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All comments by Wayne Burrows
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“If ”what we consider an honor“ is anything other than any ace, king, queen, jack, or ten, we're wrong.”

What is defined as an honour in the laws of bridge is not necessarily synonymous with the usage of honour in the vernacular?

Both definitions can exist simultaneously.
11 hours ago
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“as in even bidding over 3H”

This auction is much more nuanced than that. It can easily be wrong to bid over 3H for some pairs and wrong to pass 3H for others.

For example, one pair might have an understanding that 3H here shows or even is likely to show club support (Robson and Segal's fit non-jumps) and another have no such agreement. You can't treat them all the same. And you can't decide for any of them without investigating carefully their agreements.
11 hours ago
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“Suppose your regular partner is on lead against the auction above, holding …”

1. Obviously you do not have to disclose to declarer what you know about partner's holdings that has been deduced from declarer's bidding and play and your own cards.

2. So I know what partner would lead from a particular holding but there might be thousands of holdings that could be relevant. How precisely do I disclose all of that information?

3. Declarer has additional information that I do not have - their own cards and inferences that they have been able to deduce from that information and the play and defence. How does declarer ask such a question without tipping off their own hand or misleading the defence? It seems wrong to ask about leads from holdings (in four suits) that include cards in your own hand and also wrong to only be able to ask about holdings that do not include cards in your own hand.

While I think such questions are legitimate it seems much less clear to me how precisely the question should be asked and how precisely it should be answered.
11 hours ago
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It is a huge leap from an answer that causes declarer to misguess a suit to the defender has lied.
Oct. 20
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I would think that most serious partnerships discuss lead tendencies. If not then those partnerships are not exploiting very strong inferences from knowing your partner's lead style.
Oct. 20
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I was just being dismissive of those results as I wanted to concentrate on the play at this table.

At four of the five tables where 4 was the final contract, that is except the Chinese table where the auction was uncontested, north had made a takeout double and not bid after their partner had bid spades at the three level or below.

Not sure if that is enough to deduce for sure that spades are 2=2 but it points in that direction.
Oct. 20
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Double shows extra values but not the extra distribution or enough strength required to bid at the five level or above.

It does not bar partner from bidding. It is encouraging if he has extra distribution or enough extra values to think there might be a slam.

It does not show a heart stack or indeed any particular holding in their suit.

Similarly, if we were weaker and passed then double from partner would show extra values with no clear direction. We would bid out with distribution and spade support or pass for example.
Oct. 19
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whoops thanks
Oct. 19
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I think the queen is right. Partner can't really have the A singleton and defend this way.

If partner has the jack it does not matter what you do.

If declarer has the jack then if he has the ace he was probably going to take the finesse. If partner has the ace (not singleton) then the queen is just winning.
Oct. 19
Wayne Burrows edited this comment Oct. 19
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Yes there are other lines for the defence. The J only contributed to the misdefence there were other factors.

Thanks.
Oct. 19
Wayne Burrows edited this comment Oct. 19
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Damn. Thanks.
Oct. 19
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“3C was alerted as showing both majors. (which apparently was a correct systemic explanation, however West forgot the system and actually had a seven card club suit)

4C was explained as showing a very big hand, still with the majors

5C was explained as looking like he actually has had clubs all along. ”

I always think there is a problem with these sort of explanations. The agreement can't magically change from both majors to clubs mid auction.

If the agreement is clubs after three club bids then the first 3C bid was a two way bid that was either majors or clubs (by implicit agreement). Its majors if he doesn't do anything unusual or it could be clubs if he keeps bidding clubs. That explanation should have been given in response to the first query.
Oct. 18
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It certainly seems like north would raise if south had shown diamonds.
Oct. 17
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Only Nab for Netherlands opened (and won the board) versus Sweden.
Oct. 17
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Whatever your methods are you should play the reverse.
Oct. 13
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Most simulations do not give you single dummy results.
Oct. 13
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I use dealer.

Dealer is used by BBO when you customise hands at a bidding or teaching table. So you might be familiar with how to write constraints.
Oct. 11
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Interesting. For me there is no unbalanced hand that raises 1M to 4M, where unbalanced is defined as a hand with a singleton or void. I would always splinter with the unbalanced hand. Therefore my 4M raises are always some 4432 or rarely 4333 or 5422 shapes.
Oct. 9
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Paul I looked at only one hand for each deal in the simulation.

Looking at all four hands would not alter the expectation over multiple trials. In each deal there would be some dependency but if looking at one hand over n trials would give an accurate estimation then looking at any one of the four hands would give the same expected accuracy of estimation. So totaling the four estimates would be equally accurate.
Oct. 9
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I didn't do my Monte Carlo simulation that way. I dealt 10 million hands using the dealer program. I put no constraint on the hands, so the program looked at every hand dealt.

For each hand I checked the north hand for hits. A similar technique to what you would do at the table. Here is the code


HitS0 = spades(north)==0?1:0
HitS1 = spades(north)==1 and hascard(north,AS)?1:0
HitS2 = spades(north)==2 and hascard(north,2S)?1:0
HitS3 = spades(north)==3 and hascard(north,3S) and not hascard(north,2S)?1:0
HitS4 = spades(north)==4 and hascard(north,4S) and not hascard(north,2S) and not hascard(north,3S)?1:0
HitS5 = spades(north)==5 and hascard(north,5S) and not hascard(north,4S) and not hascard(north,2S) and not hascard(north,3S)?1:0
HitS6 = spades(north)==6 and hascard(north,6S) and not hascard(north,5S) and not hascard(north,4S) and not hascard(north,2S) and not hascard(north,3S)?1:0
HitS7 = spades(north)==7 and hascard(north,7S) and not hascard(north,6S) and not hascard(north,5S) and not hascard(north,4S) and not hascard(north,2S) and not hascard(north,3S)?1:0
HitH0 = hearts(north)==0?1:0
HitH1 = hearts(north)==1 and hascard(north,AH)?1:0
HitH2 = hearts(north)==2 and hascard(north,2H)?1:0
HitH3 = hearts(north)==3 and hascard(north,3H) and not hascard(north,2H)?1:0
HitH4 = hearts(north)==4 and hascard(north,4H) and not hascard(north,2H) and not hascard(north,3H)?1:0
HitH5 = hearts(north)==5 and hascard(north,5H) and not hascard(north,4H) and not hascard(north,2H) and not hascard(north,3H)?1:0
HitH6 = hearts(north)==6 and hascard(north,6H) and not hascard(north,5H) and not hascard(north,4H) and not hascard(north,2H) and not hascard(north,3H)?1:0
HitH7 = hearts(north)==7 and hascard(north,7H) and not hascard(north,6H) and not hascard(north,5H) and not hascard(north,4H) and not hascard(north,2H) and not hascard(north,3H)?1:0
HitD0 = diamonds(north)==0?1:0
HitD1 = diamonds(north)==1 and hascard(north,AD)?1:0
HitD2 = diamonds(north)==2 and hascard(north,2D)?1:0
HitD3 = diamonds(north)==3 and hascard(north,3D) and not
hascard(north,2D)?1:0
HitD4 = diamonds(north)==4 and hascard(north,4D) and not hascard(north,2D) and not hascard(north,3D)?1:0
HitD5 = diamonds(north)==5 and hascard(north,5D) and not hascard(north,4D) and not hascard(north,2D) and not hascard(north,3D)?1:0
HitD6 = diamonds(north)==6 and hascard(north,6D) and not hascard(north,5D) and not hascard(north,4D) and not hascard(north,2D) and not hascard(north,3D)?1:0
HitD7 = diamonds(north)==7 and hascard(north,7D) and not hascard(north,6D) and not hascard(north,5D) and not hascard(north,4D) and not hascard(north,2D) and not hascard(north,3D)?1:0
HitC0 = clubs(north)==0?1:0
HitC1 = clubs(north)==1 and hascard(north,AC)?1:0
HitC2 = clubs(north)==2 and hascard(north,2C)?1:0
HitC3 = clubs(north)==3 and hascard(north,3C) and not
hascard(north,2C)?1:0
HitC4 = clubs(north)==4 and hascard(north,4C) and not hascard(north,2C) and not hascard(north,3C)?1:0
HitC5 = clubs(north)==5 and hascard(north,5C) and not hascard(north,4C) and not hascard(north,2C) and not hascard(north,3C)?1:0
HitC6 = clubs(north)==6 and hascard(north,6C) and not hascard(north,5C) and not hascard(north,4C) and not hascard(north,2C) and not hascard(north,3C)?1:0
HitC7 = clubs(north)==7 and hascard(north,7C) and not hascard(north,6C) and not hascard(north,5C) and not hascard(north,4C) and not hascard(north,2C) and not hascard(north,3C)?1:0
Hits = HitS0 + HitS1 + HitS2 + HitS3 + HitS4 + HitS5 + HitS6 + HitS7 + HitH0 + HitH1 + HitH2 + HitH3 + HitH4 + HitH5 + HitH6 + HitH7 + HitD0 + HitD1 + HitD2 + HitD3 + HitD4 + HitD5 + HitD6 + HitD7 + HitC0 + HitC1 + HitC2 + HitC3 + HitC4 + HitC5 + HitC6 + HitC7

Where HitS3 for example is a calculation that I have precisely three spades - spades(north)==3 - and that I have the three of spades - hascard(north,3S) - and that I do not have the two of spades - not hascard(north,2s).

For each suit only at most one of the variables can be 1 so adding them all up gives the total number of hits for a bridge hand.

To check here are five hands it produced when I specified that the total Hits must be 4.

n QT875.9764.A.AQ3
n A9875.QJ3.953.T2
n A2.KJT85.Q964.K2
n A73.653.K864.853
n A954.8654..KJ985

In each case the total hits in the north hand is 4.

Here are five hands with three hits:

n A954..J9843.8754
n QJ3.A.AKJ4.QJ986
n AQ94.AQJT.T2.A43
n AQT4.A96.AQ84.82
n 953.K83.42.AQJ97

Two hits:

n K984.JT74.K9.J62
n AQJ42.9854.A2.Q4
n 95.QJ965.A83.Q82
n 983.32.2.AKJ9542
n Q63.AJT.KQ3.KQ65

One hit:

n K73.AQ6.KJ32.KJ5
n .KJ86.975.A87652
n K872.J964.432.A4
n K64.J9.AT72.QJ64
n T9.QT8752.93.A83

No hits:

n 8763.43.876.AJ32
n KQJ6.A5.K9.Q7532
n 842.A3.QT6.AKQ42
n QT92.QT752.Q7.63
n JT4.K52.Q87.T853

This method and its logic is quite different than the theoretical calculation where I calculated the number of combinations of cards for each suit length and the number of such combinations that were hits. From which the probabilities could be determined.

For each hand I multiplied the appropriate probabilities for hits and non-hits and totaled those according to how many hits.

Both the theoretical probability and the frequency from Monte Carlo were similar. I am confident that variation in the Monte Carlo numbers from the theoretical numbers are due to randomness and relatively minor flaws in the random number generator used.
Oct. 9
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