Will we often be missing 2 cashing tricks?

[pescetom]

A recent discussion hinged partly on the probability that opponents will hold two Aces (or AK in same suit, or in different suits) when we have at least 32 HCP combined between NT opener and balanced responder.

 

It's better to have a number to back up an argument rather than just an opinion or experience, so I hacked the script below to investigate.

It lets you define the point range of an NT opening (NMin Nmax) and of balanced responder (Smin Smax).

To keep things realistic and minimise argument, the NT opening is balanced but may include 5cM, the responder is purely balanced (no 5 card anything).

 

A relevant set of settings for the scope of my question is 15-17 for opener and exactly 17 for responder (that's already 6NT for most of us I think).

You can see one typical run of that below.

Producing 1000 such hands pushes us near the 1 million generation limit of BBO but still causes the number of double Aces to vary widely, so multiple runs are needed to get more significant results.

I will provide the average of ten such runs in a later post.

 

In the meantime, please point out any obvious errors.

 

 

Here is the code:

#
# Will we often be missing 2 cashing tricks?
#

produce 1000

Nmin=15
Nmax=17
Smin=17
Smax=17

Open1NT = shape(north, any 4333 + any 4423 + any 5332) and hcp(north)>=Nmin and hcp(north)<=Nmax
Raiseto6NT = shape(south, any 4333 + any 4423) and hcp(south)>=Smin and hcp(south)<=Smax

EtopS = top2(east,spades)
EtopH = top2(east,hearts)
EtopD = top2(east,diamonds)
EtopC = top2(east,clubs)
WtopS = top2(west,spades)
WtopH = top2(west,hearts)
WtopD = top2(west,diamonds)
WtopC = top2(west,clubs)
EWaces = aces(east) + aces(west)
NSaces = aces(north) + aces(south)
EWkings = kings(east) + kings(west)
EWhcp = hcp(east) + hcp(west)

OffAKany = ((EWaces == 1) and (EWkings == 1))
OffAKsame = (EtopS + WtopS == 2) or (EtopH + WtopH == 2) or (EtopD + WtopD == 2) or (EtopC + WtopC == 2)

action frequency "EW hcp" (EWhcp,0,9), frequency "EW Aces" (EWaces,0,4), frequency "EW have an A and a K" (OffAKany,0,1), frequency "EW have AK in same suit" (OffAKsame,0,1)


condition Open1NT and Raiseto6NT

#### for debug or skeptics, uncomment below and reduce Produce to a small number
#action printes (" EtopS: ", EtopS, " EtopH: ", EtopH," EtopC: ", EtopC," EtopD: ", EtopD," WtopS: ", WtopS," WtopH: ", WtopH," WtopC: ", WtopC," WtopD: ", WtopD, " OffAKsame:", OffAKsame, " NSaces:", NSaces, " EWaces:", EWaces, \n), printew

 

 

 

Here is a typical run:

Frequency EW hcp:
   0	       0
   1	       0
   2	       0
   3	       0
   4	       0
   5	       0
   6	     146
   7	     299
   8	     555
   9	       0
Frequency EW Aces:
   0	     706
   1	     291
   2	       3
   3	       0
   4	       0
Frequency EW have an A and a K:
   0	     949
   1	      51
Frequency EW have AK in same suit:
   0	     985
   1	      15
Generated 9234892 hands
Produced 1000 hands
Initial random seed 1667411112
Time needed   17.790 sec

[smerriman]

A relevant set of settings for the scope of my question is 15-17 for opener and exactly 15 for responder (that's already 6NT for most of us I think).

Small typo, since you meant 17.

 

A simpler way of looking at the aces part is:

 

produce 10000
condition hcp(north)+hcp(south)==32
action frequency "aces" (aces(north)+aces(south),0,4)

Frequency aces:
   0	       0
   1	       0
   2	      50
   3	    3472
   4	    6478

Even if you knew you had exactly 32 HCP combined (rather than 32-34 like here), the chance that the 8 missing points are two aces is about half a percent, so pretty negligible. (Shape doesn't have an impact on that figure.)

 

What is quite fascinating from your numbers is when you do have a 17 count opposite a 1NT opener, partner will have the minimum 55% of the time..

[pescetom]

Small typo, since you meant 17.

 

A simpler way of looking at the aces part is:

 

produce 10000
condition hcp(north)+hcp(south)==32
action frequency "aces" (aces(north)+aces(south),0,4)

Frequency aces:
   0	       0
   1	       0
   2	      50
   3	    3472
   4	    6478

Even if you knew you had exactly 32 HCP combined (rather than 32-34 like here), the chance that the 8 missing points are two aces is about half a percent, so pretty negligible. (Shape doesn't have an impact on that figure.)

 

What is quite fascinating from your numbers is when you do have a 17 count opposite a 1NT opener, partner will have the minimum 55% of the time..

 

Typo fixed directly to avoid future confusion, thanks.

