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Since Nafiz Zorlu became the head of the Turkish Bridge Federation (TBF) about eight months ago, he and his team have been doing an excellent job. It has really been a period of “transition in bridge” so far. Although the TBF doesn't seem to need any extra incentive to work hard and to do good, probably and hopefully, success will bring more success in every bridgerelated area.
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Although Steve hasn't voted, it should be OK to count him in for either LINE1 or 2. However, that does not change the fact that the majority's choice is LINE3. As you stated somewhere above, perhaps one reason is that a double squeeze is “cooler”.
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We both are using the same calculator, but with different inputs.
You are entering as “missing cards”: ccDDK, with 98 vacant places all the time. (cc refers to any two ♣s, DD to ♦QJ, and K to ♠K, in your notation.)
I am entering as “missing cards”: KsssssssQJddddddd, with different vacant places depending on a specified distribution of ♣s. (Ksssssss refers to all missing ♠s, QJddddddd to ♦s.)
Here is an excerpt from the “Instructions” in the website of our calculator:
“… You can check combinations in more than one suits. For example you can see the odds of the actual combined break of four clubs to the Queen and five diamonds to the Jack by entering as missing cards the following: QcccJdddd. … … The above capability is very important because, in a bridge hand, the lie of the cards in one suit definitely affects the break odds in other suits (remember vacant spaces?). For example the odds for both hearts and spades to break 32 are not 67.826 x 67.826 = 46.004 but 46.746. Big deal eh? But in some cases these differences are either bigger or make the difference.”
I think: Firstly, you should enter all missing cards in ♠s and ♦s.
Secondly, the number of vacant places should be adjusted according to the suits whose EXACT distribution is known. For instance, in the case at hand, it is not correct to reduce the number of empty places from 1211 to 1110 just because we cash a ♠ and it's followed. West, who supposedly followed the 1st round of ♠s, had already discarded a ♠ on the 2nd ♥. What to do about it? Or, what if W discarded a ♦ instead of a ♠ on the 2nd ♥? These should not matter at all, not affect the the number of vacant places. Deal all the missing ♠s and ♦s and then accurately spot the layouts yielding success. (See the example 6♥ hand given in “Instructions”.)
Thirdly, to calculate the correct success chance of LINE2 in the poll, we need to consider the change of plan when West shows out on the 2nd ♣. In that case, it's better to cash the third ♣ and go for the ending in LINE3. To get a meaningful result for LINE2, we should take into account all ♣ splits (42, 24, 51, 15, so on) as parts of the same whole in advance (in such a way that they all should add up to 100%). Please note the significant variation in the number of vacant places from one split to another.
Finally, if we assume that ♣s are not worse than 42 or 24, of course the probabilities will change. Yet then, we can get only a comparative result, not the correct overall success chance. (By the way, for 42 and 24 ♣ splits, the difference is not 4.7 but 4.4 in my results.)
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Benoit: We know that the probabilities of 42 and 24 ♣ distributions are DIFFERENT, and that's in our pocket.
Now, the problem is how we can distribute the other suits without any information on how ♣s are dealt. If ♣s are 42, there is a scenario (as you agree above, 27% scenario); if they're 24, there is a different scenario (21.6% scenario).
In conclusion, for the other suits to come on the scene accurately, we need to specify in advance which scenario is in effect.
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I certainly agree with the following:
“You know spades are 62 because of a preempt. You have 9 hearts and can finesse for the Q or play H 22. Before playing any hearts you know the odds are Qx—xx = 18%, x—–Qxx = 28% (ratio is 39%–61%)”
Since we think we know the exact distribution of ♠s, this is true.
Yet, the rest of what you say seems irrelevant to the subject here. It's a different and complicated issue how probabilities change during the play. What we are concerned here is how to calculate probabilities in advance (regarding vacant places). And it is certain that we should take into account the suits whose exact distribution is known.
By the way, you have not answered my question. Since I think it is crucial, I need to copy it:
“Knowing that W has one ♥ and E has two ♥s, W has 12 and E has 11 vacant places. So, the probability that W will have four ♣s = 27.0%, E will have four ♣s = 21.6%.”
