Protect?

[Walddk]

So spades will statistically break 6/2/1/4 or 6/1/2/4 on average.

Everything being equal (a priori probablities). You forget that those probabilities change as far as partner and RHO are concerned when RHO responds 1NT, showing 0-2 spades.

 

Roland

[hrothgar]

Not really. Pard is marked with short spades and thus should have some support for diamonds.

I disagree. It is much more likely that responder is short in spades after his 1NT. It is perfectly normal to see spades divided the way they were at the table. 3 with North, 1 with East.

Well, let's give this issue some numbers. The nr. of spades each hand has is a conditioned probability problem. There are 34 unknown cards (26 pard/RHO, and 8 LHO) and 4 undistributed spade cards.

 

LHO will have 8/34 * 4 = 0,94 extra spades on average

RHO and pard will both have 13/34 * 4 = 1,53 spades on average

 

So spades will statistically break 6/2/1/4 or 6/1/2/4 on average. (Notation: LHO/pd/RHO/we)

 

We can do the same thing for diamonds and come up with the following conditioned probability odds of...

 

2/3/3/5

 

E.g. statistically, the expectation is that pard will have 3 card support for diamonds on average.

Another fine case of Garbage In / Garbage Out. If you don't ask the right question, all your fancy little calculations don't mean squat. All you end up doing is pretending a level of precision that just isn't there.

 

In this case, you really need to consider the fact that RHO should have some values, yet he was unable to scrape up any kind of a Spade raise. Furthermore, in many cases, responder needs less values to raise Spades than to introduce a new new suit.

 

In short, responder's expected Spade length is shorter than you calculated. Partner's Spade expected Spade length will be longer, and his Diamond holding shorter.

[whereagles]

Richard,

 

> In this case, you really need to consider the fact that RHO should have some

> values, yet he was unable to scrape up any kind of a Spade raise.

 

This is nonsense unless you make one of those "garbage in/garbage out little fancy calculations" that can tell what the odds are that responder has, say,

 

Kxx

xxxx

xx

JTxx

 

as compared to all the other hands he might have when he passed 1♠. Oh, and while you're at it, compute the odds responder would effectively make a bid with that holding. I believe they're quite small.

 

 

> In short, responder's expected Spade length is shorter than you calculated.

> Partner's Spade expected Spade length will be longer, and his Diamond holding

> shorter.

 

In short, while I agree there might be a slight tendency for responder to hold less spades than the theoretical figure, I very, very much disagree that this tendency is strong enough to significantly distort the odds of pard holding 3 diamonds.

[hrothgar]

Richard,

 

> In this case, you really need to consider the fact that RHO should have some

> values, yet he was unable to scrape up any kind of a Spade raise.

 

This is nonsense unless you make one of those "garbage in/garbage out little fancy calculations" that can tell what the odds are that responder has, say,

 

Kxx

xxxx

xx

JTxx

 

as compared to all the other hands he might have when he passed 1♠. Oh, and while you're at it, compute the odds responder would effectively make a bid with that holding. I believe they're quite small.

Using the right set of analytical tools can hardly be considered nonsense. Pretending that you have the correct answer because you're too lazy to do the necessary work involved to solve a problem properly is pathetic.

 

If you actually care about this type of problem, you really don't want to to try to solve this using arithmetic. The problem space is way to complex. Monte Carlo simulation is the way to go. There are alot of good Dealer programs out there that permit you to define a 1S opening, hands that won't overcall a 1S opening, as well hands that won't advance over a 1S opening. Once you have your scripts defined you can spit out answers in a matter of seconds.

 

I have a fairly good set of scripts that model MOSCITO. I haven't bothered to design a comprehensive set for Acol or SAYC or whatever system the players in question are using. Then again, I never claimed that I knew the "right" answer to this problem. I am simply stating that the answer that you are providing may very well be wrong.

[whereagles]

Using the right set of analytical tools can hardly be considered nonsense. Pretending that you have the correct answer because you're too lazy to do the necessary work involved to solve a problem properly is pathetic.

 

(...)

 

I am simply stating that the answer that you are providing may very well be wrong.

LOL.. if anyone here is lazy it's YOU, who didn't bother to back up your claim that 'pard's diamond expectancy is significantly lower than 3' with serious data.

 

In fact.. how can someone give any credibility to attitudes like "You might very well be wrong, but I can't be bothered to check it out. You do it!"

