Stoppers

♠ ♥ ♦ ♣

This is one of a series of Declarer Play articles. These articles build upon each other, so I recommend that you study them in order.

Good declarer play relies on counting. Most of the counting we have already practiced focuses on the number of tricks each side can cash.

For most hands, neither side starts out with enough winners to succeed. So both sides will work to establish more.

That's why we view no trump contracts as a race. You must establish and cash enough tricks before they do.

Losing the lead while establishing tricks is a normal part of declaring no trump contracts. After you lose the lead to establish tricks, the defense will (normally) lose the lead back to you while they try to establish some tricks of their own.

Doing split assumptions and counting stoppers helps you figure out when this back-and-forth of losing the lead can turn into one side or the other cashing enough tricks to make or break your contract.

On this page, we're going to see what stoppers are, and how counting them will improve your declarer planning.

example 1

Dummy
♠ A Q T 2
♥ Q 8
♦ K T 3 2
♣ T 7 5

You
♠ J 9 6
♥ A K 6
♦ A Q J 8 4
♣ J 6

Contract: 3N The opening lead is the ♣4.

How many cashable tricks do you have?

You have 9 – 1 spade, 3 hearts, and 5 diamonds. You have enough cashable tricks to make your contract.

What is the split assumption for clubs?

We start with the assumption that their 8 clubs split 5-3.

Our next step is usually to review the bidding and look for missing spot cards lower than the opening lead. But on this hand I'm just going to tell you that the 5-3 assumption is correct.

They win the opening lead, and lead another club, and another, and another, and another. 5 tricks. Ouch!

Our contract is already defeated (down one), because we couldn't STOP them from cashing their entire club suit.

If we could have peeked into all four hands before choosing to play 3N, we would have seen that we have no club stopper, and we would have known it was a bad contract. Playing with a trump suit would have allowed us to trump their clubs, stopping them from cashing their entire suit.

A stopper is a high card winner that can capture the lead and stop the other side from cashing tricks in that suit.

Notice that there are two parts to our definition of a stopper:

winning a trick
preventing them from cashing the trick(s) they establish

example 2

Dummy
♣ T 7 5

You
♣ J 6

These are the clubs from example hand 1.

You have no club stopper because you cannot win a club trick.

example 3

Dummy
♦ K T 3 2

You
♦ A Q J 8 4

These are the diamonds from example 1.

You have no diamond stoppers.

Yes, you can easily win 5 tricks in this suit.

However, your opponents don't have any diamond winners. And they can't develop any for later. So your terrific diamond holding doesn't satisfy the second half of our definition – to stop them from cashing or developing diamond tricks. Lots of diamond winners for you, yes. But no stoppers.

example 4

Dummy
♥ T 8 7

You
♥ A 6

You can capture the lead with your ♥A.

They can develop high card winners and skaters in this suit.

Is your ♥A a stopper?

Yes. It satisfies both parts of our definition. It wins a trick, and it stops (delays) them from cashing their winners.

example 5

Dummy
♠ 7 5 2

You
♠ K 8 4

The opening lead is a spade against your 3N contract.

Is your ♠K a stopper?

Yes. It satisfies the first part of our stopper definition – it will win a trick. Your ♠K is promoted by playing last, so it either wins this first spade lead, or else they play their ♠A and then your ♠K wins a future trick.

And it satisfies the second part of our stopper definition – they can develop tricks in spades.

If they lead a different suit for their opening lead, will you have a spade stopper?

Maybe. If the ♠A is on sides, your ♠K will be a stopper. But if the ♠A is off sides, your ♠K will only be a stopper if the lead comes from your left so that your king plays last.

example 6

Dummy
♣ Q T 8 7 2

You
♣ J 9 6 4

Do you have any club stoppers?

No. They have 2 club winners, but you cannot do anything to take the lead away and delay when they cash them.

Furthermore, they cannot develop any club skaters because they only have 5 clubs and you have all the promotable honors and big spot cards.

So your clubs don't satisfy our definition of a stopper.

Now let's look at this suit from the defenders' point of view. Do they have any club stoppers?

