Suit combination

[fred]

J108

 

K65432

 

You need 5 tricks. Entries are not a problem. Nor are extra undertricks.

 

You lead the Jack from dummy and East follows with a small card.

 

If you chose "sometimes", please explain how to determine when it is appropriate to play the King.

 

By "appropriate" I mean "the percentage play" :)

 

If you feel inclined to ruin the problem for yourself by asking a computer program, please do not ruin the problem for other people by posting what your computer program has to say.

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[ArtK78]

I voted "sometimes."

 

Whenever the suit breaks 2-2, it is a pure guess whether to play the King or duck UNLESS (1) the auction has given you an indication of who holds the ace, or (2) the lie of the rest of the hand requires that you place the ace in one hand or the other. In the latter case, your play is dictated by that restriction.

 

If the suit is 3-1, then, once you lead the J from dummy, the only cases that matters is singleton 9 in LHO's hand. If RHO has A9x or Q9x, you must lose 2 tricks. If you were playing for a 3-1 break, you would not have led the J from dummy unless you knew that the AQ was on your right, as you would then be forced to play for a singleton 9 on your left after RHO follows small.

 

4-0 breaks are irrelevant, as you cannot win 5 tricks in the suit if the suit breaks 4-0.

[matmat]

hmmm... are all 2-2 really equivalent?

 

what's the "proper" defensive play holding Q9 with J coming off of dummy?

[kenrexford]

"A small card" seems to mean a specific card, because the 9 is not contextually small. So...

 

Missing: A/Q/9/x

Seen: x

Possible RHO holdings:

 

x

9x (your play does not matter)

Qx (guess)

Ax (guess)

AQx (riding works best)

A9x (your play does not matter)

Q9x (rising costs all three instead of two)

AQ9x (rising costs three instead of two)

 

On its face, it appears that you must finesse. However, you also clarified that you need five. On the last three, you lose two tricks no matter what. So, the remainder will be:

 

Qx

Ax

AQx

 

Slight lean toward riding. The only caveat seems to be when RHO cannot have four points or when an assumption must be made as to another card(s) that will reduce RHO's max to four points.

 

--------------------------------------------

 

So, what if "a small card" includes the 9?

 

The play of the 9 makes little sense if it accompanies the "x". So, it seems to have to be from one of the following:

 

9

Q9

A9

AQ9

 

With AQ9, a cover would be 100%. So, we eliminate that.

 

With the stiff 9, LHO has AQx. We have to lose two tricks. Needing five, this layout is irrelevant.

 

With A9, RHO will be forced to duck, making the rise better.

 

With Q9, RHO could duck and hope for a rise. Or, he could cover and hope for a finesse.

 

If the Jack is covered with the Queen, Opener losing his King to the Ace will have a "guess" at the next play. He will see a small card from LHO and will have to decide if the Queen was from Q9 or stiff Queen. That seems 50-50 to a degree, except that RHO will duck with Q9 to gain the advantage of the forced duck from A9 tight. So, Opener's odds will be to hook. But, if the odds are to hook, then RHO can also cover and win. So, RHO always wins with Q9. Because RHO always wins, we give up that scenario as a no-win for declarer.

 

This means that if "a small card" can include the 9, we have:

 

A9 (rise)

Qx (duck)

Ax (rise)

AQx (duck)

 

So, I think that you rise if the "small card" is the 9 but duck if it is the true small card. If RHO cannot have the Ace, mathematically or for an assumption, then you duck.

[han]

Now this post I think is funny.

 

It will take quite some effort to study mixed strategies, let me consider deterministic strategies first.

 

Strategy 1. Play low when 7 or 9 is played, play for the drop on the second round when the queen is played. Repeat the finnese when the ace is played and the 9 drops on the left.

 

Wins when RHO has Q7, Q9 or AQ7.

 

Strategy 2. Play the king when the 7 or 9 is played, finesse on the second round when the queen is played, play for the drop when the ace is played and the 9 appears on the left.

 

Wins when RHO has A7 A9 or Q.

 

Strategy 3. Play the king when the 9 is played but low when the 7 is played. When the queen is played finesse and when the ace is played play for the drop.