Have you got the same? I mean, in your calculations, what is the probability that West will have 4 ♣s? Is it different from the probability that East will have 4 ♣s?"
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Benoit: While considering vacant places, we should only take into account the suits the distributions of which are EXACTLY known.
Then, before distributing ♣s, we can say W has 12 and E has 11 empty places (since the exact distribution of ♥s is known). So, first we should have the following:
“Knowing that W has one ♥ and E has two ♥s, W has 12 and E has 11 vacant places. So, the probability that W will have four ♣s = 27.0%, E will have four ♣s = 21.6%.”
Have you got the same? I mean, in your calculations, what is the probability that West will have 4 ♣s? Is it different from the probability that East will have 4 ♣s?
Then, without giving the exact distribution of ♣s, if we assign 98 empty places, we will not be able to get the correct results for other suits. The number of empty places varies because, in each case, we have to assign a different (and EXACT) number of ♣s to opponents' hands…
I guess that is where the main difference in our calculations occurs, though the outcome is not affected conclusively.
NOTE: I couldn't get what you meant by “4.7% instead of 4%” in your comment.
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Thank you all for your comments and votes.
Initially let's ignore the slight difference between LINE1 and LINE2 and compare LINE1 and LINE3 (disregarding 60 or 06 ♣ split).
LINE1 wins when; (1) the same opponent holds 4 or 5 ♣s and ♠K, (2) East has 4 or 5 ♣s and ♦QJ, and West has ♠K.
LINE3 wins when; (1) E holds 4 or 5 ♣s and W has ♠K, (2) E holds 4 or 5 ♣ and ♠K and ♦QJ, (3) W holds 4 or 5 ♣s, and E has ♠K and ♦QJ, (4) W holds 4 or 5 ♣s and ♠K and ♦QJ.
Firstly, assume that ♣s ARE 42 or 24.
Knowing that W has one ♥ and E has two ♥s, W has 12 and E has 11 vacant places. So, the probability that W will have four ♣s = 27.0%, E will have four ♣s = 21.6%.
For LINE1:
When W has 4 ♣s; W has 8 vacant places, and E has 9. The probability that W will have ♠K is 47.1%. Then, 47.1 x 0.27 = 12.7. When E has 4 ♣s; E has 6 vacant places and W has 10. The probability that E will have ♠K is 41.2%. Then, 41.2 x 0.216 = 8.9. When E has 4 ♣s; the probability that E will have ♦QJ and W will have ♠K is 10.3%. Then, 10.3 x 0.216 = 2.2.
The total for LINE1 = 12.7 + 8.9 + 2.2 = 23.8%.
For LINE3:
When E has 4 ♣s; the probability that W will have ♠K is 58.8%, and the probability that E will have ♠K and ♦QJ is 5.2%. The total is 64.0%. Then, 64.0 x 0.216 = 13.8.
When W has 4 ♣s; the probability that E will have ♠K and ♦QJ is 12.4%, and the probability that W will have ♠K and ♦QJ is 8.2%. The total is 20.6. Then, 20.6 x 0.27 = 5.6%.
The total for LINE3 = 13.8 + 5.6 = 19.4%.
For 51 and 15 ♣s, a similar approach results in, 5.9% for LINE1, 5.7% for LINE3.
If we add 33 ♣ distribution (36%) to these, we obtain the TOTAL success probabilities: LINE1 = 23.8 + 5.9 + 36.0 = 65.7%, LINE3 = 19.4 + 5.7 + 36.0 = 61.1%.
I know this has already been too long, but I have to add the following.
NOTES: (1) 33 ♣ split is 36.0% here because W has 12 and E has 11 vacant places. It is 35.5% only when each has 13. (2) LINE2 adds 1.3% to the success chance of LINE1; because, if W shows out on the 2nd ♣, it is better to cash the third ♣ and go for double squeeze (to reach the ending in LINE3). (3) In calculations, stiff ♠K and tight ♦QJ in the wrong hands (meaning where we don't want them to be while calculating but they yield success) was not taken into account, which is not expected to affect the results significantly. (4) I used the “calculator” in Theodore T. Triandophyllopoulos' site. (5) Well, I have just noticed that it's not over yet. I need to present the original deal, but first I've got to go through what I've written above.
