david_c

Only in certain sequences. (Yes, No, No) The reason we play any variant of Leb is to give us two different ways to show a suit, with one being stronger than the other. For example, over a double of a weak two, we distinguish hands in the 0-7 range (approximately) from those with more like 7-10 HCP. But Transfer Lebensohl only works when the strong hands are game-forcing. So it would be useless over a double of a weak two, because neither of the two strengths are good enough to bid game. In practice, we only play Transfer Lebensohl in specifically defined sequences. If Lebensohl applies in other sequences it will be traditional Lebensohl.

:D Again, this is all true, but if you think this is the explanation for the paradox then you're missing the really beautiful part. OK, let's actually play this damn game. I'm going to put the amounts of money in the envelopes for you. I'm going to do it in such a way that the probability of the envelopes containing $M and $2M is at least 9/10 of the probability of the envelopes containing $M/2 and $M, for all possible values of $M. (Let's say the amounts will always be a power of 2, for simplicity.) Trivial calculation: this means that if you open the envelope and find that it contains $M, then the expected amount in the other envelope, given this information, is at least $(6M/5). [in fact the best possible bound involves 19s, but I want to keep the numbers simple.] But what does this actually mean? Does it mean that it's always right to switch? Well, yes and no. It turns out to be difficult to pin down exactly what we mean by "the right thing to do". But we can get some insight into what's going on by looking at what happens if you adopt various strategies. Let's play this game a large number of times, and keep track of how much money you've collected in total. We'll also keep track of how much money you would have won if you'd never switched. We'll say you're "in profit" if the amount of money you've collected is bigger than the amount you would have got by never switching. Sounds fair enough? OK, the first strategy we'll look at is if you only switch if the amount in your envelope is $1. This guarantees you a profit of $1 every time it happens (assuming you can't get any smaller amounts of money). So, in the long run you will be making a profit, albeit slowly. Now, let's instead try the strategy where you switch only if the amount in your envelope is $2. From the calculation we did earlier, we expect to make a profit on average when this comes up, of over 40 cents per time. Again, in the long run you are going to make a profit. It's not a guaranteed profit any more, but the probability that you will be in profit after playing, say, 1000000 times is very close to 1. I hope there is nothing paradoxical about this so far. But we can do the same thing for any other amount. No matter what amount we choose, we expect to make a profit in the long run, provided that we only switch when that amount comes up. Now here's another strategy: let's switch provided that the amount in our envelope is at most $64. Now, this means that we will be switching on a lot of times where the amounts in the envelopes are $1 and $2. These will cancel out in the long run, since of those occasions, we expect to open the envelope with the larger amount exactly half the time. The same goes for when the envelopes contain $2 and $4, or when they contain $4 and $8, and so on - these situations all give us zero profit. BUT we are also switching someimes when the envelopes contain $64 and $128, and here our cleverly-chosen strategy means we only switch when it is right for us to do so! So again, by following this strategy we will be making a profit in the long run. More generally, we can choose any finite subset of the possible amounts, and if we only switch when we get one of those amounts, we will make a profit (which you can calculate in terms of expected gain per play). If we have a fixed strategy like this, then the probability that we are in profit after N plays approaches 1 as N gets large. And the more often you switch, the larger your expected profit per play is. But now let's try and be greedy and switch every time. Now, we are no longer steadily accumulating profit at a rate we can calculate. [in fact the expectation is undefined, because the money that you expect to make is infinite whether you choose to switch or not.] If you track what happens to your profit as time goes on, you will find that at times you will be making huge profits: if you look at your maximum profit of all times up to the present, then this will get larger and larger as time goes on. But at other times you will be making huge losses, and the size of these losses will also approach infinity. In fact, you will find yourself in profit exactly half the time in the long run. There will be infinitely many points in time when you are making a profit, but also infinitely many points in time when you are making a loss. So, if you want to actually make a steady profit, you should only switch when you get a particular amount [or amounts] in your envelope. This guarantees ("almost surely" as the experts say) that whatever profit you want to make, you will eventually reach that level and stay there for the rest of time. If you switch all the time, then you will eventually reach that level but you won't stay there.