[awm]

Having done the montecarlo simulation, here are the first twenty hands for partner:

 

♠86 ♥J8754 ♦A8 ♣KJ65

♠83 ♥KT842 ♦AJ8 ♣T64

♠74 ♥AKJ32 ♦86 ♣KQ65

♠73 ♥K96432 ♦QT ♣KQ5

♠8 ♥Q8652 ♦AT85 ♣KJ7

♠7 ♥QJT654 ♦QT7 ♣K64

♠Q4 ♥T7 ♦AQ76 ♣Q8754

♠64 ♥AQJ7 ♦QJ8 ♣JT64

♠Q5 ♥KJT82 ♦T87 ♣KQ7

♠J ♥Q873 ♦AT85 ♣QT65

♠65 ♥JT7 ♦AJT5 ♣KT84

♠J8 ♥AQ542 ♦AQ76 ♣85

♠Q ♥J632 ♦AJT7 ♣Q764

♠865 ♥Q83 ♦A75 ♣QJT6

♠Q7 ♥J42 ♦AJ75 ♣Q865

♠84 ♥AK764 ♦QJ85 ♣J6

♠3 ♥QJ9642 ♦QJ ♣QJT7

♠QJ ♥A976 ♦QJ5 ♣K654

♠A8 ♥A9543 ♦A7 ♣JT76

♠Q3 ♥JT983 ♦AQJ ♣K75

 

My preference was to pass on this hand. Usually partner has two spades (three spades is actually quite rare). The issue is that if partner has fewer than five hearts balancing will often push the opponents into a better spot. When partner does have five hearts, the odds of holding four diamonds go down substantially, and the chance of partner holding potentially useless heart cards goes up. In any case, feel free to draw your own conclusions from the data. This simulation process was partly done by hand (I have a program, but it doesn't 100% accurately determine which hands would/wouldn't pass 1♠ as responder or as overcaller so some inspection is necessary, making it difficult to generate thousands of hands at a time).

[cherdano]

Not really. Pard is marked with short spades and thus should have some support for diamonds.

I disagree. It is much more likely that responder is short in spades after his 1NT. It is perfectly normal to see spades divided the way they were at the table. 3 with North, 1 with East.

Well, let's give this issue some numbers. The nr. of spades each hand has is a conditioned probability problem. There are 34 unknown cards (26 pard/RHO, and 8 LHO) and 4 undistributed spade cards.

 

LHO will have 8/34 * 4 = 0,94 extra spades on average

This computation is just wrong, even a priori.

[Robert]

Hi everyone

 

You are suggesting that 6412 or 6421 deals are more common than 54??

 

That conflicts with most of the Frost data is my guess.

 

That fact that the bidding went 1S-p-p suggests that the number of spades in RHO may or may not total 3(4?) cards. He was too weak to bid so the number of spades in his hand does not greatly change the fact that he did pass.

 

Even 'if' the methods permitted a weak 3M raise, holding 4333 is a big leap of faith in the bidding.

 

I tend to bid less often 'holding' three cards in the opened suit 'rather' than two cards.

 

Regards,

Robert

[hrothgar]

I'm suffering from Turkey coma.

 

I approached this problem from two different directions.

 

1. Monte Carlo simulation. I wrote a very simple script.

 

condition

 

spades(north) >= 5 and

spades(west) == 4

 

action

 

average spades(north),

average spades(south),

average spades(east),

average spades(west)

 

The output was as follows

 

average spades(north) = 5.232

average spades(south) = 1.884

average spades(east) = 1.884

average spades(west) = 4

 

This suggests a glitch in Whereagles math.

 

2. I then went back and tried to solve this analytically. I ended up with the same answer as Whereagles. I approached this as a hypergeometric distribution.

 

N = the total number of members of the population

n = the number of draws from a finite population (without replacement)

D = the number of cases that you care about (typically, you're looking at defect rates or some such)

 

in this case we're looking at

 

N = the total number of cards left to deal. We hold 13 cards. We know that LHO holds 5 Spades. So, N = 52 - 13 - 5 = 34

 

n = the number of cards that LHO still needs to draw. We know that LHO holds 5 spades. He has 8 cards left to draw.

 

D = the number of Spades that are outstanding. LHO holds 5 Spades. We hold 4 Spades. There are 4 Spades left.

 

The mean for a hypergeometric distribution = (n * D) / N

 

= (8 * 4) / 34 = .94118...

