All comments by Ping Hu
You are ignoring the author of this comment. Click to temporarily show the comment.
David, I could understand your feeling and I think we are getting to the bottom it on if the first round of play changes the probability or not. Since there are only 15 limited combinations I'll list them here:
1) W: J654, E: 32
2) W: J653, E: 42
3) W: J652, E: 43
4) W: J643, E: 52
5) W: J642, E: 53
6) W: J632, E: 54
7) W: J543, E: 62
8) W: J542, E: 63
9) W: J532, E: 64
10) W: J432, E: 65
11) W: 6543, E: J2
12) W: 6542, E: J3
13) W: 6532, E: J4
14) W: 6432, E: J5
15) W: 5432, E: J6

What does the first round of heart play do? If you see West plays 2 and East plays 3. You know the original distribution must be from
3), 5), 8) and 12). Other possibilities are ELIMINATED.

You could try you 20 different spot card combination. In each case, you will find 11 were eliminated and only 4 are possible.

If you could not agree that fact that the first round of play means East and West must be dealt with the corresponding cards from initial position, then we are not speak the same language.
May 29, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
Monty Hall problem is different. It is not completely random. There is a restrictive choice. If you read my early comment about this problem. I also think it could be a restrictive choice problem.

However we are just discussing a pure probability problem here. Let me see if I could try a different way to explain it.

With West having 4 and East having 2, initially there are
C(6,2) = 15 ways to distribute the card. Let's think about what the first round a card play means. Let's assume West plays 2 and East plays 3. Now it means in initial card distribution West must have 2 and East must have 3. If you apply these conditions, you'll find only 4 out of 15 satisfy this condition.

What about the rest? The rest cases are
1) East dealt with 3 and 2. There is only 1 way and J is in West.
2) West dealt with 3 and 2. There are 6 ways and it is 50% chance West has J.
3) East dealt with 2 and West dealt with 3. There are 4 ways and 75% chance West could have J.
So in these cases eliminated by first round play, the chance West having J is
1/11*100% + 6/11*50% + 4/11*75% = 7/11 = 63.6%

If you are to calculate initial probability, you will find it is

63.6% * 11/15 (cases eliminated by first round play) + 75% * 4/15 (cases remained after first round of play) = 10/15 = 2/3!
May 29, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
Wei-Bung, we are not talking about the same question. You are correct there were 20 possible ways opponent could play first round of heart. This is before the cards were played. We are in a multiverse where all possibility exist. Once the first round of cards are played, you are in one of these 20 and it is a universe. For example, if you see west played 3 and East played 2, then you eliminate the possibility West could have 2 or East could have 3 in your initial set of combinations. What the probability I ask for is after seeing these card from first round of play what is the chance West or East having J. If this is not what you talked about, we are comparing apple with orange.
May 29, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
David, it looks like you still did not grasp the point of what the one round of heart play means in probability calculation.

If you don't want to do what I suggested but insist in what you deals randomly, it is fine. Now I'm going to instruct West to play 2 and East to play 3 or whatever 2 spot card combination you choose. Now you count how many time you could see West played 2 and East played 3. If you could not get to this step, you cannot go to next step.

This is same as in your later example, the question is what card West played. Your missing card is K32, so the possibility is
1) W: K2, E:3
2) W: K3, E:2
3) W: 32, E:K

Before you see what card West played, it is 2:1 West having the K. Once West play a card, for example you see it played 2. Then 2) is eliminated because it is improbable any more.

If you don't believe this, think about what you would play in a suit where you only miss Kx. Finesse or drop? I believe we were all told you should play for drop because it has 52% of chance. Why 52%? Each hand has 13 cards. When you play a card, if your LHO plays K, you don't have a decision to make. If he shows out, you don't have a decision to make. If he plays x, the information you gained is he has this card. Now there are still 12 unknown on the left and 13 unknown on the right. So the possibility RHO has K is 13/(12+13) = 52%. If you ignore this fact and asking what the initial odds are for 1-1 distribution. It is 50%. You would have draw the conclusion that finesse and drop has the same percentage rate.
May 29, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
Lior, even though I don't have direct comparison for rating by board vs by session. Christ Champion and I did some comparison for our rating results in the past. We took a sample of set of games (so we have a controlled data set) and ran our rating program based on same data set. His rating is using session data also adjusted by field strength. My rating is based on board results.

The comparison is tricky because his rating is a percentage and my result is a number. In addition he dropped off certain percentage of lowest rated players. So I compared results by just comparing players from top to bottom. The result is we have an agreement of 50%. What it means is that if I count the top 10 players we have 5 overlapped (50% agreement), if I count top 20, 10 of them overlap.