By all means, teach me. You have an envelope with 2N times (2 to the -X power) money in it, where N is the amount of money in the largest amount of money in it, and X is an integer ranging from 1 to the number of envelopes, so the 2nd largest envelope will have an amount of money equal to 2N times (2 to the -2 power). You can switch that envelope with an adjacent envelope: ie., a 50% chance of switching for X'=X+1, and 50% that X'=X-1. Should you make the switch? That's the actual formula that's been crudely switched into English. When you actually use the formula, instead of just English, it should become clear that if N is infinite, the question becomes meaningless, and if N is not infinite, then the expectation cannot be infinite. The way you've chosen to set this up (having a fixed number of envelopes to choose from) means that there are only a finite number of different amounts that can be in the two envelopes. But this needn't be the case. If someone asks you to put two amounts of money into the envelopes, with only the condition that one must be twice as much as the other, there are (in theory) infintely many different ways you could do it. Here's one way you might go about it. Pick a number at random between 0 and 1. Let's call it x. Now, If 1/2 < x <= 1, then let the amounts in the two envelopes be $1 and $2. (And toss a coin, say, to decide which envelope gets the larger amount.) If 1/4 < x <= 1/2, then let the amounts in the two envelopes be $2 and $4. If 1/8 < x <= 1/4, then let the amounts in the two envelopes be $4 and $8. ... and so on. Finally, if x=0, let's say (for the sake of it) that the amounts are $5 and $10. That covers all possible cases, and no matter what x you've chosen, the amounts in the two envelopes are always finite. But, if you do it this way, the expected amount of money in the less valuable envelope is (1/2 x 1) + (1/4 x 2) + (1/8 x 4) + ... which is 1/2 + 1/2 + 1/2 + ... which is infinite. This is basically the same situation as the one Helene looked at. If you want an example where it is always "right" to switch, then you could take something like the one I gave before, where the probability of the smaller amount being 2^n was k.(1-t)^(|n|) for some very small number t. (Anything sufficiently small will work.) Here k is whatever constant is required to make the sum of the probabilities equal 1.

So, you open the envelope, and it has infinity dollars in it. No it doesn't. The amounts in the envelopes are always finite, but they can be chosen in such a way that the expectation before you open an envelope is infinite. If you don't understand how this is possible you need to learn some basic probability theory.

No, you didn't. And like I keep saying, it is perfectly possible for it to be right to switch no matter what amount you find in the envelope. Does that sound strange too?

This is all true, but it only gives you useful information in the case where the expected amount of money in the envelopes is finite. Since we are told that the amounts in the envelopes can "approach infinity", it is perfectly possible that the expectation is infinite. In this case, it may well be the case that once you've opened the envelope, no matter what amount of money you find, the expectation from switching is higher than the amount you are currently looking at.

I don't believe you because you're restricting the outcomes of the experiment to {k*2^n} sequence and you don't a priori know what "k" is. Well, two points: (i) You've misread the example. k is a normalising factor for the probabilities, to make sure they sum to 1. (ii) Yes it's a discrete distribution. That's the easiest example to write down. However, you could find a continuous distribution with the same properties if you like.

Indeed - and so we can't calculate the expectation in the original problem. The fallacy in many people's approaches is to assume that using 1/2 must work, when in fact the probabilities are completely undefined. (Though one thing we do know, as you say, is that the probabilities can't all be 1/2.) But you can resolve this problem, by specifying a distribution which we use to choose the amounts in the envelopes, and still be in the situtation where switching is best no matter what amount you found in the envelope. I was making the assumption that, after the amounts in the two envelopes had been decided, you would pick which one you were going to open at random.