 

Wins when RHO has A9, Q7 or Q.

 

So these three strategies are all equal, and I don't think any other strategy is better. Maybe a mixed strategy is better but I'll look at that later.

[fred]

I seem to have blown it once again.

 

"Small card" means 9 or 7.

 

There are no inferences to be drawn from either the bidding or play. You know absolutely nothing about the EW hands.

 

You can assume the defenders will always do the right thing.

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[Echognome]

Thinking through the holdings and not the percentages, we are missing A, Q, 9, and 7.

 

Here are the possible holdings for RHO:

 

---

A, Q, 9, 7

AQ, A9, A7, Q9, Q7, 97

AQ9, AQ7, A97, Q97

AQ97

 

We can rule out several possibilities as we need 5 tricks. We can never make with AQ sitting on the left or if there is stiff A or AQ9 or AQ97 on our right. This leaves us with:

 

Q, Q7, A7, Q9, A9, AQ, Q97, A97, AQ7

 

So now we can consider several lines:

 

1) Lead the 8 from dummy, cover the Q with the King, duck the A (obv) and there are two alternatives if you see the 7 or 9. Already we pick up Q and AQ.

a. Play low - to pick up Q97 in addition

b. Play the K - to pick up A7, A9, or A97 in addition

 

So it seems that 1b. is better and picks up Q, A7, A9, AQ, and A97

 

2) Lead the J (or T) from dummy. Here there are various alternatives. If righty plays the A we can play him for AQ7, AQ, A9, or A7. We don't need to worry about A97 because we will see the Q. In the latter 3 cases, we play the K next. In the first case, we will see the 9 and have a choice of finessing the Q. However, this seems inferior as there is only the one case and LHO will also play the 9 from Q9. If RHO plays the 7 or 9, then we can either run the Jack or rise with the King.

a. Run the J - picks up Q7, Q9, or AQ7

b. Rise with the K - picks up A7 or A9

 

So it seems that 2a. is better than 2b. The difference being that we waste an honour and lose out to A97 by leading the J initially. So let's consider 2a. as part of our overall strategy.

 

Now if Righty plays the Q, we could duck hoping for Q97, but covering will pick up more possible holdings (as will be shown). So we cover with the K and it will either lose to the A and then we can either:

c. Finesse the 8 on the way back if we see the 7 - to pick up Q along with Q7 and AQ7

d. Play the J if we see the 7 - to pick up Q9 along with AQ7

 

So it seems that 2c. is better than 2d. so we should add that to our strategy.

 

Finally, what if the K wins? Then we will either see lefty play the 7 or the 9. If lefty play the 9, then T and 8 are equals and it's irrelevant. If lefty plays the 7, then we lead back towards the J8 and lefty will have to either play the 9 or the A (in any case that matters) and we can always get that right. So basically LHO never gains to duck and RHO should never play the Q from AQ7. The fact that we can pick up AQ I guess is also in the mix.

 

Summing up the 2nd line, I'm going to go with play the J, then if RHO plays the 7 or 9, then run the J. If RHO plays the Q, then I'm going to cover. If it wins, there's nothing to think about. If it loses, then I will next lead towards dummy and finesse the 8. Finally, if RHO plays the A, I will play the K next. I believe this line picks up AQ, A9, A7, Q, and Q7.

 

So, if we believe the above (and I'm not sure I've done it all correct), then the difference between the lines is that the optimal line 1 picks up A97 and the optimal line 2 picks up Q7. Q7 should be more likely than A97. So I'll go with optimal line 2.

[lamford]

An interesting problem, and one that reminds me of a similar one which I looked at before. Firstly if second hand plays the seven, I think you should play low all the time, as A7 and Q7 doubleton cancel out, and rising loses to AQ7 with second hand, and on A97 you cannot succeed.

 

If second hand plays the nine, I think you rise all the time. The two relevant holdings are A9 or Q9 doubleton, but with the latter, second hand should cover at least some, but not all, of the time. If the queen appears you are comparing Q9 (6.78%) with Q (6.22%), but as second hand does not cover all of the time with Q9, you finesse on the way back. Basically you succeed against A9, lose to Q9 (whichever he does) and win against Q. In practice it does not matter what he does with Q9; the % success for your best line is the same.