 

As a sanity check, the average number of Spades held by RHO (who has 13 cards to draw) is (13 * 4) / 34. In a similar fashion, the average number of Spades held by partner (who also has 13 cards to drawn) is (13 * 4) / 34

 

(8 * 4) / 34 + (13 * 4) / 34 + (13 * 4) / 34 = 136 / 34 = 4

 

Not sure where I'm going wrong. I think I'll wait till the turkey is out of my system.

[gwnn]

edit: but RHO would have replied 3sp with 4 spades and nothing and would have replied 2sp with 3 spades and next to nothing so in reality the odds tilt a tad towards pd having more spades. I don't think it goes much over 2, if at all, though.

 

edit ^2: ugh, random nonsense

Edited February 17, 2008 by gwnn

[cherdano]

Richard, you are doing the same mistake as whereagles, just in a more fancy version.

 

Lets forget about opening requirements, passes etc., just focus on the fact that we know LHO has >= 5 spades. There are two problems:

1. Deal me my cards, then deal LHO 5 spades, then deal the remaining cards randomly.

2. Deal me my cards, then deal the remaining 39 cards, and throw out all deals where LHO doesn't have at least 5 spades.

 

Only no. 2 will give the correct answer, but you and whereagles are computing the answer for problem 1. It is pretty easy to convince yourself that in model 2, LHO will have exactly 5 spades a lot more often.

 

Incidentally, I had an e-mail exchange with Richard Pavlicek about this, as his random dealer suffers from the same problem (he agrees). It seems quite a tricky problem to generate random deals with constraints without brute force (= deal completely randomly and throw out all deals that don't match the constraints).

[awm]

Basically this is just the restricted choice problem in another guise. To give a simple example, say there are only four cards, of which two are spades. They are dealt in two piles of two. You're told that my pile contains at least one spade. What's my expected number of spades?

 

The calculations here basically come down to: there are three spaces left after removing my spade. One of these is filled with a spade. So my expected number of spades is 1.33 (my one spade, plus a one in three chance my other card is the other spade).

 

This would be correct if I had specified that I hold the ace of spades. But I didn't. In fact there are 6 possible deals, 1 of which involves me having no spades and 1 where I have both spades. My indicating I have a spade eliminated only one possibility, so I will have two spades 1/5 of the time and one spade 4/5 of the time, for only 1.2 spades on average.

[Finch]

This computation is just right. Really. :)

 

edit: but RHO would have replied 3sp with 4 spades and nothing and would have replied 2sp with 3 spades and next to nothing so in reality the odds tilt a tad towards pd having more spades. I don't think it goes much over 2, if at all, though.

Actually, no. Don't impose your methods on RHO. If you were to ask your opponents about their methods, you would discover that

 

At the table where RHO passed over 1S

 

- LHO's 1S opening showed 4+ spades, denied a weak NT, but they would open 1H with 4-4 in the majors

 

- RHO's pass denied 6+ HCP any shape, or possibly 5 HCP with 4 spades.

[whereagles]

Richard, you are doing the same mistake as whereagles, just in a more fancy version.

 

Lets forget about opening requirements, passes etc., just focus on the fact that we know LHO has >= 5 spades. There are two problems:

1. Deal me my cards, then deal LHO 5 spades, then deal the remaining cards randomly.

2. Deal me my cards, then deal the remaining 39 cards, and throw out all deals where LHO doesn't have at least 5 spades.

 

Only no. 2 will give the correct answer, but you and whereagles are computing the answer for problem 1. It is pretty easy to convince yourself that in model 2, LHO will have exactly 5 spades a lot more often.

 

Incidentally, I had an e-mail exchange with Richard Pavlicek about this, as his random dealer suffers from the same problem (he agrees). It seems quite a tricky problem to generate random deals with constraints without brute force (= deal completely randomly and throw out all deals that don't match the constraints).

Ah, this is getting very interesting now...

 

Sometime ago, me and a friend were evaluating odds for a new system we were designing. We came up with some rather high suit length expectancy numbers which we felt weren't correct at the table, but couldn't find any error in our calculations. (My friend lectured combinatorics at the local university at the time, so this wasn't lack of skill.. lol)

 

I see now we were doing calculations in fashion 1, while real life seems to be using fashion 2. At the moment I don't see why they should be different, but I'll have a look at it as soon as I have some spare time.

