You are ignoring the author of this comment. Click to temporarily show the comment.
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
You are ignoring the author of this comment. Click to temporarily show the comment.
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
You are ignoring the author of this comment. Click to temporarily show the comment.
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?1:0HitS1 = spades(north)==1 and hascard(north,AS)?1:0HitS2 = spades(north)==2 and hascard(north,2S)?1:0HitS3 = spades(north)==3 and hascard(north,3S) and not hascard(north,2S)?1:0HitS4 = spades(north)==4 and hascard(north,4S) and not hascard(north,2S) and not hascard(north,3S)?1:0HitS5 = spades(north)==5 and hascard(north,5S) and not hascard(north,4S) and not hascard(north,2S) and not hascard(north,3S)?1:0HitS6 = 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)?1:0HitS7 = 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)?1:0HitH0 = hearts(north)==0?1:0HitH1 = hearts(north)==1 and hascard(north,AH)?1:0HitH2 = hearts(north)==2 and hascard(north,2H)?1:0HitH3 = hearts(north)==3 and hascard(north,3H) and not hascard(north,2H)?1:0HitH4 = hearts(north)==4 and hascard(north,4H) and not hascard(north,2H) and not hascard(north,3H)?1:0HitH5 = hearts(north)==5 and hascard(north,5H) and not hascard(north,4H) and not hascard(north,2H) and not hascard(north,3H)?1:0HitH6 = 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)?1:0HitH7 = 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)?1:0HitD0 = diamonds(north)==0?1:0HitD1 = diamonds(north)==1 and hascard(north,AD)?1:0HitD2 = diamonds(north)==2 and hascard(north,2D)?1:0HitD3 = diamonds(north)==3 and hascard(north,3D) and not hascard(north,2D)?1:0HitD4 = diamonds(north)==4 and hascard(north,4D) and not hascard(north,2D) and not hascard(north,3D)?1:0HitD5 = diamonds(north)==5 and hascard(north,5D) and not hascard(north,4D) and not hascard(north,2D) and not hascard(north,3D)?1:0HitD6 = 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)?1:0HitD7 = 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)?1:0HitC0 = clubs(north)==0?1:0HitC1 = clubs(north)==1 and hascard(north,AC)?1:0HitC2 = clubs(north)==2 and hascard(north,2C)?1:0HitC3 = clubs(north)==3 and hascard(north,3C) and not hascard(north,2C)?1:0HitC4 = clubs(north)==4 and hascard(north,4C) and not hascard(north,2C) and not hascard(north,3C)?1:0HitC5 = clubs(north)==5 and hascard(north,5C) and not hascard(north,4C) and not hascard(north,2C) and not hascard(north,3C)?1:0HitC6 = 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)?1:0HitC7 = 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)?1:0Hits = 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
You are ignoring the author of this comment. Click to temporarily show the comment.
I am relieved thanks for the corrections.
Oct. 9
You are ignoring the author of this comment. Click to temporarily show the comment.
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
You are ignoring the author of this comment. Click to temporarily show the comment.
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
You are ignoring the author of this comment. Click to temporarily show the comment.
I think your strongest option AQTx is at best marginal.
Oct. 8
You are ignoring the author of this comment. Click to temporarily show the comment.
Sorry Alan I have updated again. Never trust my calculations at 2:30am.
Oct. 8
You are ignoring the author of this comment. Click to temporarily show the comment.
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
You are ignoring the author of this comment. Click to temporarily show the comment.
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
You are ignoring the author of this comment. Click to temporarily show the comment.
I will have to check in the morning I am not getting out of bed again.
Oct. 8
You are ignoring the author of this comment. Click to temporarily show the comment.
There are 4^13 = 67108864 hands with one card of each rank.

There are 52 choose 13 = 635013559600 total 13 card hands.

The probability of one card of each rank is therefore 0.000105681 which is about 1 in 9462 hands.
Oct. 8
You are ignoring the author of this comment. Click to temporarily show the comment.
I think you are right. I actually had a formula but somehow I didn't copy it into that cell in the spreadsheet and I am not sure now where 1/12 came from - just that somehow I didn't put the formula there. The formula is:

14-n choose n-1 divided by 13 choose n

14-n gives the number of cards greater than n. We need to choose n-1 of those for a suit of length n. The probability is then calculated by dividing by the number of ways of choosing n cards from all thirteen in the suit.
Oct. 8
You are ignoring the author of this comment. Click to temporarily show the comment.
The probabilities of a given number of hits for each distribution.