Yes, they have two stoppers – the ♣A and ♣K. They will win tricks with their two master cards, and winning them stops you from cashing your future club tricks.

When a stopper captures a trick, it's permanent. Nothing can undo a captured trick.

However, stopping the opponents from cashing their tricks may be only temporary. If they win a future trick in another suit, they can return to leading the original suit and possibly cash their tricks then.

Both sides can have stoppers in the same suit.

example 7

Dummy
♦ Q 7

You
♦ K 9

Do they have a stopper?

Yes. Their ♦A stops (prevents... delays...) you from cashing a high card winner in diamonds.

Do you have a stopper?

Yes, your ♦K and ♦Q combined will win a trick, stopping (preventing... delaying...) them from cashing their diamond skaters.

example 8

Dummy
♠ K Q 7

You
♠ 5 4

Do they have a stopper?

Yes. Their ♠A stops (prevents... delays...) you from cashing a high card winner in spades.

If the ♠A is on-sides (on your left), you may have 2 winners. Of course you must lead from your hand toward your losing honors rather than leading one of them. When the ace is on-sides, you will see it play before you risk one of your honors. Play your spot card when you see the ace, winning later with your honor(s). That's two winners when the ace is on-sides.

Do you have a stopper?

Yes. Your ♠K and ♠Q cannot both be squished by their ♠A, so you will win at least one trick, stopping them from cashing their spade skaters.

If the ♠A is on sides, you will win two tricks, and have 2 stoppers.

example 9

Dummy
♥ K Q 7

You
♥ J 3

You can develop two high card winners.

Do they have a stopper?

Yes, their ♥A can take the lead away and stop (prevent... delay...) you from cashing your high card winners.

They can develop skaters. The split assumption for their 8 hearts is a 5-3 split, so we expect them to be able to develop 2 skaters.

Do you have a stopper?

Yes. Actually, you have 2 stoppers. You can play your three honors on three separate tricks, winning 2 of them, and twice stopping your opponents from cashing their skaters.

Dummy
♥ K Q

You
♥ J 3

If instead this was your hand, then your 3 honors could not be played on 3 separate tricks, so you would have only one winner, and only one stopper.

example 10

Dummy
♠ 9 8 3

You
♠ A J T 5 2

Do they have a spade stopper?

Yes, maybe 2 stoppers. If you capture one of their honors with your one ace, their other honor will be a stopper.

But if you cannot capture one of their honors with your ace, both of their honors will win tricks. Both will be stoppers.

When would that happen? If LHO holds ♠KQx, he will play his spot card when you play your ace, and play an honor each time you don't play your ace. You have to play before him, so he will look at your card before he decides what to play.

Do you have a spade stopper?

Yes. Your ♠A stops (delays...) them from cashing their honor card winner(s).

example 11

Dummy
♣ 2

You
♣ A K 6

What is the stopper situation?

You have 2 stoppers.

When your stoppers are gone, how many skaters do you expect them to have?

The split assumption is 5-4. That would give them 3 skaters.

example 12

Dummy
♦ J T 9 8

You
♦ 5 3

How many stoppers do they have?

They have 3 stoppers preventing you from cashing a 4th round diamond winner.

How many stoppers do you have?

That was a sneaky question. If their diamonds split 4-3, they have no possible skater, so you would have no stopper. But if the diamonds split 5-2 (or worse), your fourth-round winner will stop (delay...) them from cashing their skater(s).

Who cashes first?

example 13

Dummy
♦ K J T 9 3
♣ 6 2

You
♦ Q 2
♣ A K 3

How many club stoppers do you have?

That was an easy one. Two stoppers.

How many diamond stoppers do they have?

Another easy one. Just their ♦A – one stopper.

You have more stoppers than they do. So if they persist with club leads and you persist with diamonds, you will be able to cash your diamond tricks before they can cash their club skaters. Like this...

They start, and knock out your first club stopper.
You lead diamonds, forcing out their ♦A
They persist with clubs, forcing out your second club stopper.
You cash all your diamonds.

Whichever partnership has more stoppers in the opponents' suit(s) gets to cash their own tricks first.

Stay ahead in the stopper race!