 

I also used Pavlicek's calculator a couple of times and indeed I noticed huge distortions in conditioned probability problems as compared to what I experienced in real life. Interesting...

 

Still, in this particular deal, there are hints that spades split 5422, rather than 6421. Did you also monte-carlo for diamonds, Richard?

 

awm: hum.. interesting. I think I get it. That might be it..!

[helene_t]

Arend is correct. If LHO has five particular spades (say 23456) the probability that he has at least one more spade is very high. On the other hand, given that he has at least five spades (but you don't know which ones), the probability that he has only five spades is high.

[hrothgar]

Richard, you are doing the same mistake as whereagles, just in a more fancy version.

 

Lets forget about opening requirements, passes etc., just focus on the fact that we know LHO has >= 5 spades. There are two problems:

1. Deal me my cards, then deal LHO 5 spades, then deal the remaining cards randomly.

2. Deal me my cards, then deal the remaining 39 cards, and throw out all deals where LHO doesn't have at least 5 spades.

 

Only no. 2 will give the correct answer, but you and whereagles are computing the answer for problem 1. It is pretty easy to convince yourself that in model 2, LHO will have exactly 5 spades a lot more often.

 

Incidentally, I had an e-mail exchange with Richard Pavlicek about this, as his random dealer suffers from the same problem (he agrees). It seems quite a tricky problem to generate random deals with constraints without brute force (= deal completely randomly and throw out all deals that don't match the constraints).

Thanks for the clarification

 

Makes much more sense on an empty stomache.

[Blofeld]

Richard, you are doing the same mistake as whereagles, just in a more fancy version.

 

Lets forget about opening requirements, passes etc., just focus on the fact that we know LHO has >= 5 spades. There are two problems:

1. Deal me my cards, then deal LHO 5 spades, then deal the remaining cards randomly.

2. Deal me my cards, then deal the remaining 39 cards, and throw out all deals where LHO doesn't have at least 5 spades.

 

Only no. 2 will give the correct answer, but you and whereagles are computing the answer for problem 1. It is pretty easy to convince yourself that in model 2, LHO will have exactly 5 spades a lot more often.

 

Incidentally, I had an e-mail exchange with Richard Pavlicek about this, as his random dealer suffers from the same problem (he agrees). It seems quite a tricky problem to generate random deals with constraints without brute force (= deal completely randomly and throw out all deals that don't match the constraints).

For shape constraints, can't you do something like:

 

Pick all the acceptable shapes (relatively small finite number). You know all the absolute probabilities of these, so can work out the probability of any of them given that it's one of them easily enough. Do this, then deal cards according to these constraints, and finally throw out any that don't match your other constraints.

 

?

 

Or does that also fail?

 

Obviously it's still a bit of a hassle, but it's better than the pure brute force approach.

[Fluffy]

Well, if you are insane enough to calculate this accuratelly, I'll tell you there is still a missing argument, The ♠ expectancy for West is also modified by wich ♠s you have, since he has opening strenght :)

 

I tried to calculate a different problem, that was finding ♠Q at trick one when you know one player has 5♠ and 5-6 HCP, and the other has 2♠ and 18-19 HCP. The answer was that you needed to know wich HCP exactly were missing.

[cherdano]

It seems quite a tricky problem to generate random deals with constraints without brute force (= deal completely randomly and throw out all deals that don't match the constraints).

For shape constraints, can't you do something like:

 

Pick all the acceptable shapes (relatively small finite number). You know all the absolute probabilities of these, so can work out the probability of any of them given that it's one of them easily enough. Do this, then deal cards according to these constraints, and finally throw out any that don't match your other constraints.

Yes, that works when you have only shape constraints. I can count it as brute force to save my statement above, though :)

 

It gets a little more tricky when you also have hcp constraints, as they do influence the likelyhood of different shapes. (If you have xxx xxx AKQJ AKQ and partner opens 2N (20-21), it is quite unlikely that he has 5 clubs...) Maybe the best approach is to deal the shapes as you suggest, then deal the cards randomly, and throw everything out if it doesn't match the hcp constraints.

 

Arend

[Blofeld]

Maybe the best approach is to deal the shapes as you suggest, then deal the cards randomly, and throw everything out if it doesn't match the hcp constraints.

Sorry, yes, this is what I was trying to get at. I didn't phrase it very well.

 

I agree that it's still basically a brute force method, but it should be a good bit faster than the naïve brute force method.