Since I know my calculation better I could explain to you how using session results would differ in my calculation. For IMP game, the best data I could find from ACBL land is from Swiss team. For Swiss team games, I could extract data about how many boards played and net IMP win/loss. Even though it does not have board level data, I could use this aggregated data to calculate rating and get the same result as using board level data. However there is a condition here is that the players could not change for these boards. If player changes I must recalculate the expected score with board calculation. In case of session calculation I think you have to make some kind of average adjustment. So I could see they would produce different results.

In case of pair event, the opponent changes every round. So the calculation from board vs session would more likely yield different results. It depends on the strength of the players. If all players are in similar strength, the difference could be small. If it varies a lot, I think the results could be quite different because the simple average adjustment from session result would be inadequate.
May 29, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
David, I know why you could not understand my point. The probability Murat and I calculated is the probability West having J AFTER one round of Heart play. You are keeping talking about the probability BEFORE this round of Heart play.

If you want to do an experiment, you need to alter the experiment this way. Given a spot card one each to East and West, instruct them to play it on first round. The reshuffle and deal the rest of card and count statistics. I would assure you out of 3,000 trial, about 2,250 would have West having J.

The problem is each card in the deck is distinguishable. If West played 2 and East played 3, West must be dealt with 2 and East must be dealt 3. In your initial random distribution you have cases where East has 2 or West has 3. With those distribution West could not play 2 and East could not play 3 at the 1st round. You could pick up any other 2 spot cards but it would produce the same results.
May 29, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
To any who don't agree with me, let's just do another test. Let's just concentrate on the case West has 4 hearts and East has 2. The initial odd for West to have J is 2:1. If you think after first round of heart the odd should not change, how about we play another round of heart and both followed low. Could you agree now that the chance West has J is 100%? Or by your argument it is still 2:1?

I did misread Charles cases where opponent always play the lowest spot cards. I agree with this rule the probability will change from 75%. However once you see the cards played, your are in one of the path (“case”) and it eliminate the other possibilities that existed before the cards played. If you don't agree, just think about what happen if you play another round of heart.
May 29, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
Yes, I think you have a valid point that I should study rating results from per board result vs using session results. I could do that once I accumulate a lot of data and get the details of how other motheds using session data calculate exactly. However my understanding is other rating methods like EBU only uses session data. In ACBL, Knockout game does not keep per game IMP score so it is hard to find a lot of IMP game data.
May 28, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
Charles, let me use you scenario but restate is as following: there are 6 possible cases what spot cards are played, then let's calculate for probability.

Case 1: West plays 4 & East plays 2, the E1 would be J4xx with East having 2x, E2 would be West having 6543 and East having J2, here Pr(E1) = 75%

Case 2: West plays 3 & East plays 2, E1: J3xx vs 2x, E2:6543 vs J2, again Pr(E1) = 75%

Case 3: West plays 2 & East plays 3, E1: J2xx vs 3x, E2:6542 vs J3,
Pr(E1) = 75%

Case 4: West plays 2 & East plays 4, E1: J2xx vs 4x, E2:6532 vs J4,
again Pr(E1) = 75%

Case 5: West plays 2 & East plays 5, E1: J2xx vs 5x, E2:6432 vs J5,
Pr(E1) = 75%

Case 6: West plays 2 & East plays 6, E1: J2xx vs 6x, E2:5432 vs J6,
again Pr(E1) = 75%.

We could make more cases for what spot card played on first round, but you could find out in each case, the probability for West to have J (E1) is 75%.

The issue is once the first round cards are played, your are in one of these case. The rest of the cases could not happen any more. We are living in an universe, not multiverse. Once an event happened and you are in this “case”, you could not revert time and get into other “cases”.
May 28, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
Lior, I got your meaning of IMP score implied a comparison between different players.

However when the score is not calculated at board level, but an average per session of 0.2 IMP/board, a lot of things are indistinguishable. For example a 12 IMP score on board level would averaged out when you looked at session score. In addition you don't have a single opponent to compare and have average their effects.