 

If I am right on the above, it seems similar to K97654 opposite Q3 for four tricks, where I think the right line is running the nine, but rising if second hand plays the 8.

[Echognome]

By the way, i see the error of my ways now.

[benlessard]

To better understand this very complicated problems we have to start with the case when the Q drop is possible. When the Q drop its 3 case

 

CASE 1    A97 ------- Q

CASE 2    A7 ---------Q9 defense play the 9

CASE 3    A7 ---------Q9 defense play the Q

 

restricted choice says that finessing for the 9 is best because with Q9 defense has a choice of play. Therefore case 1 is more likely then case 3. If defense never play the Q from Q9 then when the Q drop its always a stiff and declarer will always finesse the 9. If defense always play the Q from Q9 then declarer get no edge because CASE 1 is not more likely then case 3 anymore. However here come the downside

 

CASE 4    Q7 ---------A9

 

If defense always cover with Q9 then when then 9 fall (from A9) declarer will always go up with the K. Therefore a possible defense is a mixed strategy of playing Q from Q9 to protect against case 1 but not too often since the payback is when you have A9.

 

So let see from declarer point of view when the 9 drop now.

 

CASE 2    A7 ---------Q9 defense play the 9

CASE 4    Q7 ---------A9

 

restricted choice says that hoping with the K best because with Q9 defense has a choice of play. Therefore case 4 is more likely then case 2. The problem is that perfect defender will also would have made the same reasonning and since case 4 is slightly more likely then case 1 (A97 ------- Q). They will never play the Q from Q9.

 

Another way to see it is for declarer to duplicate the holding in wich he is failing to disable deception.

 

A7-------------Q9

 

by always finessing the 9 if the Q drop and by always playing the K if the 9 appear the deceptive play of the Q from Q9 is disabled. Playing the Q or not make no difference since declarer will failed in both case. So after this duplication the problems become a simple restrictive choice of always finessing for the 9 when the Q drop.

 

 

here are some strategies to show why going up with the k and always finessing the 9 is best.

 

 

 

remember the important cases are

 

A97 -------- Q

A7 ---------Q9

Q7-----------A9

 

 

 

 

lets say we have 3000 hands

1000 each time 3 case

 

case 1

 

A97 -----------Q

 

 

case 2

A7 ------ Q9

 

 

case 3

Q7 ------A9

 

 

 

strategy number 1 VS defense 1.1

 

declarer always play the k when the 9 appear and always finesse for the 9 (and perfect computer defense know this)

 

defense will never cover with Q9.

 

 

A97----Q=+1000

 

 

 

A7 --- Q9=-1000

 

 

 

Q7 ---A9=+1000

 

so + 1000

 

 

strategy number 1 VS defense 1.2

 

declarer always play the k when the 9 appear and always finesse for the 9 (and perfect computer defense know this)

 

defense will always cover with Q9.

 

 

A97----Q=+1000

 

 

 

A7 --- Q9=-1000

 

 

 

Q7 ---A9=+1000

 

so + 1000

 

 

 

 

strategy number 2 vs defense 2.1

 

declarer play the K 50% & but always try to pick up the 9xx if the Q drop

 

defense will always go up with the Q with Q9

 

so we got

 

A97----Q=+1000

 

 

A7 --- Q9=-1000

 

 

Q7 ---A9= EVEN +500 -500 (since go up 50%)

 

 

 

 

strategy number 2 vs defense 2.2

 

declarer play the K 50% & but always try to pick up the 9xx if the Q drop

 

defense will never go up with the Q with Q9

 

so we got

 

A97----Q=+1000

 

 

A7 --- Q9= EVEN

 

 

Q7 ---A9= EVEN +500 -500 (since go up 50%)

 

 

strategy number 3 VS defense 3.1

 

declarer play the K 50% and never finesse the 9 (defense know will never cover with Q9).

