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Russ,

Congratulations on your election as ACBL president. I have a couple questions for you.

First is about MP awards for team game. Three years ago ACBL changed the MP award for pair vs team game resulting the top MP awards for pair game 25% higher than team game for same number of tables. As a result there are less interest in team game. Some of my local sectionals do not even have Swiss team any more. This question was raised in BOG meeting a year ago I believe. Does ACBL have any plan to change it?

A deeper problem is how ACBL should award Masterpoint. Experienced players all know that winning pair game needs some luck. Even you are the best player, you don't always control your destiny in pair game. The results also depend on how some worst players did in the game, especially who they played against. So the results from pair game has some inherent randomness. On the other hand, both teams played the same hands in a team game. The team game result is more determined on the skill of the players. Shouldn't a game that result is more determined by skill be awarded more Masterpoint?

At the time pair award was increased over team game, the argument was the total MP awards from pair game is less than team game with same number. This sounds to me like ACBL is just selling Masterpoint by the amount of entry fees.

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Robert,

I think this is related to what Kurt talked about when vacant space applies and when restricted choice applies.

Once you know the distribution of every suit, there is no vacant space any more. In this case the probability is fixed by initial distribution (unless you have more information about where high card is likely to be). The card play will follow restricted choice.

My methodology is only good at dealing with vacant space problem.

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Robert,

This depends on how many cards still left in play. The odds with all 13 cards in each hands would certainly be different than each hands had 4 cards left. When you have information about what other cards each one had will change that as well as we see in this example if you have information for ♠ (and ♣) distribution will change the odds for ♥K finesse as well.

It is not clear to me what you question exactly is. If “Assume you have counted out RHO for three card in that suit” means the opponent behind dummy has 3 cards, your only chance is to play LHO to have Qx and you just need to play from top down. I suspect you meant LHO has 3 cards. In that case you have a total of C(5,2) possibilities, the count for the hand with 3 cards to have Q is C(4,2), so the odd for finesse is 3/5 = 60%.

This is a case like in singularity in Quantum Mechanics. You've already known the number of cards in each hands (there is no vacate space). Once they are fixed the probability could be calculated uniquely and it is final. If you have additinal information from bidding like who is likely to have HCP, you could adjust this odd further.

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David and Kurt

I think both of you have made good points about how to start with some initial probability and adjust them based on cards be played and information that could be derived. However you have a estimate the probability opponents play a certain low spot card (like ♥6 or ♠5).

I'm trying to use a different methodology like the following:

1) Count all possible card combination. With any information about opponent hand, this start with C(26,13) with each possible hand with equal weight. Of cource if you have some information from bidding you could make adjustment to change weight for certain hands.

2) When North made an opening lead of ♠K, those hands in initial counting with South having ♠K will have their weight reduce to 0. This will change the possible hands to C(25,12). In general you need to recalculate the odds after each new card played. I'll call it renormalization (sounds like Quantum Mechanics).

3) You could also include bridge logic in above, say ♠K implies ♠Q so North has both cards that would reduce the number to C(24,11).

4) When South played ♠5 at trick 1, you could remove those hands with ♠5 in North hands. This further reduces the total counts to C(23,11).

5) At trick 2, North played ♥6. This changes counts to C(22,10). Within these hands the possible hands for North to have ♥K is C(21,9). So the finesse odd is C(21,9)/C(22,10) = 10/22 = 45.5%.

6) As I mentioned and you pointed out, we need to do some adjustment based on information from bidding. Let's just look at spade suit and say North could not have 6 spades or more. I need to subtract the hands with Spade 7-1 and 6-2. The total counts for 7-1 (at trick 2) is C(17,5), 6-2 is C(5,1) * C(17,6). The counts for North to have ♥K are C(16,4) and C(5,1) * C(16,5) respectively. Now the finesse odd is {C(21,9)-C(16,4)-C(5,1)*C(16,5)}/{C(22,10)-C(17,5)-C(5,1)* C(17,6)} = 46.7%

7) In principle we could further adjust for possbile club distribution if there is reason to think certain combinations are not possible and go through this renormalization process again.