You have more stoppers than they do, but if they force out your first club stopper and then you fiddle around with hearts or spades and lose the lead, they will cash their suit before yours.

Let's follow how that would work:

They start, and knock out your first club stopper.
You lose the lead when you lead hearts or spades. (So sad...)
They force out your remaining club stopper.
Then you lead diamonds (too late), forcing out their ♦A .
They cash their club skaters.

example 14

Dummy
♦ K J T 9 3
♣ 6 2

You
♦ Q 2
♣ A 7 3

This example is almost the same as the previous one. I've taken away one of your club stoppers.

Now you have the same number of stoppers as they do. You have 1 club stopper, and they have 1 diamond stopper.

Remember... if the number of stoppers is the same, whoever starts knocking out the opponents' stoppers first gets to cash their tricks first.

How does the play go?

They start, and knock out your only club stopper.
You lead diamonds, forcing out their ♦A .
They cash their club skaters. They started first, so they cashed first.

If you are lucky, they might make an opening lead in hearts or spades, failing to knock out your club stopper. If you can win their heart or spade lead, you can be the one to win the club-diamond race by starting first.

They lead a heart or a spade.
You win, and seize the opportunity to lead diamonds.
They stop you from playing more diamonds (with their ♦A), and belatedly switch to clubs.
You play your club stopper, and cash your diamonds.

The Bridge Bears all clap... quietly, because furry paws don't make much noise.

example 15

Dummy
♠ 5 4
♦ J T 9 8 2

You
♠ A Q 2
♦ Q 3

Spades are led from your left. Good, your ace-queen play last.

If they are able to force out your second spade stopper, they will have 3+ spade skaters to cash.

And if you are able to knock out their diamond stoppers, you will have 3 diamond skaters to cash.

Who cashes first?

You both have 2 stoppers, and they started first. So they cash spades before you can cash diamonds.

example 16

Dummy
♥ A 5
♣ Q T 9 8 4 2

You
♥ K Q 3
♣ K J 3

They lead hearts.

The split assumption is 5-3. If correct, they have enough hearts to establish 2 future skaters. And you have enough clubs to establish 5 club winners.

Who cashes first?

You have 3 heart stoppers, and they have only 1 club stopper. So if you play clubs and they continue hearts, you cash first.

But suppose you don't play clubs right away. Instead, you try to do something with spades or diamonds, and unfortunately you lose the lead. They return to leading hearts, and...

Now who's ahead in the heart-club race?

You have 2 heart stoppers remaining, and they have only one club stopper, so you are still ahead.

But if you want to cash your clubs before they cash hearts, you must lead clubs as soon as they knock out your second heart stopper... while you are still ahead.

A Counting Shortcut

example 17

Dummy
♠ A J 3
♥ 7 6 5
♦ K J T 9 3
♣ 6 2

You
♠ K 8 5 2
♥ K Q J T
♦ Q 2
♣ A K 3

This hand has the same diamonds and clubs as one of our earlier diagrams. This time I'm showing you the other suits as well so I can make an important point about the race to cash tricks.

We've learned that counting stoppers can tell us who gets to cash their tricks first.

But this "cashing first" only applies to the suits we are comparing. It does not mean you will be able to establish and cash all suits for the entire hand.

Let me explain...

Previously, we compared their clubs to our diamonds. We have 2 club stoppers and they have one diamond stopper. We have more stoppers, so we can cash diamonds before they can cash clubs.

But what about hearts? We'd like to drive out their ♥A and have 3 cashable heart tricks to go with our 4 cashable diamonds. But unfortunately we "used up" all our club stoppers to establish and cash our diamonds. So, if they have too many ready to cash club tricks, then we must not lead hearts.

There's a counting shortcut to figure this out:

You cannot lead anything until after you capture the lead. After capturing the lead, count your remaining stoppers in their critical suit – the suit they cannot be allowed to cash. That number of remaining stoppers tells you how many times you can afford to lose the lead.

After they make an opening lead in clubs, you will play one of your club stoppers to capture the lead and stop them from cashing club tricks. Then you have only one more club stopper left. So you can only afford to lose the lead one more time.