 

A97----Q= -1000

 

 

A7 --- Q9= even since play low 50%

 

 

Q7 ---A9= EVEN since play low 50%

 

 

strategy number 3 vs defense 3.2

 

declarer play the K 50% and never finesse the 9 (defense will always cover with Q9).

 

A97----Q= -1000

 

 

A7 --- Q9= +1000

 

 

Q7 ---A9= EVEN since play low 50%

 

 

 

Strategy number 4 VS defense 4.1

 

declarer always play low and never finesse the 9 (defense will never cover with Q9)

 

 

A97----Q= -1000

 

 

A7 --- Q9= +1000

 

 

Q7 ---A9= -1000

 

 

Strategy number 4 VS defense 4.1

 

declarer always play low and never finesse the 9 (defense will always cover with Q9)

 

 

A97----Q= -1000

 

 

A7 --- Q9= +1000

 

 

Q7 ---A9= -1000

 

 

Strategy number 5 VS defense 5.1

 

declarer play high 50% and finesse the 9 50%

defense will never cover with Q from Q9)

 

A97----Q= even

 

 

A7 --- Q9= even

 

 

Q7 ---A9= even

 

 

Strategy number 5 VS defense 5.2

 

declarer play high 50% and finesse the 9 50%

defense will always cover with Q from Q9)

 

A97----Q= even

 

 

A7 --- Q9= even

 

 

Q7 ---A9= even

[helene_t]

I voted for "sometimes" because I think it is very close and I would in practice do what I "feel" to be right.

 

But assuming that opps always do the right thing, there must be an answer. Since opps have full information it cannot be a mixed strategy. Could it depend on whether it was the 9 or the 7? If RHO started with AQ9 it cannot be wrong for him to cover but if he started with AQ7 it cannot win to cover. If he can assume that we always do the right thing it may not matter for him in either case, but if he is allowed to hope for a misplay by us, the 9 should deny AQ. Also, with Q97 he should not play the 9 unless he knows we will play the K.

 

Good grief, after Fred explained what "appropriate" and "small card" mean, I find myself ranting about the meaning of "the right thing" :)

[Fluffy]

playing the king is never apropiate, you can lose 3 tricks on the suit that way, low to the king is another thing.

[jdeegan]

:) The correct answer is 'sometimes, it depends" because:

 

the relevant cases (I don't see how the nine can be a small card since it is promotable into a trick) are for RHO to hold:

 

a. AQ7 - playing small wins

b. A7 - playing the king wins

c. Q7 - playing small wins

 

With a) RHO might have played the ace, or at least exhibited a 'tell'

With c) RHO might have played the queen or at least exhibited a 'tell'

 

If I can rule out a) and c) because RHO always covers, OR at least gives off a 'tell', then the king is automatic.

 

If RHO is an automaton, then the a priori probabilities apply, and a)+c) is more likely than b. - small would be correct.

Correctly applying probabilities is a lot harder than just memorizing a few tables.

 

Thanks for a nice problem.

[helene_t]

playing the king is never apropiate, you can lose 3 tricks on the suit that way, low to the king is another thing.

Sounds correct. Even if RHO did not have full information, I don't see how the jack could tempt him to cover when it would be wrong.

[whereagles]

Agree.. if you fetch the singleton Q, you're gonna lose 2 tricks anyway. If the K holds and the suit breaks, you seem to have made the anti-percentage play (which seems to be to run the J).

[drinbrasil]

J108

 

K65432

 

You need 5 tricks. Entries are not a problem. Nor are extra undertricks.

 

You lead the Jack from dummy and East follows with a small card.

 

 

making fast response as if i was in one table:

You need 5 tricks, means can loss 1 only, after player in right played x (7 or 9), we only care about:

a ) K or run J is guess

in your left with

Ax

Qx

b ) duck is better in 3 of 5 cases.

but in our right we need care about

AQ7 (look that player cant have AQ9 or we would have played the Q.)

Q9

Q7

A7

A9

 

so i duck...not know if right, but fast thinking at table would be my play:-)

 

ps: will read now all other answers to see if i am big wrong :-)

[Halo]

Sometimes.

 

Run the jack if x = 7, play the King if x = 9.

 

I think Lamford has given a relevant analysis of reasons.

