CF card fatality

On 12/26/2014 5:36 PM, Tony Cooper wrote:
On Fri, 26 Dec 2014 16:42:06 -0500, PeterN wrote:

On 12/26/2014 12:14 PM, Tony Cooper wrote:
On Fri, 26 Dec 2014 11:13:56 -0500, PeterN wrote:

On 12/25/2014 9:34 PM, Tony Cooper wrote:


snip


Doesn't make a difference. The underlying principle is the same. The
probability of there being defective rounds in the universe of all
rounds doesn't change the fact that the gun will either fire or not
fire. Two outcome possibility.

Let me put it this way.
Take a pistol with six chambers: put one round in one chamber; spin the
cylinder; point pistol downrange and pull the trigger.
Now follow your theory of two possibilities, there is a bullet in the
chamber, or there isn't.
So is the gun fires I will give you a dollar. If the gun does not fire,
you give me a dollar. I'll make it better. I will give you $3 if the gun
fires.
Would you take that bet?

Only if you state the bet the way I have stated my example: If the
gun fires or does not fire, you give me $1.00. If neither of those
two outcomes is the result, I will give you $3.00.

Look above: fire or not fire.


Fire or not fire, is an irrelevant response to the underlying question.

I don't know what your underlying question is. I don't see a question
in your post.

My example used two different results: work or fail. It has to be
one of those two. In this case, "fire" is work and "not fire" is
fail.

So, to bet following my example, I have to pay if one of those two
results does not occur. I would never have to pay.


the underlying question is: What is the statistical probality, for each
of the two possibilities.

--
PeterN

On 12/26/2014 9:05 PM, Bill W wrote:
On Fri, 26 Dec 2014 20:09:36 -0500, PeterN wrote:

On 12/26/2014 2:50 PM, Bill W wrote:
On Fri, 26 Dec 2014 11:17:12 -0800, John Navas
wrote:

Nope. That's a misuse of "odds", as I've already explained. Ignoring
that, the correct calculation is actually 2:0, which is meaningless.

And one reason that it's meaningless is that you can substitute any
number, including infinity, for the 2. Odds expressions are derived
from the probability of an event, and are mostly useless for anything
except determining the value of a bet.


Are you saying that probability is useless for determining the size of a
sample?

I think you might have worded that question wrong, and I'm not sure
what you meant to ask. By "odds expression", I meant stating the
probability of an event in the form of odds (x:y), as opposed to
probability, a value between 0 and 1.

A further point of confusion for lots of people is that they think 3:1
odds also means one chance in three, when it's actually one in four.
If you're evaluating a proposed bet, it's easier to use the odds form,
but probability is easier to understand for everything else.

It would be an fun exercise to try to determine who has a better
approach to camera settings for moving objects - the issue that Floyd
and Savageduck are discussing. The question might be which approach
would get you a greater number of usable shots when you have X seconds
of access to the object. It's "certain to be in focus" shots, vs.
total number of shots in the continuous burst, minus those that are
OoF. It's going to depend on the continuous burst rate, and the
ability of the AF system in that situation.

I think it's an interesting question - some of those continuous burst
shots are going to be good enough, even if the camera didn't determine
that they were in focus. Think of a situation where you twitched, and
the target slipped outside of the focus point area. If you have the
camera set so the lens doesn't start immediately hunting, the shot
might still be in perfect focus.



A high burst mode doesn't apply to my D800. It applies to my D300, which
I only use as an emergency backup. Having said that, I agree it is an
interesting topic.
When shooting critters one does not usually have time to change settings
in the field.
e.g. in Alaska on a foggy day I was shooting Denali using manual focus.
A timber wolf appeared, carrying a hare in his mouth. There was no time
to focus or switch to manual focus. Of course I missed the shot.



