3 of the time.

AQx as two tricks: what odds?

If your side has more points than theirs, the odds improve.

Suppose you have 14 HCP and partner about 13. That gives the opponents 13 HCP, too.

The odds are now 50 per cent that partner has the K and 50 per cent that an opponent does. If RHO has the K (25 per cent) you have 2 tricks. That brings it up to 75 per cent.

Of course partner need not have the king for you to make 2 tricks.

If partner has the jack, two tricks are sure.

If partner has 10-9-x your chances for two tricks are improved. If LHO leads the suit, you have two tricks. All these extra chances boost the odds for two tricks to over 80 per cent, close to 85.

A832 6 94 K85432

Partner opens 1, RHO passes.

What action should you take?

You have spade support, a weak hand but with 7 losers you are too good for 2. The jump to 4 expresses this hand type.

It is too weak for 2 (which also has no preemptive effect).

The two hands might be like this:

A832 6 94 K85432

KQ954 1095 K8 A97

4 is respectable, succeeding if the diamond finesse Ace is onside or if clubs are 2-2, all in all about a 70 per cent chance.

KQ942 Q10 AQ873 2

You open 1, LHO passes, 2 from partner, RHO passes.

What action should you take? You have 5 losers. The 2 raise, 6-9 points, is usually 9 losers.

5 plus 9 = 14. 24 minus 14 = 10 tricks. Bid 4.

A763 32 K54 8754

KQ942 Q10 AQ873 2

4 is great and should make easily. Of course, partner's King of diamonds is a jewel. Partner's cards might not always be so useful.

The two hands could be like this:

A763 K54 62 8754

KQ942 Q10 AQ873 2

4 is still all right. On a lucky day, you might make 11 tricks if the diamonds finesse is on.

The two hands might be:

A763 954 62 K754

KQ942 Q10 AQ873 2

Now 4 may well fail. Too bad. If you need to be right all the time, bridge is not your game.

Like the lass who said, 'I thought I was lucky to find Mr. Right, until I found out his first name was Always.'

It pays to be optimistic. You win some, you lose some. Play well and the wins will exceed the losses. Do not expect eternal bliss. . . certainly not at bridge. . . 60 per cent bliss is pretty good.

AK62 A6 AQ9753 2

Partner passes. You open 1 - 3 from partner. What action should you take?

You have 4 losers. Partner's 3 as a passed hand (or as a limit raise) should be 8 losers.

4 plus 8 = 12. 24 - 12 = 12 trick potential. Head for slam.

AK62 A6 AQ9753 2

Usually you would ask for aces but with three aces yourself and a singleton in the other suit, you can simply jump to 6 over 3. It never hurts to have a reputation as a flamboyant player.

The sequence will have been:

Pass	1	Pass	3
Pass	6

It would be OK to ask with 4NT but you will finish in 6 later anyway

The two hands might be like this:

754 K4 K862 A643

AK62 A6 AQ9753 2

6 is laydown and 7 virtually hopeless. 7 might make if partner had two spades. Be satisfied to reach 6 opposite a passed hand.

A974 Q9 AK6 J753

1	Dbl	Pass	2
Pass	?

What action do you take?

You have 8 losers. Partner's jump reply, around 10 points or so, should be 8 losers.

8 plus 8 = 16. 24 - 16 = 8 tricks. Pass 2.

Expectancy for a takeout double is 7 losers. You have 8. That makes you sub-minimum for the double. There is no case for bidding on after partner's 2 reply.

The two hands might be like this:

A974 Q9 AK6 J753

K853 J7 QJ73 A86

Most of the time you will make just 8 tricks, losing 1 spade, 2 hearts and 2 clubs. 3 is living beyond your means. 4 is madness.

We estimate 3 points = 1 trick. Here the partnership has 3 points in hearts for no tricks. Notice how the LTC counts nothing for the hearts and keeps the partnership at a sensible, low level.

A974 Q9 AK6 J753

K853 J7 QJ73 A86

Even 2 might fail if the spades are not 3-2. You will lose less in 2 than those who have climbed higher in spades.

Even if they compete to 3, you should not bid 3.

