But is there a formula or criterion for them?

At this point I've found two examples (153 and 407) in addition to the ones you gave and an upper bound on the range within which others can occur (2,916). Based on my trial-and-error searching, I doubt that there are any more three-digits examples, but I expect that there are some with four.

I await enlightenment from other sources.

Here's a different problem, one that I *have* managed to solve:

I think you've got your facts mixed up. The Dvorak keyboard is a later optimized design intended to remedy the defects and inefficiency of the QWERTY layout. https://en.wikipedia.org/wiki/Dvora...

I too have heard that the QWERTY layout was chosen to prevent jamming of the keys. Pray tell, why would they be less likely to jam in that configuration than in another, randomly chosen one? Jamming of the keys in the original mechanical typewriters occurred whenever two or more keys were struck at once. That can happen with any layout.

Thanks for the information and the link.

<al wazir: I too have heard that the QWERTY layout was chosen to prevent jamming of the keys. Pray tell, why would they be less likely to jam in that configuration than in another, randomly chosen one?>

My pleasure. I am reminded

"That would be a butcher's way, not an artist's"

[ Search Kibitzing ]

Still,

#36. You gotta go with what works.

My post was poorly phrased.

Keys can jam with any configuration, whether chosen with a view to speed in typing or randomly.

But if the aim was to minimize jamming, putting all of the most-frequently used keys (ETAIONSHRD) on one row would be the way to go. QWERTY doesn't do that. Linotype does.

I think you are saying that the QWERTY layout was deliberately chosen to be inefficient, to slow down typists.

What QWERTY does do is cause the left hand to make about 75% of the keystrokes. For right-handed people (which is most of us), this is indeed inefficient. It would have made more sense to reverse the arrangement, giving the right hand the majority of keystrokes. This would have been just as effective in reducing jamming.

To my mind, that is evidence that the design was not thought out carefully or with any consideration of typing speed.>

<Rain likely, mainly after midnight. Mostly cloudy, with a low around 51. South wind 9 to 14 mph. Chance of precipitation is 60%. New precipitation amounts of less than a tenth of an inch possible.>

And for Tuesday, I see this: <A 40 percent chance of rain before noon. Mostly cloudy, with a high near 57. New precipitation amounts of less than a tenth of an inch possible.>

I once asked an NWS employee what these percentages mean. His answer, as I recall it, is that if the probability is given as 60%, then six times out of ten someone living *anywhere* within the area covered by the forecast will get rain *sometime* during the period in question. Of course the higher the percentage, the more likely it is to rain. But if the chance of rain next Wednesday is given as 90% and I don't get any rain, that doesn't prove that the Weather Service goofed. A 90% probability means I can expect that one time out of ten, it *won't* rain.

For a long time I've wondered how, at least in principle, I can check the accuracy of these probabilities. How can I tell if, e.g., they average too high or too low? If they were always the same percentage it would be easy. For example, if the probability were always stated as 50%, then I would keep a record of the times rain is predicted and find the fraction in which I actually got some rain. How well that fraction approximates 1/2 would be a measure of the accuracy of the predictions. Over a long time it should be pretty close.

But the stated probabilities vary. Sometimes they're only 20% or 30%, sometimes as high as 90%, and very rarely, even 100%.

How can I determine the accuracy for all these predictions taken together? I want a single figure of merit that says how closely the NWS predictions approximate reality.

<If "prototype" can be verbed, then someone who prototypes is a prototyper, unless he's a prototypist. >

Suppose an insurance company is providing insurance to the shipping industry. Suppose they have a model that says 10% of shipments through a certain region of the South Pacific are lost, and also 10% of shipments through a certain region of the Caribbean are lost. They set their insurance rates accordingly.

In the following year, suppose they insure 9,000 shipments through the dangerous region in the South Pacific, of which 500 are lost, and 1,000 shipments through the dangerous region in the Caribbean, of which 500 are lost.

So, out of 10,000 shipments through the dangerous regions, 1,000 are lost, which is indeed 10% as their model provided. Success! The insurance company makes money.

The insurance company in the toy example has a model that takes one observation (the route of the shipment) and applies two simple rules (through Caribbean, through South Seas) to estimate the probability of a shipment loss event. If you like, those rules could be based on a very simple simulation of the activities of the pirates of the Caribbean and South Pacific storms.