The beauty of this methodology is I don't need to estimate what the probability opponent plays a certain spot card (♥6 or ♠5) is. I only need to know what cards were played.

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Yes. I agree that if you know opponent's carding and opponents always play according to their carding the odds are going to change. So the discussion we could make so far is in absence of that information.

As I stated, Kurt's statement about the probability ♥K is in North vs South is correct BEFORE North played 2nd trick. So if you ask the question of finesse success rate before North plays 2nd trick, it is correct. However that is not what happened in a real game. In the real game North played ♥6, this placed it in North hand and ♥6 could not be in South hand.

You could argue that North will always play his low spot cards (2 or 6) at trick 2. This is fine but we change it to a theoritical question now. The question is changed to “when North played a spot card (2 or 6), what is the probability for finesse to succeed?”

The practical question on this hand is North played ♥6 at trick 2, what is the probability for finesse to succeed?

I agree even this calculation (♥K is in North vs South) is probably incomplete because we have not put any restriction on ♠ and other suit distribution except North had ♠KQ. As others pointed out North most likely will not have 6 or more ♠. So the odd will still change a little bit taking into account of that.

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Kurt,

Thanks for your response. This is a question worth to discuss. If you ask the question what is the odds North has ♥K and not have ♥K before North played 2nd trick. I agree with you “Given that North has the ♠KQ, there are 1144066 different hands where he has the ♥K, and 1352078 hands where he does not”.

However this 1352078 is all inclusive for South has ♥K. It includes those hands North had void in ♥ (352716). Once North played 2nd trick, does the odd change? If you have watched Poker game, you could see every card played changes odd.

I view this a little similar to quantum mechanics. It has a starting state with certain probability (or a combination of states each had certain probability). Once you take a measurement (played a card), it changes the state and the probalility changes.

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David:

I think there is a question about how to handle low spot cards. Are they really indistinguishable?

For example North played ♥6 at trick2, is it same as ♥2? In your list, you clearly removed the case N had ♥K2 and S had ♥6.

Now the question is why S played a low ♠ spot card should be treated differently? It should at least ruled out that specific card is in the North.

Maybe the problem is really theoritical vs. practical. If the question is South played a low ♠ at trick 1 and North played low ♥ spot at trick 2, it is more thoeritical. In a real game, South played ♠5 at trick 1 and North played ♥6. It is a practical question.

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How did you get North doubleton ♠KQ and played 1♥ count as 23724? In that case N has 10 unknown cards, S has 7 unknown card (6♠ is known). So the combination should be C(17,7) = 97,240.

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If you play ♥Q and North did not play ♥K, the case for N to have singleton K or void in ♥ could not be true. This changes odds a little bit. I have finesse probability as 10/22 = 45.5%

Playing ♥A will win in following cases: 1. S has singleton K. Chance is about 26%. 2. either N/S has Kx and 3 or more ♦. 3. N has 3 trumps with 4 or more ♦.

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John,

You could choose any number of rounds you want but there are a couple factors you want to consider.

First it takes extra time to move from one round to next especially without electronic scoring device. Players needs to compare scores, calculate IMP and report to you. More rounds would require extra time.

Second, less boards per round means the results could be more random because it is easier to have an upset with less boards. Will your players like it?

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Here is a link to youtube video for BOG meeting. Based on what I heard. Baze was cancel for the Fall NABC, but it might be considered if a Senior Swiss could be put into Spring or Summer NABC schedule in the future.

Club Swiss team game typically runs 4 rounds with 6 boards per round. You let players shuffle every rounds, play all boards (relay the boards between 2 tables) and compare scores. Then report scores to you and you manually input to ACBLscore. You also need to input player names manually. This is what most tournament TDs do today.

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This program is to work with the existing ACBLscore that includes other functions like calculating ACBL masterpoints. It is not an independent scoring software.

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I just read the new chart. It looks like it outlawed 1NT for lower limit less than 10 HCP. However it also has a rule exception says any bids requires Average Strength could be made with near Average Strength in 3rd/4th seat. If this applies it is still legal to open 1NT with 8 HCP on 3rd seat if the range is 5 or less.

