Re: relating controls to hcp

"Frances" <fhinden@xxxxxxxxxx> writes:

> Eddie Grove wrote:
>> I've been touting an approximation here for a long time.
>>
>> approximately: E[controls | hcp] = hcp * .4 - 1

[regarding exact values versus sims]

> <sigh> you don't need mathematica. Excel, or any other spread***
> does admirably.

You're clearly better at that than I am. I guess it is time for me to
learn this skill.

>>>From hcp in the range 5 to 29, the best fit line for expected controls

Your newsreader added an extra '>' making it appear that's a quote
from me.

> is given by
> no. controls = 0.402 * hcp - 1.0308 which is an almost perfect fit.

I don't see why you call this an "almost perfect fit". Consecutive
entries in your table differ by .38 and .43, which are significantly
different from .402 [even accounting for rounding], for common values
10, 11, and 12. The relation isn't linear [affine] to two decimals.
An approximation A * hcp - B cannot be accurate to two decimals.

> the exact figures, to 2 dp, are shown below. If you want any other
> figures (such as ranges, or groups) happy to oblige.

If you think it is sufficient to report the number of expected
controls to 2 dp, why bother to use a formula involving 4 dp?
Given the variance and the accuracy, I don't think it makes sense to
bother with multiple digits of precision when my suggested
approximation works pretty well. Perhaps my opinion is biased.

I did the sim in part so I could estimate the variance. If you feel
like wasting more energy on this, exact measures of the variance or of
the probability that controls differ by more than something from
whichever approximation might be interesting. I don't know what the
best measure is. A 20 hcp hand averages 7.05 controls. Are hands
with 6 or 8 controls rare enough to be worth downgrading/upgrading, or
are they common enough to be considered typical? That's the sort of
question it would be nice to have a rule-of-thumb to answer.

I'm not focusing upon exact answers. I think what's useful is to have
simple tools that can be used to provide helpful insights at the table.


> HCP Expected Controls

> 10 2.94
> 11 3.37
> 12 3.75

> p.s. yes it's true that the distribution of number of jacks is
> virtually independent of point count.

That's pretty neat, isn't it?


Eddie
.

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