[benlessard]

Im surprised most dont understand the problem at all.

 

1 first of all you need 4 tricks. so all case when you are sure to lose 2 trick are pointless. and extra undertricks are also pointless.

 

2 if RHO played the 7 you have to let run the J because of (9----AQ7)

 

3 playing LOW (the 8) to the king is inferior because of (9----AQ7)

 

 

So the whole point of the problem is

 

A7-------Q9

Q7-------A9

 

THESE ARE THE ONLY CASE WHERE A SMALL CARD (the 9) IS PLAYED THAT MATTERS.

 

i repeat when a 7 is played on 1st round you have to play the J its obvious.

 

 

Now we have a restricted choice problem because from Q9 its possible to play the Q or the 9.

 

http://en.wikipedia.org/wiki/Principle_of_...choice_(bridge)

 

 

Why is it possible for defense to play the Q from Q9 ?

 

 

 

Because of this case

 

CASE (1)   A97 ------- Q.

 

If the Q pops up on round 1 its could be from

 

CASE (1)   A97 ------- Q

 

CASE (3)    A7 ---------Q9 and RHO decided to play the Q

 

Again we have a restricted choice problem. Since from Q9 the defense could play the 9 or the Q. So when the Q appear on round 1 its more likely the initial position was A97 ----Q and its less likely the initial position was A7----Q9. So declarer best line is to finesse the 9 each time the Q come down on round 1. So With Q9 the defense can play the Q knowing declarer (because of restricted choice) will finesse the 9 and lose.

 

So what the problem of always playing Q from Q9 if it will always succeed ?.

 

The problem is that when you play a 9 its now way more likely to be from A9 then from Q9 (because defense will sometimes or always play Q from Q9). So declarer will go up with the king knowing that RHO holding of A9 is more likely then Q9.

 

So if we put the pertinent cases together we have

 

CASE (1)   A97 ------- Q

CASE (2)   A7 ---------Q9 defense play the 9

CASE (3)   A7 ---------Q9 defense play the Q

CASE (4)   Q7 -------- A9

 

All other cases are pointless.

 

 

So declarer strategy is

 

VS a Q on first round

 

putting the K and finessing the 9 on the 2nd round

putting the K and playing the T on 2nd round hoping RHO had Q9.

 

VS a 9

 

Putting the K hoping RHO had A9

Letting the J ride to win if RHO had Q9

 

 

So the 4 non-mixed strategies are

 

Finessing the 9 if a stiff Q appear and putting the K VS the 9 it will win in case 1+4 and will always fail in case 2 &3

 

Finessing the 9 if a stiff Q appear and letting the J ride VS the 9 will win in case 1+2 and will always fail in case 4

 

To play for D (2-2) by playing the T on 2nd round if a stiff Q appear and putting the K VS the 9 will win in case 3+4 and will always fail in case 1 & 2

 

To play for D (2-2) by playing the T on 2nd round if a stiff Q appear and letting the J ride VS the 9 will win in case 2+3 and will always fail in case 1 & 4

 

Because of restricted choice CASE 1 & 4 are more likely then case 2 & 3 therefore Finessing the 9 if a stiff Q appear and putting the K VS the 9 is better

 

The worst play is to not finesse the 9 is the Q drop on round 1 and to let the J ride if the 9 is played on round 1.

 

 

For the defense since Q7 -----A9 is slightly more probable then A97-----Q the mixed defense should not play the Q from Q9. But against less then excellent declarer putting the Q is probably better.

[Edmunte1]

Benlessard is right. Winning cases (RHO hand), nad their percentages:

1)AQ7 - 6.2%

2)A7 -6.8%

3)Q7 -6.8%

4)A9 -6.8%

5)Q9 and RHO decides to put 9 - 3.4%

6)Q9 and RHO decides to put Q - 3.4%

7)AQ -6-8

8)Q -6.2%

 

Cases 6-8-9 are irrelevant for our decision at this time. In the other cases when it's important to take the decision now:

-if 7 appears - small wins in 13%(AQ7+Q7) of cases, K wins in 6.8% (A7) of cases

-if 9 appears - small wins in 3.4% (Q9 and RHO decides to put 9 - 3.4%), K wins in 6.8% of cases.

