PeterAlan

Possibly this piece Ken?

You think that's a problem? I have a twin brother. His birthday is the day before mine. He shares his with Donald Trump; I share mine with Xi Jinping.

I speculate that the bug (or whatever) is causing your clipboard's contents to be displayed - had you copied the postcode?

In the example of EBU that I gave you earlier they're to be found at the start of section 3 on page 10 of the Blue Book that I linked then. EBU publishes both blank system cards of their approved form and some sample cards pre-filled with some common basic systems here.

RAs publish their regulations, typically always on their websites (I haven't made any sort of comprehensive check to validate this assertion, but every time I've had occasion to look a website search has worked), and frequently as printable PDF version. Here in EBU's jurisdiction, for example, the regulations that players need to be aware of are in the Blue Book, now supplemented for online play by the SkyBlue Book. There is also a simple Announcing and Alerting Summary and the Best Behaviour code, both of which many clubs will have on display. These cover all that players normally need. Further comprehensive guidance for TDs, and some more specialised material such as the screen regulations, are published in the White Book. There are also some other special regulations, for example regarding matches played privately, and other supplementary material: a full list is here.

https://www.washingtonpost.com/politics/2021/08/10/con-is-winding-down/

AL78, if they were on RealBridge you can usually log back into the session (use the original link) and run through the auction and play of any hand.

Suit below the singleton was the traditional mantra, but now, for example, the EBU's teaching material has to open 1♥ with a singleton ♣, 1♣ with a singleton ♦ and 1♦ with a major suit singleton: your hand would therefore open 1♦. I'm not expressing a view one way or the other on the merits of this approach, but it has moved on from "suit below".

I'm a Tom Lehrer fan too. His song about von Braun is one of his best and happily that story is an urban myth:

I'm glad it was easily solved.

Assuming you're entering your user name correctly, it sounds like you need to look harder at the password. Start with my suggestions above. If you have any uncertainty about a stored password, you can always use your login on the desktop to change it.

It shouldn't be a problem - I routinely log in on more than one laptop. What failure message do you get, and are you sure (if you use a stored password on your main device) that you're using the correct password on the laptop? Are you getting upper / lower case correct throughout?

Guilty, and I'm glad you pointed this out. In your Bayesian realm, I presume that one would start with a null hypothesis concerning the fairness of the dealing algorithm. Making this well-defined may not be straightforward: see below. I probably have much more interest in these questions than most; if you're not interested in further discussion I quite understand. However, and at the risk of digging a still deeper hole, there are a couple of things still to be said about implicit hypotheses in everyone's answers to the original question (independence has already been mentioned): I and the CLT folk are giving probabilistic answers for the mathematical universe of all deals and its HCP distribution, or models of it. The BBO dealing software in question might not conform to that, and to give a better answer to the question originally posed (what's the probability in BBO etc) we'd really need to know the single-hand HCP distribution that results from it - of course, we don't have that information. There's also the possibility that the BBO software could have precisely the correct overall HCP distribution for single hands but nevertheless have bias as to which hands get which points; we'd need information about the full deals to get a better handle on that; and A simulation approach doesn't actually answer the question posed: what it gives is answers about a third HCP distribution, namely that of the dealing algorithm underlying the simulation software. This seems to be important. Of course, if all the dealing algorithms are pretty close one-to-another and to the theoretical universe then the results are sufficiently good for the everyday question. All this is really saying is that a complete answer to just the original question is not straightforward; we all simplify the problem by adopting one assumption or another in order to come up with an answer, and in doing so pick up our calculation tool of choice.

Quite frankly, you single yourself out possum. You pose a bunch of questions, some clearly expressed, others less so, but if someone takes you seriously and engages with you, you then accuse them of being patronising, rude, etc. As far as I am concerned, all the rudeness is coming from your side. You can engage in civil debate, or be ignored: which would you prefer?

Let's say that the probability of some outcome (total HCP <= 187, exactly 187, whatever) over a set of 25 boards is p. Then the a priori chance of it happening twice in two sets of 25 boards (assuming independence throughout) is (p^2). But once it has happened in the first such set, the chance of it happening in the second set is still p, and not something less.

I can't tell whether or not you're interested in the answer, possum, but the probability of exactly a total of 187 HCP in 25 independent hands is 0.0001538068 (1 in 6502).