EDIT: Mistakenlywritten “LINE2”s was replaced with LINE3.
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Well, since you made 7♥ at the table, it is not easy to argue with success, though Line2 offers the best chance. And the result of the poll suggests that more players have a feel for that play.
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I hardly followed your calculation, but the results seem compatible with mine, with a few percent of difference, probably because you ignored long ♣s, ♠K and ♦QJ in the same hand. Note that this is automatically handled in Line2 (and 1) since it basically succeeds when long ♣s and ♠K are in the same hand (no need for ♦QJ). However, it should be separately considered in Line3 because it mainly wins when long ♣s are in East but ♠K in West.
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You are right. Line2 is around 1% superior to Line1. I've done the math; yet, it's a bit messy and I need time to put it into order and write down here (probably this evening).
NOTE: If East has long ♣s and ♦QJ, all lines (with endings I've given below) will succeed.
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If East shows up with five ♣s, adding two ♥s to that, 6 vacant places remain in E, but 11 in W. It is now much more likely that West has ♠K; so, cashing a third ♣ (Line 3) may be a good idea.
If West has five ♣s (7 vacant places in W and 10 in E), there are two choices: (1) West has ♠K. (2) East has ♠K and ♦QJ.
If (1) is more likely, do not cash third ♣, go for ♣♠ squeeze. If (2) is more likely, cash third ♣, go for ♠♦ squeeze on East.
If both opponents follow two rounds of ♣s, now what? To get a quantitative result, similar probabilities need to be estimated when E or W has four ♣s (taking into account the numbers of vacant places in each case).
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Yes, the slight difference between Line 1 and 2 is that Line2 offers a change of plan if East shows up with 5 ♣s. And, in this case, one may consider cashing the third ♣ and go for double squeeze in ♦, to reach the ending in Line 3.
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Ruffing a minor does not look right to me. As Benoit Lessard diagrammed above, in the 5 card ending, say, we have the following.
QT x x x
  AKT Ax
If East has both minors guarded (and ♠K with West), we should not ruff anything but play the last ♥. However, in that case, we will lose some bigger chances. So, it seems better to decide earlier in which minor suit to keep an entry to the hand, and play accordingly.
Okan Zabunoglu
Okan Zabunoglu
Okan Zabunoglu
Okan Zabunoglu
At least, we both agree on the line offering the best chance.
Thanks for your comments.
See you on another deal.
Okan Zabunoglu
Okan Zabunoglu
You are entering as “missing cards”: ccDDK, with 98 vacant places all the time. (cc refers to any two ♣s, DD to ♦QJ, and K to ♠K, in your notation.)
I am entering as “missing cards”: KsssssssQJddddddd, with different vacant places depending on a specified distribution of ♣s. (Ksssssss refers to all missing ♠s, QJddddddd to ♦s.)
Here is an excerpt from the “Instructions” in the website of our calculator:
“… You can check combinations in more than one suits. For example you can see the odds of the actual combined break of four clubs to the Queen and five diamonds to the Jack by entering as missing cards the following: QcccJdddd. … …
The above capability is very important because, in a bridge hand, the lie of the cards in one suit definitely affects the break odds in other suits (remember vacant spaces?). For example the odds for both hearts and spades to break 32 are not 67.826 x 67.826 = 46.004 but 46.746. Big deal eh? But in some cases these differences are either bigger or make the difference.”
I think:
Firstly, you should enter all missing cards in ♠s and ♦s.
Secondly, the number of vacant places should be adjusted according to the suits whose EXACT distribution is known. For instance, in the case at hand, it is not correct to reduce the number of empty places from 1211 to 1110 just because we cash a ♠ and it's followed.
West, who supposedly followed the 1st round of ♠s, had already discarded a ♠ on the 2nd ♥. What to do about it? Or, what if W discarded a ♦ instead of a ♠ on the 2nd ♥? These should not matter at all, not affect the the number of vacant places. Deal all the missing ♠s and ♦s and then accurately spot the layouts yielding success. (See the example 6♥ hand given in “Instructions”.)
Thirdly, to calculate the correct success chance of LINE2 in the poll, we need to consider the change of plan when West shows out on the 2nd ♣. In that case, it's better to cash the third ♣ and go for the ending in LINE3.