Now take a look at these two hands:

K85 AKQ 943 A862

AQJ543 965 8 753

South has 8 losers, North 7. 8 plus 7 = 15. 24 - 15 = 9 tricks, yet 4 is virtually foolproof. What has gone wrong?

The LTC formula (Count your losers, add partners, deduct from 24) works well when each player has some ruffing potential.

When you have no ruffing potential, a better approach exists:

Calculate partner's losers, deduct this from 13, then add on your winners. The final answer is the number of tricks expected.

This is worth repeating. When you have no ruffing potential:

Calculate partner's losers (in the usual way)
Deduct the answer from 13. That gives you partner's winners.
Add your winners.

The answer is the number of tricks expected.

This is the usual approach after partner's pre-emptive opening. It is also sensible after partner's weak two opening if you have no ruffing value. Indeed, you should use it any time that you have support for partner but no ruffing value.

AK6 876 AK53 874

Partner opens 2, a weak two, RHO passes.

What do you do?

The weak 2 has 7 or 8 losers. Deduct from 13 = 6 or 5 tricks. You have 4 tricks. Opposite 6 tricks that would produce game. You should invite game. The Ogust 2NT response is used to do that.

Opposite a minimum, stop in 3. Opposite a maximum, bid 4.

The two hands might be like this:

AK63 876 AK5 874

92 AKQ532 762 105

How embarrassing not to reach game with these values.

The auction could go:

2	Pass	2NT	Pass
3NT*	Pass	4**

* AKQ suit, ** It would be very risky to pass 3NT, though it might work.

Likewise, how poor not to reach 4 with these hands:

AK63 876 AK5 874

92 KQ10432 762 A2

The auction could go:

2	Pass	2NT*	Pass
3**	Pass	4

* Ogust, ** 3 shows a maximum weak two with a good suit.

4 makes if hearts are 2-2 or if Responder has the Ace of hearts or if the Jack of hearts is singleton. All in all, pretty good odds.

Of course, the hands might be:

AK63 876 AK5 874

92 KQJ432 762 92

Now the auction might go:

2	Pass	2NT*	Pass
3**	Pass	3	Pass
Pass

* Ogust, ** 3 shows a minimum weak two with a good suit.

It would be unlucky for 3 to fail, but it could happen.

AKQ43 K2 A32 876

Partner opens 1, you bid 1 and Opener rebids 3.

Where would you head now?

Partner's rebid shows a 6-card or longer suit and about 16-18 points. Because of the 6-card suit, expectancy for 3 is 5 losers.

13 - 5 = 8 tricks with partner. You have 5 winners. Aim for 7.

The two hands might be like this:

AKQ43 K2 A32 876

92 AQJ643 K85 AJ

The auction could go:

1	Pass	1	Pass
3	Pass	4NT*	Pass
5**	Pass	7

* RKCB, ** two key cards plus the trump queen

Opener is minimum for the 3 rebid (15 HCP, 6 losers) but 7 is still a great contract.

This brings up an important guideline for bidding a grand slam:

If the loser count indicates 13 tricks potential AND You have an adequate trump fit AND Your side has all the aces PLUS the K and Q of trumps Then bid seven.

K832 KQ63 AKQJ 3

Partner opens 1.

How many losers do you have?

What is your potential? What do you do?

You have 4 losers.

Partner opened the bidding. Expect 7 losers.

4 plus 7 = 11. 24 minus 11 = 13 tricks potential.

Bid 4NT. You gain nothing by bidding more slowly.

Opposite one ace, sign off in 5. Opposite two aces, bid 6. Opposite three aces, bid 7. Potential is 13 tricks and you know the partnership has all the aces and the K-Q of trumps.

The two hands might be like this:

K832 KQ63 AKQJ 3

A5 AJ542 853 A84

The play in 7 is routine. 6NT is there but not 7NT. You had an easy auction since you had the K-Q of trumps yourself.

KQ32 J863 AKQJ A

A5 AKQ42 853 842

Responder still has 4 losers and 7 is still a great contract.

4NT RKCB makes it easy to locate the K and Q of trumps. Without RKCB, it is much tougher to find the trump honors.