Hmm, I believe that last sentence is incorrect, and I think people are missing the point of what's really paradoxical about this. Owen here makes the good point that the conditional probability P ( our envelope contains the smaller amount | our envelope contains $M ) is not necessarily the same as the a priori probability (which is 1/2 of course). We're trying to work out the expectation from switching given that our envelope contains $M, and in order to do this we need to know the probability that our envelope contains the smaller amount given that it contains $M. That is, the a priori probabilities are useless, what we need to know are the conditional probabilities. And Han's very first post proved that it is not possible for all the conditional probabilities to be 1/2. (Or to put it another way, we can't arrange things so that all possible amounts of money are equally likely.) BUT Even though the conditional probabilities can't all be 1/2, we can make them as close to 1/2 as we like. For example, we can put amounts into the envelopes according to a probability distribution in such a way that all the conditional probabilities, for every possible amount $M, are between 0.49 and 0.51. [For the smart math people: just take a very slowly decaying distribution. For example, let the probability that envelopes contain $(2^n) and $(2.2^n) be k.(1-epsilon)^(|n|).] In this case, when analysing whether it's right to switch for a particular amount M, the calculations are so close to what they would be if the probabilities were 1/2 that it makes no difference. The conclusion is For every amount M you see in the envelope, your expectation if you switch is greater than M. (In fact, greater than 1.2M, say. We can get any multiple less than 1.25.) Is this a paradox? It shouldn't be - it's true. Do you believe me?

I've recently been playing a Polish Club variant with Mike (mickyb). I've written up the system notes for the 1♣ opening and you can find them on this page. Important / distinctive features of the system: Opening bid: - We open 1♣ on all balanced hands in the 12-14 HCP range without a 5-card suit (so a 1♦ opening will not be 4-3-3-3 or 4-4-3-2). - Some hands in the 18+ HCP range are opened 1♦ or 1♥ rather than 1♣. Response to 1♣: - 1NT is natural and invitaitonal. - 2♣/2♦ are non-forcing. We have artificial sequences to deal with strong minor-suit hands. Other agreements: - Some use of transfers by responder after an overcall. - Many other artificial sequences, both in uncontested auctions and in competition. Opener's diamond bids are very often artificial. It's quite serious stuff, but the aim is to have enough agreements to deal with any situation that might come up. There's a particularly long section on competitive auctions.

But we do need to regulate conventions based on the strength of the hand somehow. To give an extreme example, we want a 1♦ bid which shows "17+ HCP any shape" to be allowed, but we don't want a 1♦ bid which shows "0-8 HCP any shape" to be allowed. So there has to be some mention of strength - not necessarily in terms of HCP, but enough to differentiate between things like those two examples. More generally, players should be given more flexibility for bids which show strong hands than for bids which might be weak. Agree with Helene and Gerben that a rigid boundary based on HCP does not mean players are not allowed to use judgement: it just means that players must set their agreements sufficiently high that all the hands they judge to be worth that bid meet the HCP requirement.

I think you're right - the only reason is familiarity. An interesting question is whether a lack of familiarity is a legitimate reason for banning a particular method. I think it is: it makes the method more difficult to defend against, and being difficult to defend against is why things get disallowed.

Simplicity is a good thing, but it's not as important as making sure you allow the right methods. In this case, IMO the right point to start is to have a discussion about whether assumed-fit pre-empts and suchlike should be allowed. If we think they should be allowed, then maybe the suggestion might work. If we decide they should not be allowed, then some modification will be needed. Only if it is a very close decision should we let the potential simplicity of the regulations affect which methods we want to allow. Personally I don't think that assumed-fit pre-empts should be allowed on the GCC. But I don't mind if you disagree - the point is that we need to decide this first, and then work out how to write regulations to achieve it. It is perfectly possible to write clear regulations that don't allow assumed-fit pre-empts; for what it's worth I would like to allow any 2-of-a-suit opening which: (i) Shows 5 or more cards in that suit; or (ii) Shows 4 or more cards in that suit and constructive values. [by "constructive" I mean basically an opening hand - I'd use the same definition as for artificial 1m bids, whatever we decide that should be.] [Edit: obviously this list is not exclusive.]