--
PeterN

On Tue, 23 Dec 2014 16:43:00 -0500, Tony Cooper
wrote:
: On Tue, 23 Dec 2014 16:21:04 -0500, nospam
: wrote:
:
: In article 2014122312054283784-savageduck1@REMOVESPAMmecom,
: Savageduck wrote:
:
: true, but the more cards you have the more likely one of them will fail.
:
: No, the odds remain the same. A card will fail or won't fail. With
: each additional card purchased, the odds remain at 1:1.
:
: nope. the odds of a failure is higher for the more cards you have.
:
: go back to math class. you failed the last time around.
:
: Not quite. The odds do not change for individual cards, in much the
: same way the odds don't change if you buy more lottery tickets.
: The potential for failure is the same for a single card as it is for 10
: cards. Each stands alone. Buy another card of the same type, size, and
: quality, from the same manufacturer and you have two cards with exactly
: the same potential risk for failure. The odds have not change one bit.
:
: the odds for a *particular* card to fail does not change, but the odds
: of *any* card failing is higher with the more cards you have. simple
: math.
:
: Nope. You still haven't managed to phrase your point correctly.
: "Any card" is still subject to the 1:1 odds of being good or bad.
:
: What you are so abjectly failing to do is get across that the more
: cards you own, the more likely you are to have a card that will fail.
: I know this is over your head, but if you can't articulate what you
: are trying to point out, you fail.
:
: If you were a reasonable person, this is where you'd say "I was wrong.
: I misstated my point". However, this where nospam says "you're
: twisting things".
:
: exactly the point. the more cards you have the more chances of a card
: to fail, just like a lottery ticket.
:
: No, the more cards you have the more chances you will have a card that
: will fail. That is not the same as saying "the more chances of a card
: to fail". The odds remain at 1:1.

There's no way I'm going to slog through this whole thread, but I'll make one
point in case no one else has brought it up:

If all of your cards have the same probability of failure, and if you
distribute your pictures evenly across any number of cards, what stays the
same is the expected value of your loss. What changes is how close you're
likely to come to that expected value. Suppose the probability of failure is
10% and you have 1000 pictures. The expected value of your loss is 100
pictures, no matter how many cards you use. The difference is that if you put
all 1000 pictures on the same card, there's a 10% probability that you'll lose
everything and a 90% probability that you'll lose nothing; while if you put
one picture on each of 1000 cards, there's a high probability that you'll lose
100 pictures. (You can calculate that probability exactly, as well as the
probabilities that you'll lose 99 or 101 or whatever. But why bother? It has
little effect on the argument.) So by using more cards, you're spreading the
risk, increasing the probability that you'll lose something, but decreasing
the probability that you'll lose everything.

Bob

On 1/3/2015 2:35 AM, Robert Coe wrote:
On Tue, 23 Dec 2014 16:43:00 -0500, Tony Cooper
wrote:
: On Tue, 23 Dec 2014 16:21:04 -0500, nospam
: wrote:
:
: In article 2014122312054283784-savageduck1@REMOVESPAMmecom,
: Savageduck wrote:
:
: true, but the more cards you have the more likely one of them will fail.
:
: No, the odds remain the same. A card will fail or won't fail. With
: each additional card purchased, the odds remain at 1:1.
:
: nope. the odds of a failure is higher for the more cards you have.
:
: go back to math class. you failed the last time around.
:
: Not quite. The odds do not change for individual cards, in much the
: same way the odds don't change if you buy more lottery tickets.
: The potential for failure is the same for a single card as it is for 10
: cards. Each stands alone. Buy another card of the same type, size, and
: quality, from the same manufacturer and you have two cards with exactly
: the same potential risk for failure. The odds have not change one bit.
:
: the odds for a *particular* card to fail does not change, but the odds
: of *any* card failing is higher with the more cards you have. simple
: math.
:
: Nope. You still haven't managed to phrase your point correctly.
: "Any card" is still subject to the 1:1 odds of being good or bad.
:
: What you are so abjectly failing to do is get across that the more
: cards you own, the more likely you are to have a card that will fail.
: I know this is over your head, but if you can't articulate what you
: are trying to point out, you fail.
:
: If you were a reasonable person, this is where you'd say "I was wrong.
: I misstated my point". However, this where nospam says "you're
: twisting things".
:
: exactly the point. the more cards you have the more chances of a card
: to fail, just like a lottery ticket.
:
: No, the more cards you have the more chances you will have a card that
: will fail. That is not the same as saying "the more chances of a card
: to fail". The odds remain at 1:1.