(Corrected I hope they are right now - third attempt)

`4333	0.105361303	0.001193617	0.020957934	0.13773749	0.401631993	0.4384789654432	0.215511756	0.000833362	0.016347785	0.119901409	0.389638561	0.4732788834441	0.029932188	0.000363649	0.009773064	0.091732325	0.366181143	0.5319498195332	0.155168465	0.000557021	0.012875154	0.10690277	0.381704549	0.4979605065422	0.10579668	0.000388902	0.009789689	0.090161669	0.362178559	0.5374811815431	0.129307054	0.000243064	0.007382488	0.079162014	0.353521287	0.5596911475440	0.012433371	0.002757671	0.052756867	0.319779687	0.624705775	05521	0.031739004	0.00011343	0.004075372	0.05369803	0.306497533	0.6356156365530	0.008952027	0.001843235	0.04170977	0.299162645	0.657284351	06322	0.056424896	0.000148539	0.006660916	0.07828327	0.355490587	0.5594166876331	0.034481881	9.28369E-05	0.004645825	0.067135414	0.345592846	0.5825330796421	0.047020747	6.4817E-05	0.003377045	0.053052753	0.314739522	0.6287658636430	0.013262262	0.001053277	0.040868402	0.307877258	0.650201063	06511	0.007053112	1.8905E-05	0.001188313	0.025516214	0.229709298	0.743567276520	0.006510565	0.000491529	0.021802834	0.239302028	0.738403608	06610	0.000723396	8.19215E-05	0.005839836	0.130265706	0.863812537	07222	0.005129536	1.48539E-05	0.003871565	0.061184832	0.331573941	0.6033548087321	0.018808299	9.28369E-06	0.002468003	0.050479145	0.318756744	0.6282868257330	0.002652452	0.00015086	0.038098612	0.312044834	0.649705694	07411	0.003918396	4.05107E-06	0.001106351	0.029706342	0.263009635	0.7061736217420	0.003616981	0.000105328	0.026816558	0.271808477	0.701269638	07510	0.001085094	3.07206E-05	0.008151925	0.162508417	0.829308937	07600	5.56459E-05	0.000133123	0.036447042	0.963419836	0	        08221	0.001923576	0	        0.001820665	0.041875284	0.295402822	0.6609012298311	0.001175519	0	        0.001137915	0.032089213	0.278561675	0.6882111978320	0.001085094	0	        0.029585799	0.286982249	0.683431953	08410	0.000452123	0	        0.012910167	0.218934911	0.768154922	08500	3.13008E-05	0	        0.097902098	0.902097902	0	        09211	0.000178109	0	        0.000910332	0.026854802	0.251251707	0.7209831599220	8.22041E-05	0	        0.023668639	0.26035503	0.715976331	09310	0.000100472	0	        0.014792899	0.23964497	0.74556213	09400	9.66074E-06	0	        0.167832168	0.832167832	0	        010111	3.95798E-06	0	        0.000455166	0.016385981	0.196631771	0.78652708210210	1.09605E-05	0	        0.01183432	0.207100592	0.781065089	010300	1.54572E-06	0	        0.192307692	0.807692308	0	        011110	2.49103E-07	0	        0.00591716	0.142011834	0.852071006	011200	1.14971E-07	0	        0.153846154	0.846153846	0	        012100	3.19363E-09	0	        0.076923077	0.923076923	0	        013000	6.29908E-12	0	        1	        0	        0	        0	1       	0.00056272	0.012841576	0.106381877	0.380222953	0.494935986[\code]`
Oct. 8
Wayne Burrows edited this comment Oct. 8
You are ignoring the author of this comment. Click to temporarily show the comment.
I know of other players who would prefer to sit facing a wall as they get distracted when facing other tables.
Oct. 2
You are ignoring the author of this comment. Click to temporarily show the comment.
RFID could be used for playing cards and bidding cards. This would capture all bids made and cards played and record timing of all actions.

For far too long we have thrown away information in this game. Which could be used for analyse performance and detect irregularities.

In competitive chess scorecards are kept and signed for each game. This could also be useful. Perhaps it would be a way for players to annotate the bidding (and play) with their agreements. Such information could be used to check for anomalies in the bidding and play that might point to additional exchanges of information.
Oct. 1
You are ignoring the author of this comment. Click to temporarily show the comment.
In a rock paper scissors sort of way.
Sept. 29
You are ignoring the author of this comment. Click to temporarily show the comment.
Do you need to alert that?
Sept. 28
You are ignoring the author of this comment. Click to temporarily show the comment.
It also appears declarer failed to factor in the possibility of 4=2=5=2. The favourable vulnerability for NS makes the 4S bid (and the 1S bid) a very wide range. Dismissing 4S as offbeat is possibly not giving south enough credit and overstating the accuracy of the club finesse.
Sept. 28
You are ignoring the author of this comment. Click to temporarily show the comment.
If they are not published then they are easily calculated. I have a one line formula in a spreadsheet that will calculate an IMP score based on a butler datum.

With this much data there should be a better measures of performance, although having said that we throw away so much data to our detriment.
Sept. 28
.

Bottom Home Top