The bidding might go:

1	Pass	4NT	Pass
5*	Pass	5**	Pass
6***	Pass	7

* 0 or 3 key cards, obviously 3, ** asks for the trump queen, *** no other king to show

When you have no ruffing value, you remember from earlier:

Calculate partner's losers. Deduct from 13.
Add on your tricks.

You can use an equivalent approach when you have the long suit and partner might have no ruffing values for you. Count your losers. Deduct from 13 to calculate your tricks. Add on the winners estimated in partner's hand.

AKQJ54 92 AK2 104

You open 1, partner replies 1NT, RHO passes.

What do you do?

You have 5 losers. That gives you 8 tricks. Partner's 1NT, 6-9 points, will normally include two tricks. That takes you to ten tricks and so you should rebid 4.

The two hands might be like this:

32 AK63 8743 986

AKQJ54 92 AK2 104

A very easy game. Partner might bid on if you rebid 3 but some partners might pass 3.

4 would still be a good spot if partner had a singleton or void in spades or if the hearts were A-Q-6-2. A game around 50 per cent is OK.

At this point, you should be thinking, 'Hey, wait a moment. If partner has a singleton or a void in spades, we do not have a trump fit.'

That is true but the formula (count your losers, deduct from 13, add on partner's winners) works perfectly well if you have a suit that needs no support, a self-sufficient suit.

The AKQJ54 in spades in the preceding example is such a suit.

When is a Suit Self-Sufficient?

There is an easy test to apply for this.

Count the number of cards in your long suit. Add the number of honors in that long suit. The total is the Suit Quality (SQ).

If the quality is 10 or more, the suit is self-sufficient.

AKQJ54 : Length = 6. Honors 4. Total 10. Thus this is a self-sufficient suit.

Likewise, these suits are self-sufficient (SQ is 10 or more).

KQJ1032 or AQJ6432 or QJ1098632, and so on.

If such a suit is a major, try to insist on it as trumps.

So, if you have a self-sufficient suit:

Count your losers.
Deduct from 13 to calculate your tricks.
Add on the winners estimated in partner's hand.

The answer is the number of tricks you are likely to make.

This formula will also work for you if you have a trump fit.

AKQ432 85 AQJ 87

You open 1, partner raises to 2, RHO passes. Your move?

You have 5 losers. 13 - 5 = 8 playing tricks.

Partner's 6-9 points figures to produce 2 tricks.

Your 8 tricks and partner's 2 = 10. Therefore, rebid 4.

When you have a self-sufficient suit (or a trump fit), how can you tell how many tricks partner has? How do you translate points into tricks?

Responder's 6-9 point hands will usually produce two tricks (but it could be 3 --- AKQ in one suit — or only one (ace and 2 jacks). 2 tricks is a good guideline but we are not dealing with certainties.

1 trick = 3 points, so 10-12 = 3 tricks, 13-15 = 4, 16-18 = 5, etc.

KQ2 2 42 AQJ7643

You open 1, partner responds 3NT (16-18 points, no 4-card major).

What do you do now?

You have a self-sufficient club suit. 5 losers = 8 playing tricks.

Partner's 16-18 points = 5 tricks likely. 8 plus 5 = 13. A grand slam is possible. Do not sell out too soon.

KQ2 2 42 AQJ7643

After 1 - 3NT, a 4 rebid is sensible. This sets the trump suit.

Partner continues with 4, showing the diamond Ace.

What is your plan now?

After 1 - 3NT - 4 - 4, ask for key cards.

If you find 3 key cards only, stop in 6.

If partner has 4 key cards, ask for kings. If partner has a king as well, you can rebid 7NT.

On your lucky days, the two hands might be like this:

A95 AK3 A1086 K65

KQ2 2 42 AQJ7643

If partner had denied a king after showing 4 key cards, you would bid 6NT. Imagine the King of hearts above is the Jack of hearts and you can see 6NT is fine.

K832 KQ63 AKQJ 3

Suppose partner has opened 1.

What action do you take?

You have a good trump fit and 4 losers. Partner opened the bidding. Expect 7 losers. 4 plus 7 = 11. 24 minus 11 = 13 tricks potential.

