[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]

Quote

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]

mike777, on Dec 17 2008, 08:29 AM, said:

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]

foo, on Dec 19 2008, 09:19 PM, said:

mike777, on Dec 17 2008, 08:29 AM, said:

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]

TimG, on Dec 19 2008, 10:44 PM, said:

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]

TimG, on Dec 19 2008, 10:44 PM, said:

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

Right.

foo, on Dec 19 2008, 11:27 PM, said:

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]

655321, on Dec 20 2008, 01:32 AM, said:

TimG, on Dec 19 2008, 10:44 PM, said:

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

Right.

foo, on Dec 19 2008, 11:27 PM, said:

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]

foo, on Dec 20 2008, 03:53 AM, said:

655321, on Dec 20 2008, 01:32 AM, said:

TimG, on Dec 19 2008, 10:44 PM, said:

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

Right.

foo, on Dec 19 2008, 11:27 PM, said:

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]

foo, on Dec 20 2008, 09:53 AM, said:

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]

foo, on Dec 20 2008, 06:54 AM, said:

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]

foo, on Dec 20 2008, 02:54 PM, said:

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]

Lobowolf, on Dec 20 2008, 01:47 AM, said:

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. (: