Card Combination

[mike777]

Best way to play this Combo for 4 tricks?

 

A72

 

QT983

[TimG]

I think we should designate Ax(x) opposite QT(98)x the forum combination.

[sceptic]

Q to ace if it holds

 

then Ace and low to 10

 

if Q is covered Ace then low to 10

 

Bridgemaster is great

[Stephen Tu]

Q to ace if it holds

 

then Ace and low to 10

Your wording is confusing here. Perhaps you mean run the Q unless it is covered. Saying "Q to A" sounds like play the Q and overtake the A even if not covered.

 

And in any case, although running Q first is equivalent for chances of taking 4 tricks, it's better to run the T first since you can pick up all 5 tricks if LHO has stiff K. The other crucial cases, RHO with stiff honor, cancel out, running the Q works if it's the J, but running the T works if it's the K.

 

Suitplay is better than bridgemaster for these kind of problems.

 

I think you have confused this with the bridgemaster combo A7 vs. QT98x, a significantly different combo with only the 7 card fit and only ability to take one hook. Here running the Q first is better.

[foo]

Best way to play this Combo for 4 tricks?

 

A72

+

QT983

There are two lines. The critical holdings that differentiate them are

K+Jxxx

vs

Kxxx+J

Obviously, they have the same apriori odds of occurring. ~3%

 

The big difference is that if you play for K+Jxxx and you are right, you get 5 tricks. OTOH, if you play for that and the situation is actually Kxxx+J, you will only get 3 tricks.

 

Therefore, if you want to maximize your chances of taking 4 tricks, you sacrifice any chance of taking 5.

 

The line that does that starts with Q -> A72, running it unless it is covered.

 

If the Q is covered by 2nd hand but both opponents follow, you are guaranteed 4 tricks. (if the suit is 5-0 you were always only getting 3 tricks.)

 

Whether the Q wins or loses to the K, your next play in this suit is T -> A7; again intending to run it unless it is covered.

 

odds are ~74% that you will take 4 tricks on this line.

[TimG]

Best way to play this Combo for 4 tricks?

 

A72

+

QT983

There are two lines. The critical holdings that differentiate them are

K+Jxxx

vs

Kxxx+J

Obviously, they have the same apriori odds of occurring. ~3%

 

The big difference is that if you play for K+Jxxx and you are right, you get 5 tricks. OTOH, if you play for that and the situation is actually Kxxx+J, you will only get 3 tricks.

 

Therefore, if you want to maximize your chances of taking 4 tricks, you sacrifice any chance of taking 5.

 

The line that does that starts with Q -> A72, running it unless it is covered.

 

If the Q is covered by 2nd hand but both opponents follow, you are guaranteed 4 tricks. (if the suit is 5-0 you were always only getting 3 tricks.)

 

Whether the Q wins or loses to the K, your next play in this suit is T -> A7; again intending to run it unless it is covered.

 

odds are ~74% that you will take 4 tricks on this line.

I think you've missed a few things.

 

Leading the Queen results in only 3 tricks when the opponents hold Jxxx/K; leading the Ten results in only 3 tricks when the opponents hold Kxxx/J.

 

Leading the Ten does pick up 5 tricks when 2nd hand plays the singleton King; leading the Queen against this holding wins 4 tricks.

 

Against all other holdings leading the Queen and leading the Ten produce the same number of tricks. You don't have to sacrifice your only chance of 5 tricks in order to maximize your chances of 4 tricks.

 

I think (and suitplay agrees) that leading the Queen or leading the Ten will result in 4 tricks about 71% of the time. Perhaps you forgot to subtract out the ~3% for the Jxxx/K holding.

[foo]

I think (and suitplay agrees) that leading the Queen or leading the Ten will result in 4 tricks about 71% of the time. Perhaps you forgot to subtract out the ~3% for the Jxxx/K holding.

Let's see.

 

If you start with the T to maintain your chances of possibly taking 5 tricks, you take =at least= 4 tricks in 2/3 of the holdings that matter.

As you noted, there are only 3 holdings that matter.

K+Jxxx, you take 5 tricks.

Jxxx+K, you take 4 tricks.

Kxxx+J, you take 3 tricks.

 

If you start with the Q, you take =exactly= 4 tricks in 2/3 of the holdings that matter.

K+Jxxx, you take 4 tricks.

Jxxx+K, you take 3 tricks.