Bid 4NT. You gain nothing by bidding more slowly.

Opposite one ace, sign off in 5. Opposite two aces, bid 6. Opposite three aces, bid 7. Potential is 13 tricks and you know the partnership has all the aces and the K-Q of trumps.

The two hands might be like this:

K832 KQ63 AKQJ 3

A5 AJ542 853 A84

The play in 7 is routine. 6NT is there but not 7NT. You had an easy auction since you had the K-Q of trumps yourself.

AJ63 J98 Q63 K74

Partner opens 1, you respond 1, Opener raises to 2.

What do you do now?

You have 11 HCP. Most players would invite game. You have a trump fit, so count losers. You have 9 1/2 losers but let's upgrade the Queen of diamonds as partner opened 1. So, 9 losers. Partner opened, so 7 losers. 9 plus 7 = 16. 24 - 16 = 8 tricks. The LTC says 8 tricks is the likely outcome, so pass 2.

The two hands might be:

AJ63 J98 Q63 K74

K842 Q10 AK742 J2

1	Pass	1	Pass
2	Pass	Pass

Note that the Queen of diamonds is full value but 4 is still a poor contract. You should lose 2 hearts and 1 club. You might lose a second club and you might easily have a spade loser (Queen of spades offside or spades 4-1). 4 is about 20 per cent. If you go on at all, it might be best to try 2NT. Trouble is partner might take you back to spades. Partners are like that . . . put on earth to test your tolerance. I'd rather pass 2 and gain when things go badly (4-1 trumps, etc.)

K9754 AK832 82 7

LHO opens 1, 1 from partner, double on your right (negative).

What do you do?

You have 6 losers. Partner's overcall = 8. 6 plus 8 = 14. 24 minus 14 = 10 tricks. Bid 4.

The two hands might be:

AQ832 75 K94 862

K9754 AK832 82 7

4 is a great spot and may well make 11 tricks if the Ace of diamonds is with the Opener. N-S may have a decent save in clubs but they haven't found it yet.

KQ985 AKQ862 4 7

Partner opens 1. How do you feel about this hand?

What do you do?

After partner's 1, an optimist is one who thinks, 'Great. I have an opening hand. We have a game here, maybe a slam."

A pessimist is one who thinks, 'Oh, just great. I bet partner is 6-5 in the minors and the misfit leads to a terrible disaster."

OK, so we know what an optimist is and what a pessimist is.

Do you know what a realist is?

The wise ones tell us, "A realist is one who KNOWS the pessimist is right."

OK, back to reality. Partner opens 1. What do you respond?

KQ985 AKQ862 4 7

You have an excellent hand, only 3 losers, but it is too soon to count losers. There is no evidence yet of a trump fit.

Bid just 1. There is no need to jump to 2 and no benefit either.

Partner rebids 1.

What do you do now?

What great news! You have a trump fit, so count losers. You have 3. Partner opened and has not promised extras, so count 7. 3 plus 7 = 10. 24 minus 10 = 14 tricks potential!

OK, so a grand slam is possible and you are the one who knows it. The one who knows should take control. Check on aces/key cards. If partner shows three aces, bid 7. (Not 7NT: the hearts may not run.)

The two hands might be:

A632 7 AJ643 A42

KQ985 AKQ862 4 7

7 is a great spot. Win the lead, draw trumps (remember to play the Ace of spades first to guard against J1074 with North). Then set up hearts.

AKQ632 K8652 5 3

You open 1, partner responds 2.

What action do you take?

You have a (great) trump fit, so count losers. You have 4 losers, partner should have 8 or better (10 points up) 4 plus 8 = 12 tricks potential, so check on aces.

AKQ632 K8652 5 3

1	Pass	2	Pass
4NT	Pass	5*	Pass
?

* 2 key cards (2 aces here) but no heart Queen.

What do you do now?

Since the 2 response shows five or more hearts, chances are (78 per cent) that hearts are 2-1 and no loser. Bid 6.