 

I intentionally gave North a real 1NT opening and South a truly balanced response rather than just produce two generic hands with the necessary total 32 HCP, to be closer to the original diatribe but also to real life quantitative bidding.

I'll take your word for it that shape does not influence the Aces issue.

[pescetom]

The averages of ten runs given 15-17 1NT opposite 17 truly balanced are as follows:

 

- opps have just 8 HCP: 53%

- opps have 2 Aces: 0.22% (0.42% of when they have 8 HCP)

- opps have AK in same suit: 1.3%

- opps have any A any K: 6.1%

[gszes]

SIGH I hope I am reading this right. It appears that south is getting dealt 15 - 17 not 17 (of course the intent may have changed)

I see you like variables vs just plugging in numbers thus and HCP(south) == 17 do you work for a coding company that gets paid by the line by the government?:)))

 

I am assuming top2 is reserved code asking for the A or K in the given suit since it is undefined elsewhere.

 

good job:))

[gszes]

deleted duplicate

[LBengtsson]

Interesting statistics. Thanks for sharing. It would be good to know also - sorry for asking - whether different point ranges, for example 20-21 balanced opposite 12 skew the results, or do they 'tighten' the criteria as the no-trump range is now only 2HCP as opposed to 3HCP with a 15-17NT.

 

Just a complete guess - though I think my math is right - but is 15-17 opposite 17 not the most likeliest high card point distribution of the 32-34 range, and 12-14 opposite 20 would come up more often?

 

12-14 point hands 20.63% opposite a 20 point hand 0.64% = 21.27%

 

15-17 point hands 10.09% opposite a 17 point hand 2.36% = 12.45%

 

Just from bridge experience at the table, I cannot recall many hands where I have had 17 balanced opposite a strong NT, but it is more likely that the bidding goes 1♣ - 1♥ - 2NT (18-19 balanced) - and then you have to make a decision with a balanced 14HCP hand. (This is why I asked about the 2HCP point range above.)

 

I realise that the 32-33 combination is different from the 32-34 but smerriman calculated that opener will have a minimum 55% of the time with the 3HCP point range, so I guess with the 2HCP range having a minimum would be far greater.

[pescetom]

SIGH I hope I am reading this right. It appears that south is getting dealt 15 - 17 not 17 (of course the intent may have changed)

I see you like variables vs just plugging in numbers thus and HCP(south) == 17 do you work for a coding company that gets paid by the line by the government?:)))

 

I am assuming top2 is reserved code asking for the A or K in the given suit since it is undefined elsewhere.

 

good job:))

South is currently getting dealt 17 - 17, in other words exactly 17. It's North that is getting 15-17 (and opening NT).

 

Smart governments and employers know that it's worth paying more for cautiously structured code with variables :)

 

Top2 is reserved code in the sense that it's a function listed in the description of Dealer script. There are also Top3, Top4 etc. plus other useful toys like 'hascard' and 'tricks' (sadly disabled by BBO).

[pescetom]

Interesting statistics. Thanks for sharing. It would be good to know also - sorry for asking - whether different point ranges, for example 20-21 balanced opposite 12 skew the results, or do they 'tighten' the criteria as the no-trump range is now only 2HCP as opposed to 3HCP with a 15-17NT.

I'm no expert on probability, that's why I do simulations :)

But I think you are right that all else being equal (same boundary of range) a narrower point range will tighten things: you can test any range you choose with that script.

 

Just a complete guess - though I think my math is right - but is 15-17 opposite 17 not the most likeliest high card point distribution of the 32-34 range, and 12-14 opposite 20 would come up more often?

 

12-14 point hands 20.63% opposite a 20 point hand 0.64% = 21.27%

 

15-17 point hands 10.09% opposite a 17 point hand 2.36% = 12.45%

 

Just from bridge experience at the table, I cannot recall many hands where I have had 17 balanced opposite a strong NT, but it is more likely that the bidding goes 1♣ - 1♥ - 2NT (18-19 balanced) - and then you have to make a decision with a balanced 14HCP hand. (This is why I asked about the 2HCP point range above.)

 

I suspect your math is wrong about the two combinations you cite: you should be multiplying two probabilities, not adding them, and the second should take account of dependencies on the first situation (the probability of 12-14 opposite 20 is not the same as the probability of 12-14 opposite unknown). I expect 17 + 15to17 will be more frequent than 12to14 + 20, as it is close to 16 + 16 (think about the frequency of the opposite extreme 0 + 32).

From bridge experience at the table, I wouldn't be surprised by either combination you compare: they are not everyday occurences, but not freaks like also being off two aces :)

[thepossum]

Thx for the thread :)

 

I am interested in another aspect of the 32 point slam scenario

If you have all the aces chance of 6NT is around 80%

With only three aces its around 70%

With only 2 its insignificant chance of that so don't worry about it

 

EDIT Other assumptions was a strong NT (15-17) opening bid