Kxxx+J, you take 4 tricks.

 

Crunching the numbers, your Suitplay program should give an expected and Max tricks of 3.74 if you follow the line that starts with the T.

IOW, you have a 74% chance of taking =at least= 4 tricks if you start with the T.

[655321]

I think (and suitplay agrees) that leading the Queen or leading the Ten will result in 4 tricks about 71% of the time. 

Right.

 

IOW, you have a 74% chance of taking =at least= 4 tricks if you start with the T.

Wrong.

[Lobowolf]

Am I playing against human beings in a typical stratified game, an A-flight event, or a computer? I'll take my chances with the 2 off of dummy in a real world club game.

[foo]

I think (and suitplay agrees) that leading the Queen or leading the Ten will result in 4 tricks about 71% of the time. 

Right.

 

IOW, you have a 74% chance of taking =at least= 4 tricks if you start with the T.

Wrong.

Neither statement is wrong.

 

If you start with the T, you will take 4 tricks ~71% of the time and 5 tricks ~3% of the time.

 

Therefore, you will take at least 4 tricks ~74% of the time if you start with the T.

 

Starting with the Q nets you exactly 4 tricks ~71% of the time because it only take 4 tricks instead of 5 against the ~3% holding that the other line can play for 5 tricks.

 

Therefore, starting with the Q gets you exactly 4 tricks ~71% of the time.

 

This means that the line that start with the T is the one to play for max tricks and that max tricks is ~3.74

 

I'm sure Suit Play or other such tools say exactly this.

[655321]

I think (and suitplay agrees) that leading the Queen or leading the Ten will result in 4 tricks about 71% of the time. 

Right.

 

IOW, you have a 74% chance of taking =at least= 4 tricks if you start with the T.

Wrong.

Neither statement is wrong.

 

If you start with the T, you will take 4 tricks ~71% of the time and 5 tricks ~3% of the time.

 

Therefore, you will take at least 4 tricks ~74% of the time if you start with the T.

 

Starting with the Q nets you exactly 4 tricks ~71% of the time because it only take 4 tricks instead of 5 against the ~3% holding that the other line can play for 5 tricks.

 

Therefore, starting with the Q gets you exactly 4 tricks ~71% of the time.

 

This means that the line that start with the T is the one to play for max tricks and that max tricks is ~3.74

 

I'm sure Suit Play or other such tools say exactly this.

Your statement is wrong, and you keep repeating it. To state the same thing another way, you will lose 2 tricks around 29% of the time, whether you run the ten or run the queen. You don't get to double count the case where there are 5 tricks.

 

Please make an effort to understand the maths this time instead of just reposting your original error.

[Free]

If you start with the T, you will take 4 tricks ~71% of the time and 5 tricks ~3% of the time.

 

Therefore, you will take at least 4 tricks ~74% of the time if you start with the T.

This is a pure violation of math! 655321 is right.

[foo]

So I went and found Suitplay. Then I added the probabilities it gave for each defensive holding.

 

25.9565% of the time you will lose 2 tricks regardless of whether you start with the T or the Q.

 

2.8261% of the time you will win 5 tricks if you start with the T. Suitplay calls this "line A"

 

71.2174% of the time you will win 4 tricks regardless of whether you start with the T or the Q.

 

Quick check: 25.9565 + 2.8261 + 71.2174= 100.

 

Suitplay gives max tricks as 3.7404. By following "line A"- start with the T.

 

IOW, regardless of whether you start with the T or the Q, you will win exactly 4 tricks ~71% of the time.

If you start with the T you will win at least 4 tricks ~74% of the time.

 

There are only 3 holdings where your choice of line matters.

Each has a probability of 2.8261%

Jxxx+K line A takes 4 tricks. Line B takes 3.

K+Jxxx line A takes 5 tricks. Line B takes 4.

Kxxx+J line A takes 3 tricks. Line B takes 4.

Another quick check: 65.5652 + 5.6522 + 2.8261= 74.0435

 

Now, gentlemen. Where exactly is this mistake you keep referring to?

[TimG]

I don't know exactly what you are doing wrong, other than it appears you are double counting one of the 4-1 holdings.