With only one club stopper remaining, you cannot drive out two aces. So you will establish diamonds, because playing diamonds gives you more tricks than playing hearts. You will leave hearts untouched.

Making a Plan

How to declare a no trump hand:

The general approach is usually called "count and plan," but I'm going to divide count-and-plan into more detailed steps to help you understand what to do.

Take inventory, including counting stoppers and winners for both sides, doing split assumptions, and adjusting split assumptions based on the bidding and the opening lead.
Assess the threat, especially how many additional winners they can cash when your stoppers are gone. The number of stoppers you have tells you how many times you can safely lose the lead when developing the tricks you need.
Make a plan to win the race. This plan must be based on the information from the first two steps. Don't just play cards and hope for the best.

Let's practice...

example 18

Dummy
♠ K Q T 7
♥ 8 6 3
♦ K T 3 2
♣ A 5

You
♠ J 9
♥ A K J 5
♦ Q J 8 7 4
♣ K 6

Take inventory:

Your contract is 3N (you need 9 tricks).

The opening lead is a club. What is the split assumption for clubs?

They have 9 clubs, so we assume a 5-4 split.

You have 4 winners (♥AK and ♣AK). They have two winners (♠A and ♦A). Neither side has enough, so both sides will work to establish more.

They have good prospects to establish club skaters.
And you have prospects in all three other suits. We'll see in the "Make a plan" section which suits to play.

How many stoppers do you have?

You have 2 club stoppers.

Assess the threat:

Would it be OK if they establish and cash club skaters?

No. Three club skaters plus their 2 aces are enough to set 3N.

example 18 (repeated)

Dummy
♠ K Q T 7
♥ 8 6 3
♦ K T 3 2
♣ A 5

You
♠ J 9
♥ A K J 5
♦ Q J 8 7 4
♣ K 6

Make a plan:

You will win the opening lead with one of your club stoppers. After that, you can only lose the lead one more time because you only have one club stopper left.

You cannot play both diamonds and spades. They have aces in both suits, so you would lose the lead twice and your contract would fail.

You can drive out one of those two aces, but you'll have to leave the other suit untouched. Of course you'll play the suit that gives you more tricks. Is that spades or diamonds?

You can establish 4 winners in diamonds, but only 3 in spades. So play diamonds, bringing your total of winners to 8.

After you force out their ♦A, they will force out your last club stopper. You now have the lead and 8 winners (the four you started with, plus four new diamond winners). You still need another winner, but we've already figured out you cannot play spades.

So where can you get one more winner, without losing the lead?

Take the heart finesse. If it wins, you make your contract.

example 19

Dummy
♠ K Q T 7
♥ 8 6
♦ K T 3 2
♣ A 5 2

You
♠ J 9
♥ A K J 5
♦ Q J 8 7 4
♣ K 6

Take inventory:

Your contract is 3N (you need 9 tricks).

The opening lead is the ♣3. The split assumption for their 8 clubs is 5-3, but... why do you reject this assumption?

The ♣3 is the lowest outstanding spot card, so West has only 4 clubs. The split is 4-4.

Count winners for both sides.

You have 4 winners (♣AK , ♥AK). They have 2 winners (♠A and ♦A).

plush toy bear Little Bear says, "This hand looks like the one we just did. We still can't stop them from cashing club skaters before we have 9 certain tricks, so I guess we still take the heart finesse for our 9th trick."

Me: "I see you are looking ahead at the stoppers. We have 2 club stoppers and they started first, so we cannot force out their two aces before they cash their club skaters. But let's finish our first two steps before we jump ahead to making a plan."

Assess the threat:

Would it be OK if they establish and cash club skaters?

Yes, that's OK. Two club skaters plus their 2 aces are not enough to set 3N. So we don't have to win the race to make 9 tricks before they cash club skaters.

Make a plan:

This time club skaters are no immediate threat, so we can go slowly and drive out both of their aces. We will establish 11 tricks – 3 spades, 2 hearts, 4 diamonds, and 2 clubs.

How many of those 11 tricks will we get to cash?

We will only get to cash 9 tricks, because they will cash 4 tricks first. 4 tricks for the defense does not set our 3N contract.

Why would it be a mistake to take the heart finesse?