Ping Hu

My second question is about the new Masterpoint committee. It supposed to be formed now. Any news? What is going to be its priority?

Ping Hu

Congratulations on your election as ACBL president. I have a couple questions for you.

First is about MP awards for team game. Three years ago ACBL changed the MP award for pair vs team game resulting the top MP awards for pair game 25% higher than team game for same number of tables. As a result there are less interest in team game. Some of my local sectionals do not even have Swiss team any more. This question was raised in BOG meeting a year ago I believe. Does ACBL have any plan to change it?

A deeper problem is how ACBL should award Masterpoint. Experienced players all know that winning pair game needs some luck. Even you are the best player, you don't always control your destiny in pair game. The results also depend on how some worst players did in the game, especially who they played against. So the results from pair game has some inherent randomness. On the other hand, both teams played the same hands in a team game. The team game result is more determined on the skill of the players. Shouldn't a game that result is more determined by skill be awarded more Masterpoint?

At the time pair award was increased over team game, the argument was the total MP awards from pair game is less than team game with same number. This sounds to me like ACBL is just selling Masterpoint by the amount of entry fees.

Ping Hu

I think this is related to what Kurt talked about when vacant space applies and when restricted choice applies.

Once you know the distribution of every suit, there is no vacant space any more. In this case the probability is fixed by initial distribution (unless you have more information about where high card is likely to be). The card play will follow restricted choice.

My methodology is only good at dealing with vacant space problem.

Ping Hu

This depends on how many cards still left in play. The odds with all 13 cards in each hands would certainly be different than each hands had 4 cards left. When you have information about what other cards each one had will change that as well as we see in this example if you have information for ♠ (and ♣) distribution will change the odds for ♥K finesse as well.

It is not clear to me what you question exactly is. If “Assume you have counted out RHO for three card in that suit” means the opponent behind dummy has 3 cards, your only chance is to play LHO to have Qx and you just need to play from top down. I suspect you meant LHO has 3 cards. In that case you have a total of C(5,2) possibilities, the count for the hand with 3 cards to have Q is C(4,2), so the odd for finesse is 3/5 = 60%.

This is a case like in singularity in Quantum Mechanics. You've already known the number of cards in each hands (there is no vacate space). Once they are fixed the probability could be calculated uniquely and it is final. If you have additinal information from bidding like who is likely to have HCP, you could adjust this odd further.

Ping Hu

I think both of you have made good points about how to start with some initial probability and adjust them based on cards be played and information that could be derived. However you have a estimate the probability opponents play a certain low spot card (like ♥6 or ♠5).

I'm trying to use a different methodology like the following:

1) Count all possible card combination. With any information about opponent hand, this start with C(26,13) with each possible hand with equal weight. Of cource if you have some information from bidding you could make adjustment to change weight for certain hands.

2) When North made an opening lead of ♠K, those hands in initial counting with South having ♠K will have their weight reduce to 0. This will change the possible hands to C(25,12). In general you need to recalculate the odds after each new card played. I'll call it renormalization (sounds like Quantum Mechanics).

3) You could also include bridge logic in above, say ♠K implies ♠Q so North has both cards that would reduce the number to C(24,11).

4) When South played ♠5 at trick 1, you could remove those hands with ♠5 in North hands. This further reduces the total counts to C(23,11).

5) At trick 2, North played ♥6. This changes counts to C(22,10). Within these hands the possible hands for North to have ♥K is C(21,9). So the finesse odd is C(21,9)/C(22,10) = 10/22 = 45.5%.

6) As I mentioned and you pointed out, we need to do some adjustment based on information from bidding. Let's just look at spade suit and say North could not have 6 spades or more. I need to subtract the hands with Spade 7-1 and 6-2. The total counts for 7-1 (at trick 2) is C(17,5), 6-2 is C(5,1) * C(17,6). The counts for North to have ♥K are C(16,4) and C(5,1) * C(16,5) respectively. Now the finesse odd is

{C(21,9)-C(16,4)-C(5,1)*C(16,5)}/{C(22,10)-C(17,5)-C(5,1)* C(17,6)} = 46.7%

7) In principle we could further adjust for possbile club distribution if there is reason to think certain combinations are not possible and go through this renormalization process again.