Indeed Richard. Or, if looking for an approximate answer with a quick calculation, you can follow smerriman's and ALT78's CLT route, though that needs some technical knowledge (and the standard deviation in question). But thepossum seemed to be looking for some way to think about the problem, and I was trying to give that. Simulations have their place, but they do nothing to help you understand the underlying problem and its possible solutions.

Possum, you can think about the original problem in this way: Consider the formal polynomial p(X) = a(0) + a(1).x + a(2).x^2 + ... + a(37).x^37 where the coefficients a(i) = probability of i HCP in a single hand (a(0) = 0.003639, a(1) = 0.007884 etc). Essentially, the power of x in the formal polynomial is used to label the number of HCP. For N hands (here N=25) calculate p(x) ^ N = b(0) + b(1).x + ... b(37.N).x^(37.N); the coefficients b(i) then represent the probability of holding exactly i HCP in N hands. For the probability of holding <= M HCP in total (here M=187), sum b(i) for 0 <= i <= M. Calculating the coefficients b(i) of p(x)^N is quite straightforward - you can do it in a spreadsheet from the raw a(i) data (which obviously you need). I wrote a program instead: it accepts N & M as its input and does just that calculation.

That is exactly what my program did: it has the HCP distribution of a single hand as its raw data, and does precisely the calculation you describe. Rather than the rounded probability numbers in your link, the easiest source of raw data (for me, since I already had it) was to start with the 635,013,559,600 possible ways of dealing a 13-card hand, tabulated by the number (2,310,7889,600) with 0 HCP, etc through to (4) 37 HCP. If anyone wants the parameters to slightly more precision, its standard deviation is 4.13023287. I calculated its third (skewness) and fourth (kurtosis) standardised moments to be 0.24044733 and 2.87800178 respectively.

I was well aware of that, and that was the question I answered, since I had the tool to hand to do it without approximation. My comment about the normal distribution was badly worded; I was just trying to say that the approach that you and smerriman adopted was giving good results: I should have been clearer about that, and separated it from the first remark. I was aware of the central limit theorem, and if I weren't your first post would have educated me. As the wiki article says "The central limit theorem gives only an asymptotic distribution. As an approximation for a finite number of observations, it provides a reasonable approximation only when close to the peak of the normal distribution; it requires a very large number of observations to stretch into the tails.", but it works pretty well here. I suspect, without any further analysis, that this is largely due to (1) the value in question (74.8% of the mean) not being that far along the lower tail of the cumulative distribution, and (2) the HCP distribution of a single hand being reasonably well-behaved with only a small proportion of hands in the tail above 20 or so HCP. You may well have a better insight. In particular, I haven't looked at how well, in this application, the CLT approximation improves with the number of observations.

Congratulations. But you also have a partner; what were their HCP? Incidentally, whilst the HCP distribution of a single hand is not normal, the HCP distribution of a pair of hands from the same deal is pretty close to normal. (It is obviously symmetrical about its mean of 20: for every pair of hands totalling N HCP there is another pair of hands with 40-N HCP, namely the other two.)

The chances of <= 187 (= 7.48 x 25) HCP in 25 independent hands is 1 in 1045.160. The HCP distribution of a hand is not normal, obviously, as it has a range of 0-37 and a mean of 10, but it has a long and very flat tail, and so, as you see above, gives a fairly reasonable approximation when applied to this sort of problem. A while ago I built myself a simple program that calculates exactly the statistic you want, based on the full 0-37 HCP distribution, and that's the basis of the answer I've given you above.

The frequency of 6-5 in the majors (6-5-2-0 or 6-5-1-1) happens to be exactly one-third of the frequency of 5-5 (5-5-3-0 or 5-5-2-1) in the majors (or minors for that matter), which may give some context: uncommon but not really that rare. 6-5-2-0 or 6-5-1-1 shape occurs in 1.3564% of all hands, and so 6-5 in the majors in 0.226% (1/6 of all such; 1 in 442 hands); you might expect to hold one per 16-20 normal pairs sessions on average.

You need to edit your post to give the East hand its 13th card (I realise it's in the hidden spoiler).

I note that you're maintaining the convention that the century started with 1 January 2001, notwithstanding the celebratory events of 1 January 2000. The problem's too tough for me right now (and almost certainly hereafter)!