If the finesse loses, the ♥Q becomes the setting trick.

Are you a confused Bridge Bear?

If you are, go back and review each example until you can think your way through all the steps while only looking at the cards in the example diagrams.

This may take quite some time, but if you forge ahead without truly understanding the material, your confusion will turn into frustration. And there's no honey for Bears who get lost in the woods.

example 20

Dummy
♠ K Q 7
♥ 8 6 3
♦ Q T 9 8 7 2
♣ A 5

You
♠ J 9
♥ K Q J T
♦ J 6 4
♣ K Q 6

Take inventory:

Your contract is 2N. You need 8 tricks.

The defenders would defeat your contract if they take 6 tricks.

How many winners does each side have?

You have 3 winners – ♣AKQ. You need 5 more to make your contract.

They have 4 winners – ♠A, ♥A, ♦AK. They need 2 more to set your contract.

The opening lead is the ♣4. What is the club split assumption?

8 clubs are assumed to split 5-3.

Assess the threat:

Can you afford to let them establish and cash club skaters?

No. We assume they would get 2 club skaters to add to the 4 winners they start with, which is enough to defeat 2N.

You will use one of your club stoppers on the opening lead. You have 2 more, so you can lose the lead 2 times, and each time you will be able to recapture the lead with a stopper.

Let me repeat that. It's okay to lose the lead 2 times, but not 3 times.

Make a plan:

Now let's look at the suits and see how many tricks you can establish while losing the lead only twice.

We'll start with diamonds. They have 2 diamond stoppers, so you would lose the lead 2 times before you could enjoy diamond skaters.

Would you get enough diamond winners to make the contract?

No, you would get only 4 winners from diamonds, and you need 5.

Maybe you could use your 2 lost-leads elsewhere and get the 5 winners you need. Like this...

How about playing both hearts and spades? They only have one stopper in each suit, and you have enough club stoppers to reclaim the lead twice.

How many tricks would you establish if you drive out their aces in hearts and spades?

3 hearts and 2 spades... just enough to make 2N!

If you choose the right suit(s) to establish (starting at trick 2), you will make your contract.

example 21

Dummy
♠ K 8 3
♥ J 6 4
♦ K J 8 4
♣ K Q 7

You
♠ 7 6
♥ A Q 2
♦ A Q 2
♣ A J T 3 2

Take inventory:

Your contract is 3N. You need 9 tricks.

You have 10 winners – 1 heart, 4 diamonds, and 5 clubs. If they lead any of those three suits, you will win and cash all 10 of your winners, winning the race and making an overtrick.

But the opening lead isn't in one of the suits where you have your winners, it's the ♠5.

Assess the threat:

Is your ♠K a stopper? It's only a stopper if it wins a trick and captures the lead. If it gets squished by their ♠A, how many tricks can they cash?

The split assumption is 5-3, so they will cash 5 spade tricks and set your contract.

It seems like you need luck to have the ♠A on-sides.

If you play your ♠K on the opening lead, you will indeed need the luck of the ♠A being on-sides.
Too bad your ♠K isn't playing last. If it played last, it would be promoted to a winner and be a stopper. On this hand you don't need an extra winner, but you do need a stopper.

Make a plan:

If only RHO were the one to lead, your ♠K would play last and be a stopper. So...

...how can you pass the lead over to RHO and thereby turn your ♠K into a stopper?

Play low from Dummy on the opening lead. RHO can't allow you to win the trick with your ♠6, so he plays third-hand-high to win the trick.

At trick 2, RHO is stuck.

If he leads another spade, your ♠K is promoted, stopping them from cashing the setting tricks. You would be ahead in the race to cash your tricks before they cash theirs.
If he leads something else – anything else – win the trick and cash the rest of your 10 winners, making an overtrick.

We've solved the hand. But let's continue with some additional counting practice:

If RHO leads spades at trick 2, how many tricks would you make?

You would make 11 tricks – 1 spade, 1 heart, 4 diamonds, and 5 clubs. They could only take 2 spade tricks, the one you duck at trick 1, and the ♠A.