At session level, an average+ game would have 0.2 IMP/board and 53%, but they could be quite different at board level. I did calculate both IMP rating and MP rating. They are correlated but NOT strongly correlated. I'm not surprised that EBU find their results at session level strongly correlated because they lack of data granularity to see the difference. The problem I have now is I don't have enough IMP game data. Most of my study are MP data computed for IMP rating. Ideally you want to compute IMP rating using IMP game data, and comparing it against MP rating computed from MP game data.
May 28, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
Wei-Bung, you mixed probability with fact. When an event has not happen, it is a probability. When an event has happened, it is a fact. Fact is 100%. Before East played 2, it could be either in East hand or West hand. You could calculate each one's probability. Once he played 2, it is a fact. It is 100%. You could not take this card back and put it in West's hand. So no matter how you calculate, the probability for that card in West's hand is ZERO. If you calculation is still counting West could have this card, it is wrong. That 1/6 and 1/4 are now both changed to 1 after the card is played.
May 28, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
Charles, I'm not sure what you are trying to average. After East showed 2, the probability for West to have 2 is zero. There is nothing to average.
May 28, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
Charles, you just made a self-conflicting statement. First, you said “every card played may change the odds, yes”, then you said the probability (after playing one round of heart) “must apply in advance”. If I read it correctly, you meant the odd should not change.

I have explained in previous response why the odd changes after one round of heart. You have seen two cards and that eliminate the possibility these cards located in opposite side. If you would rather think in term of original position, you know the position of two out of six cards. The unknown is the 4 cards left. It does not matter how many choices opponents could select to play on first card. Once they played whatever the card they selected, it is 100%. Other choice do not exist any more.
May 28, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
Lior, as a former particle physicist, I concur with your comments about statistical model and variance. I disagree with your last part of comments because bridge is different from chess.

In chess, both players are starting with the same position (you could argue white and black makes a difference but the chance to draw white/black is equal). A chess game's result is definitive, win, loss or draw. So you could compute the chess rating from single game result because both player starting with equal and the result could be compared with a prediction by rating.

Bridge is different. Every board is a different hand. If you only know a score of 170, you could not judge it is a good score or bad score. It has to be compared against others who played the same board. In order to calculate rating, I chose to compare at board level where the players has exact the same position(same hand). The only difference is they face the different opponent but this is factored into the calculation. If you calculate it using session aggregated results like EBU, you introduce another systematical uncertainty that increases the variance. Finally MP and IMP has different strategy and I don't think one could be equivalent to the other in a definitive way. If you put these two types of game into one rating, it introduces another systematical uncertainty. When the variance due to these systematical uncertainty is large enough, it makes the rating result less meaningful.
May 28, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
It was a case that 10% happened. Don't blame your teammate.
May 28, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
The probability definitely changes after you played 1 round of heart, as it eliminated a lot of possible choices that were in initial position but becomes impossible once 4 cards are played.

For example, with initial 4-2 distribution, there are 5 possible choice for J with 2 cards, 10 with 4 cards. Once you played one round of heart, each side has showed a card. The possibility that these cards are in the other side are eliminated. Now you've already know the positive of 2 cards out of original 6. So the remaining possibility changed. There is only 1 choice J could be in the short side, and 3 choices otherwise.
May 27, 2015
Ping Hu edited this comment May 27, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
In finesse case, B has odd of 3:1 in favor West having J, so the chance is 3/4, not 2/3.

You had an assumption of 50% 9 is by restrictive choice. You also had an assumption of 50-50 East having 2.
May 27, 2015
Ping Hu edited this comment May 27, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
The problem here is how you read 9. Is it a restrictive choice or an “illusion”? If he started with Jxxx and T9, would he throw a instead of 9?
May 27, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
I'm not sure this is a case you could use the normal probability calculation because of the discard. If opponent discards are random and we assume you know East has Q, at this 3 card ending the odd West having the other is 3:2. In addition there is 25% chance East started with J so it favors playing for drop.

However if you read West discarding of 9 as the last (why not discard 2 if he could), then is 4-2 and the chance West having J is 3:1 and you should play finesse.

What is West discard a instead of 9? If he starts with !Jxxxx, by restrictive choice he has to discard a , it seems it is a stronger evidence for finesse.
May 27, 2015
You are ignoring the author of this comment. Click to temporarily show the comment.
Tim, as a matter a fact, USCF needs 26 games to get a regular rating. Before that it is consider “provisional”. Let's assume at their 26 games, one was rated 1220 and the other 1180 but they were really the same level. As long as they continue to play, their rating will continue to change and fluctuate around 1200. As Robin pointed out, there is an inherent measurement “error”. Everyone could have a good game or bad game in chess or bridge. So when we use game results to adjust rating, there is a built-in uncertainty that would make the rating fluctuates around the “true rating”.

The website below is a section's rating from last year's chess World Open.
http://www.uschess.org/msa/XtblMain.php?201407068692.7
As you could see, a rating change of +-20 is very normal. For players that performed very well, their rating could go up by over 200 points (they probably did improve, if you click on their name, you could find their rating history).
May 26, 2015
.

Bottom Home Top