There's no way I'm going to slog through this whole thread, but I'll make one
point in case no one else has brought it up:

If all of your cards have the same probability of failure, and if you
distribute your pictures evenly across any number of cards, what stays the
same is the expected value of your loss. What changes is how close you're
likely to come to that expected value. Suppose the probability of failure is
10% and you have 1000 pictures. The expected value of your loss is 100
pictures, no matter how many cards you use. The difference is that if you put
all 1000 pictures on the same card, there's a 10% probability that you'll lose
everything and a 90% probability that you'll lose nothing; while if you put
one picture on each of 1000 cards, there's a high probability that you'll lose
100 pictures. (You can calculate that probability exactly, as well as the
probabilities that you'll lose 99 or 101 or whatever. But why bother? It has
little effect on the argument.) So by using more cards, you're spreading the
risk, increasing the probability that you'll lose something, but decreasing
the probability that you'll lose everything.


Haven't hear from you for a while. Hope you and Martha are well.

HNY.



--
PeterN

On 1/2/2015 11:35 PM, Robert Coe wrote:

If all of your cards have the same probability of failure,...

Unfortunately, semiconductor fabrication and physics come into play
here, so that is not necessarily a valid presumption when comparing
larger versus smaller capacity cards.

Data retention and endurance decrease with process shrinks so higher
density, higher capacity, memory cards have a greater chance of losing
data than lower density memory cards.

In some cases both lower and higher capacity cards are made with the
same fab process. In that case the higher capacity cards tend to have an
advantage because of the wear-leveling performed by the controller, if
they are not used to their full capacity. A 32GB card of which only 16GB
is used will have about twice as many cycles per cell as a 16GB card
where all 16GB is used.

Practically speaking, most amateurs shoot JPEGs and transfer their
photos to their computer then delete photos or re-format the memory card
and start over. They aren't doing editing directly on the memory card.
They probably aren't filling the card unless they are shooting a lot of
video. They are never going to reach the cycle limit of the flash. The
bad thing to do is to store all your photos only on flash memory cards
and expect the flash to be a long-term storage medium.

Where flash endurance is a big issue is in industrial equipment but
there are industrial quality flash cards for these applications
(generally CF). They are about 10X the cost of CF for cameras. We are
very careful at work to design our software to minimize flash usage
because this equipment has a long life cycle and flash is soldered down.

On Sat, 03 Jan 2015 08:40:17 -0800, John Navas
wrote:
: In Article
: on Sat, 03 Jan 2015 02:35:11 -0500, Robert Coe wrote:
:
: On Tue, 23 Dec 2014 16:43:00 -0500, Tony Cooper
: wrote:
: : No, the more cards you have the more chances you will have a card that
: : will fail. That is not the same as saying "the more chances of a card
: : to fail". The odds remain at 1:1.
:
: There's no way I'm going to slog through this whole thread, but I'll make one
: point in case no one else has brought it up:
:
: If all of your cards have the same probability of failure, and if you
: distribute your pictures evenly across any number of cards, what stays the
: same is the expected value of your loss. What changes is how close you're
: likely to come to that expected value. Suppose the probability of failure is
: 10% and you have 1000 pictures. The expected value of your loss is 100
: pictures, no matter how many cards you use. The difference is that if you put
: all 1000 pictures on the same card, there's a 10% probability that you'll lose
: everything and a 90% probability that you'll lose nothing; while if you put
: one picture on each of 1000 cards, there's a high probability that you'll lose
: 100 pictures. (You can calculate that probability exactly, as well as the
: probabilities that you'll lose 99 or 101 or whatever. But why bother? It has
: little effect on the argument.) So by using more cards, you're spreading the
: risk, increasing the probability that you'll lose something, but decreasing
: the probability that you'll lose everything.
:
: That's assuming failure results in the loss of all pictures on a card,
: but there are also failures that lose only some of the pictures on a
: card, so it's actually more complicated than that, with a lower risk
: of complete loss.

True, of course. If you assume that cards have partial failures (and they do,
of course, in real life), then you'll get different probabilities. But most
plausible failure modes favor spreading a shoot over multiple cards.

Bob