At trick 2, suppose RHO switches to leading a heart. If the ♥K is on-sides you could finesse twice, winning the first finesse with Dummy's ♥J, and make 2 extra overtricks. Is this a risk-free finesse?

Risk-free?? Absolutely not. It could lose, and then your main worry, the ♠K, wouldn't be playing last any more. How bad could that be? West could win the heart finesse and switch back to spades. Your ♠K might then get squished, and you would lose not just 5 spade tricks for down one, but 5 spades and the ♥K for down 2.

If you don't finesse when East leads a heart at trick 2, what will you do?

Rise with the ♥A, and cash your 10 winners, making your contract without taking any risks.

When you take inventory, you know both sides have enough possible winners to win the race. So when they give you a chance to win the lead, seize the opportunity, cash your tricks, and make your contract.

It's true that if you are psychic and you know the ♠A and the ♥K are both on-sides, you can win the ♠K on the opening lead, take the winning heart finesse, and make 12 tricks instead of only 10 or 11.

But if you are just a normal Bridge Bear like the rest of us, you should take the safe path of turning your ♠K into a stopper (by playing low from Dummy on the opening lead) to guarantee your contract.

example 22

Dummy
♠ Q J T 9 2
♥ K 7
♦ A Q J 7 2
♣ 8

You
♠ K 4
♥ Q J T 4
♦ K T 9 8
♣ A 9 5

Take inventory:

Your contract is 3N.

A quick glance at our cards makes it seem like you can drive out the two major suit aces and have 4 spade tricks, 3 hearts, 5 diamonds, and 1 club. That's 13 possible tricks total.

Little Bear says, "With 13 possible tricks, taking 9 to make 3N should be easy, right?"

Not necessarily, my fuzzy friend...

...because those pesky opponents start with a club lead, and persist with clubs until they drive out your only club stopper. What is the split assumption for clubs?

They have 9 clubs, so we assume the split is 5-4, though it could be worse.

Assess the threat:

As soon as you lead a major suit, they play their ace to get the lead, and cash their club skaters. How many tricks can they take before you can get the lead back?

If the clubs split 5-4 as expected, they can take 4 clubs and two aces. Six tricks for them. Or perhaps seven, if the clubs split 6-3.

Little Bear looks disappointed, "What happens to my overtricks? I like overtricks."

"Well, Little Bear, you start with only 6 cashable tricks, and you only have one stopper. Your "easy" contract goes set because they cash their tricks first. You have to discard some of your planned winners on their club skaters." (Little Bear pouts...)

Make a plan:

This contract did not have a plan that works. Some contracts cannot be made against best defense. Maybe you'd be lucky and the opening lead would be some other suit, not clubs. Then you could establish more winners before your club stopper is gone.

Now let's change the hand just a little and see what happens.

example 23

Dummy
♠ Q J T 9 2
♥ K 7
♦ A Q J 7 2
♣ 8

You
♠ K 4
♥ Q J T 4
♦ K T 9 8
♣ A K 5

Take inventory:

This is almost the same hand as above... I've given you the ♣K.

Now when the defense attacks clubs, you have a second stopper. You might think this would increase your tricks by one, or maybe two (plus one for you and minus one for them because we took away their ♣K). But that's not what happens.

They drive out your first club stopper, and you drive out the ace that gives you the most tricks.

They drive out your remaining club stopper, and you cash your tricks.

How many tricks will you have (in total)?

You will have 11 total tricks – 4 spades, 0 hearts, 5 diamonds, and 2 clubs. All they get is their two aces. They have to discard all their club skaters on your diamonds and spades. (Little Bear says that's ok with him...)

On the previous hand they established and cashed their tricks first. You had to make discards on their club skaters.
On this hand you cash yours first. They have to make discards on your spades and diamonds.

That's why we say no trump contracts are a race.

The maximum number of tricks you might be able to establish is not as important as being the first to cash the critical number for you to make your contract, or for them to break it.

plush toy bear Go to the next topic:

The Dangerous Opponent

Bridge Bears is run by a retired teacher and ACBL life master who has 35 years teaching experience and who's been playing bridge for over 50 years. I don't claim to be one of the top players, but I do understand how slowly beginners need to go when they are trying to learn how to play bridge.