The two hands might be:

AKQ632 K8652 5 3

4 AJ743 A98 J952

Partner has a minimum, 8 losers as expected. 6 is an excellent slam. Draw trumps and set up spades will be the plan. By playing the King of hearts first, partner can even guard against Q109 with North.

Now take a look at these two hands:

96 AQJ54 A2 A632

8732 K762 K5 754

You have 6 losers. partner has 9. 9 plus 6 = 15. 24 - 15 = 9. Tricks expected 9, You lose 2 spades and 2 clubs only. So, the result is just as you would expect.

But suppose the hands were like this:

96 AQJ54 A2 A632

8732 K762 754 K5

Partner has the same HCP, same shape, same number of Ks, same losers (9), but now 4 is a very good contract (ruff 1 club low, 1 club high).

What has happened?

Partner's K-x in clubs is far more valuable than the K-x in diamonds

K-x opposite length is far more valuable than K-x opposite a short suit. The location of the high cards can boost the value of a hand.

After a major suit raise to the 2-level, if you have a 2-loser or 3-loser suit, you can check whether partner has useful cards opposite by making a 'long suit trial bid'.

A trial bid is a new suit after a major suit raise to the 2-level.

1	Pass	2	Pass
3*

* 3 is a trial bid, asking for help in clubs.

1	Pass	2	Pass
3*

* 3 is a trial bid, asking for useful stuff in diamonds

The typical trial bid by Opener in such auctions is a 6-loser hand and usually around 16-18 points.

The trial bid is an invitation to game and the trial suit is usually a 3-card or longer suit with 2 or 3 losers.

1	Pass	1	Pass
2	Pass	3*

* 3 is a trial bid in diamonds by Responder.

1	Pass	1	Pass
2	Pass	3*

* 3 is a trial bid in hearts by Responder.

Opener treats Responder's trial bid as a game invitation. Again the trial suit is a 3-card or longer suit with 2 or 3 losers.

If it is a game invitation, Responder will usually have 8 losers and about 10-12 points.

While Responder's trial bid will usually be a game invitation, Responder may also use a trial bid as the first move in a slam exploration. Responder may be interested in whether Opener has help in the trial suit.

In reply to a trial bid, you sign off in 3-of-the-major if you have a poor holding in the trial suit. Bid game in the major if you have a strong holding in the trial suit. In judging your holding in the trial suit, losers can be a very good guide.

Your holding in the trial suit:

3 losers (J-x-x or worse): Sign off in 3-Major.
2 losers: Sign off in 3-Major if minimum, bid 4-Major if maximum.
0 or 1 loser in the trial suit: Bid 4-Major.

This easy guide will produce the right decisions most often.

Now let's look at those earlier hands again:

96 AQJ54 A2 A632

8732 K762 Diamond K5 Club 754

1	Pass	2	Pass
3*	Pass	3	Pass
?	Pass

* trial bid

Opener has 6 losers and invites game. Responder has 3 losers in the trial suit and rejects the invitation. You stop at a sensible level.

96 AQJ54 A2 A632

8732 K762 754 K5

This time it goes:

1	Pass	2	Pass
3	Pass	4

Responder has only 1 loser in the trial suit and accepts the invitation. You reach a good game. (If it went 1 - 2 - 3, Responder would pass.)

96 AQJ54 A2 A632

8732 K762 75 K54

This time it goes:

1	Pass	2	Pass
3	Pass	3

Responder has 2 losers in the trial suit but is minimum. Invitation rejected. Opener figures lose 2 spades, 1 diamond, 1 club and just make 3

96 AQJ54 A2 A632

8732 K762 KQ5 54

This time it goes:

1	Pass	2	Pass
3	Pass	4

Responder has 2 losers in the trial suit but is maximum. Invitation accepted. Opener figures lose 2 spades and 1 club, making 4 easily.

Now take a look at these two hands:

AQ863 A6 K76 632

K975 K532 5 AQ87

Opener has 7 losers. Responder has 6. 7 plus 6 = 13. 24 - 13 = 11. Tricks expected 11, and in spades Opener figures to lose 1 diamond and 1 club. If the King of clubs is onside, 11 tricks. If not, 10 tricks. Stopping in 4 is eminently sensible.