[Codo]

If the opponents always play the smallest card, ducking is slightly better. Most cases cancle out and you win against one more holding on your right. (AQ7).

 

But good opps won´t be so kind.

Lets assume that they have the perfect count and know that the suit is diveded like it is. In this case, rho will take the queen in a significant numbers of cases from Q9, Q97, but the ace seldom (in reality maybe never) from Ax.

So the prohibility for a good rho to have Q9 is much lower compared to A9, which makes rising a much better choice then it appeard at first sign.

 

 

I cannot get the math correct, but I would use my nose and raise sometimes and let it run at other moments.

[fred]

Hints follows:

 

It is possible to come up with the answer by purely by logical deduction. No math is required. Restricted choice does not come into play. The easiest path that I have found is very concise - less than 10 short lines of text. This is not a trick question - it is just a normal suit combination problem (but an unusually interesting one).

 

The above hint is unlikely to help very many people actually solve the problem, but it might save some people from typing more very long analyses.

 

Most likely I will provide some more hints in a few hours :rolleyes:

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[ralph23]

Im surprised most dont understand the problem at all.

 

1 first of all you need 4 tricks.

Don't you need five (5) tricks per the problem statement?

 

Maybe I didn't understand the problem at all... :rolleyes:

[ArtK78]

I am sure that the 4 tricks statement was just a typo, as was explained in the second half of the sentence "so all case when you are sure to lose 2 trick are pointless."

 

You have to win 5 tricks. So you are not going to take any line that guards against 3 losers, since even 2 losers is too much.

[Apollo81]

Now that I read Fred's hint, I don't think I'm right anymore, oh well =)

 

---

Sometimes.

 

The only holdings that matter are Q7, A7, AQ7, Q9, and A9.

 

If RHO plays the seven then it is right to play low two in two out of three cases clearly making low the percentage play.

 

If RHO plays the 9 (he could not do so from A97 or AQ9 without possibly blowing a guaranteed trick) it appears that we have a 50-50 guess, but do we? Let's consider RHO's decision.

 

If RHO always plays the 9 from Q9 then declarer has a 50-50 guess when he sees the 9 but has a 100% play on the next round when RHO covers with the queen (which must be singleton). If RHO always plays the Q from Q9 then declarer has a 50-50 guess on the next round when RHO covers with the queen, but has a 100% play when he seems the 9 on the first round (which must be from A9). In each case, a deterministic play from RHO will ensure 75% success rate from declarer.

 

If RHO plays either the Q or 9 randomly, then it is right for declarer to play the K in the problem given by restricted choice (the 9 is twice as likely to be forced from A9 as chosen from Q9). Also if the Q is covered then it is right for declarer to finesse for the 9 on the next round (stiff Q twice as likely as Q9). Declarer will succeed 66% of the time in this case.

 

Note that if RHO assumes declarer has used the above reasoning, his choice of Q or 9 from Q9 is irrelevant; declarer will fail in both cases.

 

There is one more factor to consider. Perhaps RHO is unaware of how many cards in the suit are held by each player. In this case he would be hesitant to play the Q for fear of crashing a singleton honor in partner's hand. Let's say RHO plays the Q some percentage of the time P, 0<P<50. It is still right for declarer to play the K, although his chance of success will be somewhere between 50% and 66% depending on the value of P. Additionally, his chance of success when RHO does cover with the Q goes up to somewhere between 66% and 100%, depending on P. As long RHO sometimes plays the Q, the K is the percentage play.

[hotShot]

Interesting use of "appropriate".

 

Since it is impossible to get more informations from bidding and playing the side suits, this problem is very restricted.

 

It is a tiny bit better never to play the king.

This is, because the 5% of the boards where RHO is void in ♠ can be excluded. So the average ♠ length for RHO is a little above 2. So the Queen is a little more often with RHO. RHO holding AQ9 is a little more likely due to this fact, but we would have seen the Q on the J than.

 

Basically it would not matter much, if humans could decide this truly random. Playing the King based on subjective randomness, will lead to worse results most of the time.