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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.

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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.

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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.

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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.

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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.

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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.

Wayne Burrows

Wayne Burrows

Wayne Burrows

Wayne Burrows

Wayne Burrows

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.

Wayne Burrows

Wayne Burrows

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.

Wayne Burrows

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

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.

Wayne Burrows

Wayne Burrows

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

Wayne Burrows

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.

Wayne Burrows

Wayne Burrows

Wayne Burrows

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

Wayne Burrows

Wayne Burrows

Wayne Burrows

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.

Wayne Burrows

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.

Wayne Burrows

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

Wayne Burrows