In both the cases, they need a way to check how well the models perform so they can make adjustments. I'm not understanding what you view as the crucial difference between the two systems in principle. Yes, the NWS system uses many more inputs and a much more complex function, but those are differences in scale, not in principle.

So I'll go ahead and post my solution and hope that someone will comment on it.

Suppose the odds that something will happen are stated as n':n", where n' and n" are positive integers. This means that if you bet n' dollars at these odds and it happens, you win n'' dollars. The odds that it won't happen are n":n'. A person betting against you puts up n" dollars and wins your n' dollars if it doesn't happen.

This is equivalent to saying that the chance of its happening is given by a percentage p = n'/n, and the chance of its not happening is 1 – p = n"/n, where n = n' + n". (Apparently the “chances” stated by the NWS are always multiples of 10%, so we can assume that n = 10, but this isn't essential.)

Let's suppose you repeatedly bet that rain will occur assuming that the forecast is right, i.e., you always bet at the equivalent odds, and that the stakes S (the total of the amounts on the two sides of the bet) are the same each time. (I'll come back to that assumption.) Then you stand to win (n"/n)S or lose (n'/n)S. If the forecast always said that the chance is the same unchanging percentage p, and if the forecast of rain turns out to be correct m' times and wrong m" times, where m' + m" = m, the number of bets, then your total winnings or losses T would be given by

T = S(m' x n"/n – m" x n'/n) = mS[(m'/m)(1 – n'/n) – (1 – m'/m)(n'/n)] = mS(q – p),

where q = m'/m.

If q = p, i.e., if the observed frequency of rain equals the predicted chance of rain, then T = 0. This would confirm that the predictions are accurate. T may not exactly vanish, but the closer it is to zero the better the agreement between predictions and observations.

But the predicted chance of rain varies from one forecast to the next: the forecasts employ different values of the probability p. If we label them p_1, p_2, etc., the corresponding numbers of bets m_1, m_2 , etc., and the observed frequencies q_1, q_2, etc., the formula can be generalized to

T = SΣ m_i(q_i – p_i),

where the summation is over the index i.

If all the terms are close to 0, then T will again be close to 0. But T can also be close to 0 if some of the terms are positive and some are negative and they cancel. (This is the situation in your insurance example.) Saying that this confirms the accuracy of the predictions would be very misleading. The way to avoid that is to rewrite the formula using the *absolute value* of each term:

T = SΣ m_i|q_i – p_i|.

Of course, since each term is nonnegative, the more forecasts that are included in the sample, the bigger T becomes. Thus defined, T is more likely to be close to 0 for a small sample than for a large one, which makes it a misleading figure of merit. The way to correct for this is to normalize by dividing by m, the total number of cases in the sample. And since the absolute size of the “bet” is irrelevant, we can set S = 1:

T = (1/m)Σ m_i|q_i – p_i|,

where m = Σ m_i.

This is my answer.

But is it realistic? If you were trying to maximize your gains, would you vary the size of your bets at different odds n':n" in proportion to n'/n, i.e., would you bet more when both your potential loss and the chance of winning are larger? Maybe you would bet the same amount each time, or bet more when the potential payoff is bigger.

I don't really agree that "rain or no rain" is more absolute than something like "stomach cancer or no stomach cancer." Over a given region at a given time, there could be anything from heavy rain over the whole region to 1mm of rain in one small patch, and an arbitrary decision is made where to draw the line.

The typical buzzword for these functions to measure the difference between the model's distribution and the real world's distribution is "loss function," and the one you derived is "absolute loss." Another common one is "quadratic loss" where the differences are squared instead of placed under absolute value.

<Over a given region at a given time, there could be anything from heavy rain over the whole region to 1mm of rain in one small patch, and an arbitrary decision is made where to draw the line.>

Of course. If you look at the examples I posted you'll see that the bar is set pretty low, e.g., "new precipitation amounts of less than a tenth of an inch possible." What does "less than a tenth of an inch" mean? One hundredth of an inch is less than a tenth of an inch. A light fog drip is less than a tenth of an inch.