This doesn't seem important to me. I think you could afford to have completely separate rules for each opening bid if you wanted to (though it is more normal to group certain bids together, for example 1♥ with 1♠). Always remember that when you're writing system regulations, the question you're trying to answer is "is my bid allowed?". People don't want to learn all the regs, they just want to be able to tell whether they can play their particular convention. If someone wants to find out whether their 2♥ bid is allowed, it makes no difference to them what the rules for 1♥ are. It seems to me that the most common fault is trying to make the regulations too short, believing that this will make them simple. In reality you can afford to have as much detail as you like, if you feel this is the best way to express what things should be allowed, provided that the detail is clear. The EBU system regs run to almost thirty pages and we don't seem to have big problems deciding which things are allowed like you do in the ACBL. Oh, no, there are plenty of us who are not like that.

But this isn't the same as the GCC - for example he is disallowing 5-card pre-empts.

I believe the opposite of "natural" is "artificial" rather than "conventional". So, I agree, the call in question is both natural and conventional. Oh we've had this discussion so many times. "Conventional" is defined in the Law book (albeit not terribly clearly), but "Natural" means different things to different people. So not everyone will agree about whether Muiderberg is natural, but there is no right answer - it just depends whose definition you are using. See: blog: What does "Natural" mean?

How about we draw trumps pitching dummy's second diamond. Then AK and another club. If they win the third club and play a diamond back, we can pitch the blocking club on this trick while ruffing in hand.

The current system sucks only because the ACBL refuses to keep up to date and to correct the ambiguities that have been pointed out to them. Other countries regulate conventions in a similar way to the ACBL, but for them it works much better because the authorities actually care about finding ways to improve the regulations.

It would be nice to eliminate the petty politics and personal influence. But I don't like your alternative either. System regulation is difficult - you shouldn't expect it to be possible to have an axiomatic approach.

We don't have any regulations specifically about pre-alerting here in England. Many people who play unusual systems do make a point of doing it, but technically it's not compulsory. On the other hand, we do have regulations about convention cards, and if a pair playing an unusual system had failed to exchange convention cards properly you might expect the TD to take some action. The front of the convention card must contain all the things that opponents might need to know, so this basically amounts to the same thing as a pre-alert.

Yes and no. Here in England, before we had announcements the alerting rule was "Because of agreements which opponents are unlikely to expect, you must alert: ... (g) a natural 1NT opening which may be made on 4-4-4-1 or 5-4-3-1 distribution." So it was considered natural, but it was still alertable because the agreement was unexpected. Whether it is conventional is a slightly different question, since this is [supposed to be] defined by the Law book, not by SO regualtions. Well, the Law book is not very clear. But in my opinion, the Laws do not support the interpretation that a 1NT bid containing a singleton is always conventional. There is no way that "willingness to play" translates into a strict dividing line between hands with singletons and hands without. For what it's worth, I think the cause of problem is the Law which says SOs are only allowed to regulate conventional bids. They should be allowed to regulate all agreements. In my opinion, it would be perfectly reasonable to disallow 1NT openings with a singleton, but the Laws currently do not give SOs that authority. (Though unfortunately some SOs claim that they do, and there's not much anyone can do to stop them.)

Really? In the six years or so that I've been reading forums like this one, I've certainly seen plenty of complaints about ACBL regulations, but I can't remember any that have been acted upon by the ACBL. But maybe I've just missed them - can you give any examples? From the same period of time, I can think of at least half a dozen significant things that people have complained about with our English regulations which the EBU has actually changed - including opening 1NT with a singleton. (Plus a large number of individual conventions which have been permitted.) It seems to me that while the ACBL may in theory have the capability to make changes, the pace of change is ridiculously slow.

In this part of the world it's pretty much the opposite of that.

West doesn't have anywhere near enough strength to bid 2♦. The blame goes to East: I don't think 3♣ is a bad bid on the 7-card suit, but 2♠ is better, and on this occasion you see precisely why bidding 2♠ is a good idea.

[hv=d=w&v=b&s=skj876h7dkjt843c7]133|100|Scoring: IMP[/hv] Both vulnerable, three passes to you. What do you do?