To get a meaningful result for LINE2, we should take into account all ♣ splits (42, 24, 51, 15, so on) as parts of the same whole in advance (in such a way that they all should add up to 100%). Please note the significant variation in the number of vacant places from one split to another.
Finally, if we assume that ♣s are not worse than 42 or 24, of course the probabilities will change. Yet then, we can get only a comparative result, not the correct overall success chance. (By the way, for 42 and 24 ♣ splits, the difference is not 4.7 but 4.4 in my results.)
Okan Zabunoglu
We know that the probabilities of 42 and 24 ♣ distributions are DIFFERENT, and that's in our pocket.
Now, the problem is how we can distribute the other suits without any information on how ♣s are dealt. If ♣s are 42, there is a scenario (as you agree above, 27% scenario); if they're 24, there is a different scenario (21.6% scenario).
In conclusion, for the other suits to come on the scene accurately, we need to specify in advance which scenario is in effect.
Okan Zabunoglu
“You know spades are 62 because of a preempt.
You have 9 hearts and can finesse for the Q or play H 22. Before playing any hearts you know the odds are
Qx—xx = 18%, x—–Qxx = 28% (ratio is 39%–61%)”
Since we think we know the exact distribution of ♠s, this is true.
Yet, the rest of what you say seems irrelevant to the subject here. It's a different and complicated issue how probabilities change during the play. What we are concerned here is how to calculate probabilities in advance (regarding vacant places). And it is certain that we should take into account the suits whose exact distribution is known.
By the way, you have not answered my question. Since I think it is crucial, I need to copy it:
“Knowing that W has one ♥ and E has two ♥s, W has 12 and E has 11 vacant places. So, the probability that
W will have four ♣s = 27.0%,
E will have four ♣s = 21.6%.”
Have you got the same? I mean, in your calculations, what is the probability that West will have 4 ♣s? Is it different from the probability that East will have 4 ♣s?"
Would you just answer this question?
Okan Zabunoglu
Then, before distributing ♣s, we can say W has 12 and E has 11 empty places (since the exact distribution of ♥s is known). So, first we should have the following:
“Knowing that W has one ♥ and E has two ♥s, W has 12 and E has 11 vacant places. So, the probability that
W will have four ♣s = 27.0%,
E will have four ♣s = 21.6%.”
Have you got the same? I mean, in your calculations, what is the probability that West will have 4 ♣s? Is it different from the probability that East will have 4 ♣s?
Then, without giving the exact distribution of ♣s, if we assign 98 empty places, we will not be able to get the correct results for other suits. The number of empty places varies because, in each case, we have to assign a different (and EXACT) number of ♣s to opponents' hands…
I guess that is where the main difference in our calculations occurs, though the outcome is not affected conclusively.
NOTE: I couldn't get what you meant by “4.7% instead of 4%” in your comment.
EDIT: Some grammatical corrections.
Okan Zabunoglu
W: ♠K7642 ♥J ♦QJ954 ♣63
N: ♠QT9 ♥KQ9852 ♦7 ♣875
E: ♠J85 ♥T7 ♦8632 ♣JT42
S: ♠A3 ♥A643 ♦AKT ♣AKQ9
LINE3 wins.
Note that, on ♦Q or ♦J lead, even Deep Finesse cannot make 7♥.
Okan Zabunoglu
Initially let's ignore the slight difference between LINE1 and LINE2 and compare LINE1 and LINE3 (disregarding 60 or 06 ♣ split).
LINE1 wins when;
(1) the same opponent holds 4 or 5 ♣s and ♠K,
(2) East has 4 or 5 ♣s and ♦QJ, and West has ♠K.
LINE3 wins when;
(1) E holds 4 or 5 ♣s and W has ♠K,
(2) E holds 4 or 5 ♣ and ♠K and ♦QJ,
(3) W holds 4 or 5 ♣s, and E has ♠K and ♦QJ,
(4) W holds 4 or 5 ♣s and ♠K and ♦QJ.
Firstly, assume that ♣s ARE 42 or 24.
Knowing that W has one ♥ and E has two ♥s, W has 12 and E has 11 vacant places. So, the probability that
W will have four ♣s = 27.0%,
E will have four ♣s = 21.6%.