The beauty of this methodology is I don't need to estimate what the probability opponent plays a certain spot card (♥6 or ♠5) is. I only need to know what cards were played.

Ping Hu

As I stated, Kurt's statement about the probability ♥K is in North vs South is correct BEFORE North played 2nd trick. So if you ask the question of finesse success rate before North plays 2nd trick, it is correct. However that is not what happened in a real game. In the real game North played ♥6, this placed it in North hand and ♥6 could not be in South hand.

You could argue that North will always play his low spot cards (2 or 6) at trick 2. This is fine but we change it to a theoritical question now. The question is changed to “when North played a spot card (2 or 6), what is the probability for finesse to succeed?”

The practical question on this hand is North played ♥6 at trick 2, what is the probability for finesse to succeed?

I agree even this calculation (♥K is in North vs South) is probably incomplete because we have not put any restriction on ♠ and other suit distribution except North had ♠KQ. As others pointed out North most likely will not have 6 or more ♠. So the odd will still change a little bit taking into account of that.

Ping Hu

Thanks for your response. This is a question worth to discuss. If you ask the question what is the odds North has ♥K and not have ♥K before North played 2nd trick. I agree with you “Given that North has the ♠KQ, there are 1144066 different hands where he has the ♥K, and 1352078 hands where he does not”.

However this 1352078 is all inclusive for South has ♥K. It includes those hands North had void in ♥ (352716). Once North played 2nd trick, does the odd change? If you have watched Poker game, you could see every card played changes odd.

I view this a little similar to quantum mechanics. It has a starting state with certain probability (or a combination of states each had certain probability). Once you take a measurement (played a card), it changes the state and the probalility changes.

Ping Hu

I think there is a question about how to handle low spot cards. Are they really indistinguishable?

For example North played ♥6 at trick2, is it same as ♥2? In your list, you clearly removed the case N had ♥K2 and S had ♥6.

Now the question is why S played a low ♠ spot card should be treated differently? It should at least ruled out that specific card is in the North.

Maybe the problem is really theoritical vs. practical. If the question is South played a low ♠ at trick 1 and North played low ♥ spot at trick 2, it is more thoeritical. In a real game, South played ♠5 at trick 1 and North played ♥6. It is a practical question.

Ping Hu

S has ready played a ♠ at trick 1. So it has 12 unknown. If you take into account of this adjustment, I think you would get the same number as I did.

Ping Hu

Ping Hu

10/22 = 45.5%

Playing ♥A will win in following cases:

1. S has singleton K. Chance is about 26%.

2. either N/S has Kx and 3 or more ♦.

3. N has 3 trumps with 4 or more ♦.

All together playing ♥A could win 62.4%.

Ping Hu

Ping Hu

You could choose any number of rounds you want but there are a couple factors you want to consider.

First it takes extra time to move from one round to next especially without electronic scoring device. Players needs to compare scores, calculate IMP and report to you. More rounds would require extra time.

Second, less boards per round means the results could be more random because it is easier to have an upset with less boards. Will your players like it?

Ping Hu

https://www.youtube.com/watch?v=zyC7GD8jzuY

Ping Hu

Ping Hu

http://web2.acbl.org/documentLibrary/clubs/SwissTeamsGuide2015.pdf

Club Swiss team game typically runs 4 rounds with 6 boards per round. You let players shuffle every rounds, play all boards (relay the boards between 2 tables) and compare scores. Then report scores to you and you manually input to ACBLscore. You also need to input player names manually. This is what most tournament TDs do today.

If you have Bridgemate or BridgePad, you could use them to run Swiss team and save the problems to have players calculating results and manually input everything. For Bridgemate you'll need my program to do this with ACBLscore. I have a separate post. Please look at

http://bridgewinners.com/forums/read/clubs-and-teachers/running-swiss-team-with-bridgemate/

Ping Hu

Ping Hu

Ping Hu

Ping Hu