But suppose the hands were like this:

AQ863 A6 976 K32

K975 K532 5 AQ87

HCP, high cards, losers are the same but now 6 is excellent.

What has happened?

Opener's club King is far more valuable than Opener's king in diamonds.

Just as trial bids help you to diagnose whether partner's high cards are well-located for game purposes, so splinter bids aim to find out whether partner's cards are in the right place for slam.

Often it not what you have that counts, but where it's located. If you have the means to find out, you will be able to bid slams more accurately.

Splinters can be used by Responder or by Opener. A splinter is usually an unnecessarily high jump in a new suit. For example, 1 - 4 or 1 - 4 or 1 - 4

The splinter shows a singleton or void in the suit bid PLUS at least 4 trumps for partner PLUS enough points for game.

For an immediate splinter by Responder, you can expect about an opening hand.

K975 K532 5 AQ87

If partner opens 1 or 1, this is a pretty respectable splinter bid of 4. As a splinter bid will usually have 6-7 losers, the hand could actually be a bit weaker than this in HCP.

When partner makes a splinter bid, the ace is useful opposite a singleton, but the king, queen and jack are almost always wasted.

After an immediate splinter by Responder, Opener should disregard the K, Q and J in the splinter suit. If Opener has 13 or more HCP in the other suits, slam is likely to be a good bet.

After discounting the wasted cards in the short suit, if Opener's HCP fall below 12, slam is probably a poor bet. Sign off in game.

So, let's take a look at those two splinter hands again.

AQ863 A6 K76 632

K975 K532 5 AQ87

1	Pass	4	Pass
4	Pass	Pass

After deducting the 3 HCP for the wasted King of diamonds, Opener has only 10 HCP. Therefore, Opener signs off in 4. If very strong, Responder can bid on.

AQ863 A6 976 K32

K975 K532 5 AQ87

1	Pass	4	Pass
4NT*	Pass	5**	Pass
6

* RKCB, **2 key cards

Now Opener has nothing wasted in diamonds and has 13 HCP. Therefore, Opener heads for slam and bids 6 after Responder shows 2 key cards.

What if the hands were like this?

AQ863 A6 K76 K32

K975 K532 5 AQ87

1	Pass	4	Pass
4NT	Pass	5*	Pass
6

* 2 key cards, no trump Queen

Opener has 3 HCP wasted in diamonds but still has 13 HCP after deducting the wasted strength. So, Opener again heads for slam.

The 1 - 4 splinter is dangerous. A few sequences that go 1 - 4, Pass . . . Oops, I forgot . . . destroy the value of playing splinters. Still, the first debacle like that is usually enough to ensure it isn't forgotten again.

The splinters after a 1 opening are: 3/4/4

The theory is that such jumps are not as useful for other purposes. Splinters are often used after the major suits. Some also use splinters after 1, e.g., 1 - 3 or 1 - 3 or 1 - 3, or after 1 (1 - 3 or 1 - 3 or 1 - 4).

Opener can also make a splinter bid after partner's response.

1	Pass	1	Pass
4*

* shortage in diamonds, 4 spades and enough for game. As Responder has promised only about 6 points, Opener's splinter here would be around 19 points or more.

KQ3 9732 J2 AQJ4

A75 8 AK54 K8632

1	Pass	3*	Pass
4	Pass	4NT	Pass
5	Pass	6

* splinter

It is best to treat the splinter as game-forcing and so Opener's 4 is not droppable. Opener is concerned about diamonds. With good controls in all suits, Responder asks with 4NT.

KQ63 K52 Q3 QJ54

A75 8 AK54 K8632

1	Pass	3*	Pass
3**	Pass	4

* splinter, ** One stopper opposite the singleton may be enough but it is risky. 3 suggests playing in a 4-3 fit. 5 is OK but 4 is fine at MatchPoints.

Splinters can be very helpful after a 2 opening:

AKQ632 void KQJ3 A74

8754 86432 A64 2

With only 3 losers, Opener is strong enough for a 2 opening.

2	Pass	2	Pass
2	Pass	4*	Pass
4**	Pass	5**	Pass
7

* splinter, ** cuebids