For LINE1:
When W has 4 ♣s; W has 8 vacant places, and E has 9. The probability that W will have ♠K is 47.1%. Then, 47.1 x 0.27 = 12.7.
When E has 4 ♣s; E has 6 vacant places and W has 10. The probability that E will have ♠K is 41.2%. Then, 41.2 x 0.216 = 8.9.
When E has 4 ♣s; the probability that E will have ♦QJ and W will have ♠K is 10.3%. Then, 10.3 x 0.216 = 2.2.
The total for LINE1 = 12.7 + 8.9 + 2.2 = 23.8%.
For LINE3:
When E has 4 ♣s; the probability that W will have ♠K is 58.8%, and the probability that E will have ♠K and ♦QJ is 5.2%. The total is 64.0%. Then, 64.0 x 0.216 = 13.8.
When W has 4 ♣s; the probability that E will have ♠K and ♦QJ is 12.4%, and the probability that W will have ♠K and ♦QJ is 8.2%. The total is 20.6. Then, 20.6 x 0.27 = 5.6%.
The total for LINE3 = 13.8 + 5.6 = 19.4%.
For 51 and 15 ♣s, a similar approach results in,
5.9% for LINE1,
5.7% for LINE3.
If we add 33 ♣ distribution (36%) to these, we obtain the TOTAL success probabilities:
LINE1 = 23.8 + 5.9 + 36.0 = 65.7%,
LINE3 = 19.4 + 5.7 + 36.0 = 61.1%.
I know this has already been too long, but I have to add the following.
NOTES:
(1) 33 ♣ split is 36.0% here because W has 12 and E has 11 vacant places. It is 35.5% only when each has 13.
(2) LINE2 adds 1.3% to the success chance of LINE1; because, if W shows out on the 2nd ♣, it is better to cash the third ♣ and go for double squeeze (to reach the ending in LINE3).
(3) In calculations, stiff ♠K and tight ♦QJ in the wrong hands (meaning where we don't want them to be while calculating but they yield success) was not taken into account, which is not expected to affect the results significantly.
(4) I used the “calculator” in Theodore T. Triandophyllopoulos' site.
(5) Well, I have just noticed that it's not over yet. I need to present the original deal, but first I've got to go through what I've written above.
EDIT: Mistakenlywritten “LINE2”s was replaced with LINE3.
Okan Zabunoglu
Okan Zabunoglu
Okan Zabunoglu
Note that this is automatically handled in Line2 (and 1) since it basically succeeds when long ♣s and ♠K are in the same hand (no need for ♦QJ). However, it should be separately considered in Line3 because it mainly wins when long ♣s are in East but ♠K in West.
Okan Zabunoglu
NOTE: If East has long ♣s and ♦QJ, all lines (with endings I've given below) will succeed.
Okan Zabunoglu
If West has five ♣s (7 vacant places in W and 10 in E), there are two choices:
(1) West has ♠K.
(2) East has ♠K and ♦QJ.
If (1) is more likely, do not cash third ♣, go for ♣♠ squeeze.
If (2) is more likely, cash third ♣, go for ♠♦ squeeze on East.
If both opponents follow two rounds of ♣s, now what?
To get a quantitative result, similar probabilities need to be estimated when E or W has four ♣s (taking into account the numbers of vacant places in each case).
Okan Zabunoglu
Okan Zabunoglu
(1) Dummy: ♠Q ♥x ♦ ♣xxx
Hand: ♠ ♥ ♦T ♣AKQx
(2) Dummy: ♠Q ♥x ♦ ♣x
Hand: ♠ ♥ ♦T ♣Ax
(3) Dummy: ♠QT ♥x ♦x ♣
Hand: ♠ ♥ ♦AKT ♣x
Okan Zabunoglu
NOTE: Trumps are 12, since we have ten of them.
Okan Zabunoglu
the following.
QT
x
x
x


AKT
Ax
If East has both minors guarded (and ♠K with West), we should not ruff anything but play the last ♥. However, in that case, we will lose some bigger chances. So, it seems better to decide earlier in which minor suit to keep an entry to the hand, and play accordingly.