 

Line A (lead the Ten) wins 4 tricks against:

 

J - Kxxx 2.83%

Jx - Kxx 10.17%

Jxx - Kx 10.17%

Jxxx - K 2.83%

Kx - Jxx 10.17%

Kxx - Jx 10.17%

KJ - xxx 3.39%

KJx - xx 10.17%

KJxx - x 8.48%

 

and 5 against:

 

K - Jxxx 2.83%

 

Add those up and you get 71.22%

 

Line B (lead the Queen) wins 4 tricks against:

 

J - Kxxx 2.83%

Jx - Kxx 10.17%

Jxx - Kx 10.17%

K - Jxxx 2.83%

Kx - Jxx 10.17%

Kxx - Jx 10.17%

Kxxx - J 2.83%

KJ - xxx 3.39%

KJx - xx 10.17%

KJxx - x 8.48%

 

Add those up and you also get 71.22%

 

Yes, I just copied and pasted from SuitPlay.

[Stephen Tu]

25.9565% of the time you will lose 2 tricks regardless of whether you start with the T or the Q.

 

2.8261% of the time you will win 5 tricks if you start with the T.  Suitplay calls this "line A"

 

71.2174% of the time you will win 4 tricks regardless of whether you start with the T or the Q.

 

Quick check: 25.9565 + 2.8261 + 71.2174= 100.

 

Now, gentlemen.  Where exactly is this mistake you keep referring to?

 

You are misreading the suitplay output. You are double counting the stiff K onside then canceling that out in the total by forgetting about the stiff K or J offside.

 

The figures are:

run T first:

25.9565% - lose 2 tricks w/ the same holdings running the Q loses to.

2.8261% - lose 2 tricks to the stiff J offside, which running the Q wins with.

2.8261% - 5 tricks

68.3913% - exactly 4 tricks (= 68.3913+2.8261 = 71.2174% at least 4 tricks)

 

run Q first:

25.9565% - lose 2 tricks w/ the same holdings running the T loses to.

2.8261% - lose 2 tricks to the stiff K offside, which running the T wins with.

71.274% - exactly 4 tricks (same holdings T first takes 4 tricks with, but subtract the stiff K offside, add the stiff J offside, then add the stiff K onside which the T first wins 5 tricks with).

[Free]

So I went and found Suitplay.  Then I added the probabilities it gave for each defensive holding.

 

25.9565% of the time you will lose 2 tricks regardless of whether you start with the T or the Q.

 

2.8261% of the time you will win 5 tricks if you start with the T.  Suitplay calls this "line A"

 

71.2174% of the time you will win 4 tricks regardless of whether you start with the T or the Q.

 

Quick check: 25.9565 + 2.8261 + 71.2174= 100.

 

Suitplay gives max tricks as 3.7404.  By following "line A"- start with the T.

 

IOW, regardless of whether you start with the T or the Q, you will win exactly 4 tricks ~71% of the time.

If you start with the T you will win at least 4 tricks ~74% of the time.

 

There are only 3 holdings where your choice of line matters. 

Each has a probability of 2.8261%

Jxxx+K  line A takes 4 tricks.  Line B takes 3.

K+Jxxx  line A takes 5 tricks.  Line B takes 4.

Kxxx+J  line A takes 3 tricks.  Line B takes 4.

Another quick check:  65.5652 + 5.6522 + 2.8261= 74.0435

 

Now, gentlemen.  Where exactly is this mistake you keep referring to?

The percentages on the LEFT side of the screen is "Goal = 4" means at least 4 tricks.

 

Look at suitplay again, standard suit combination analysis, at the RIGHT side of the screen.

 

For line A you get 3 tricks for holdings:

void - KJxxx (1.9565%)

x - KJxx (8.4783%)

xx - KJx (10.1739%)

xxx - KJ (3.3913%)

Kxxx - J (2.8261%)

KJxxx - void (1.9565%)

 

Sum of these percentages = 28.7826%

 

Now tell me, how can you win 4 tricks 74% of the time if you can only make 3 in 28.7826%? (Btw, doing the same for the cases where line A makes 4 or 5 tricks, you'll get 71.2174%)

 

Line B will get you the same percentage for 4+ tricks, but in case K - Jxxx you'll win 1 more trick with line A, so the best line is A. However if you follow line B in that case, your goal of 4 tricks is still achieved.

[mike777]

Am I playing against human beings in a typical stratified game, an A-flight event, or a computer?  I'll take my chances with the 2 off of dummy in a real world club game.

Thank you all for your input.

 

 

btw at the table starting the 2 off the table would have won. A play I did not find. (: