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All comments by Wayne Burrows
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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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I am relieved thanks for the corrections.
Oct. 9
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I think we are correct. These are the numbers I got from a Monte Carlo simulation of 10 million deals using dealer:

0 4970764
1 3825137
2 1069591
3 128816
4 5692

which compared reasonably consistently with the theoretical calculation above 0.00056272 0.012841576 0.106381877 0.380222953 0.494935986
Oct. 9
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Your calculation is the same as mine. I calculated the probability of each suit length being a hit. The four suits in any hand are independent so it is relatively easy to calculate the probabilities of zero, one, two, three, or four hits for each hand shape.

Then I multiplied each of these probabilities by the a priori probability of each hand shape. I calculated these rather than cribbing them from some other source. Those calculations are also easy. The probability of a 4-4-3-2 is the number of combinations of four cards from thirteen in two suits and three cards in thirteen and two cards in thirteen in the other two suits. Multiply all of those combinations and divide by the number of combinations of 13 cards from 52 cards and finally multiply by 12 since there are 12 different 4-4-3-2 hands - its a very easy formula in excel.
Oct. 8
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I think your strongest option AQTx is at best marginal.
Oct. 8
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Sorry Alan I have updated again. Never trust my calculations at 2:30am.
Oct. 8
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My editing skills are:

1. better at 730am than 230am
2. better when I am not already late to pick up my wife - the first edit not at 230am
Oct. 8
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What reason would hand dealing make hands with one cards of every suit “rather higher” than the theoretical calculation without bias? I am sure hand dealing is biased but I am not sure why it would be biased in that direction.
Oct. 8
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