How dangerous is soaring?

On 31 Oct, 11:32, 1LK wrote:
On Oct 31, 5:36 am, Ian wrote:
How old are you? A 1 in 80 chance of dying today means that you
have a

79 in 80 chance of making it to tomorrow, which is a (79/80)^365 = 1%
chance of making it through a year. I'll bet even 100 year olds have a
better survival rate than that ...

Ian

Well under 100. Ian (although not as far under as I'd like). It's a
multifactorial analysis and some of the factors that go into it are
personal, but, given the assumptions, it's probably a decent
characterization. The risk isn't additive, BTW; it remains about the
same day to day as long as the factors used to calculate it are
stable.

So you really do only have a 1% chance of being here next year? What
are you doing posting to Usenet - go flying. It's probably not worth
buying a flarm, by the way. Or a parachute.

Ian

On Oct 31, 8:09 am, Ian wrote:
On 31 Oct, 11:32, 1LK wrote:





On Oct 31, 5:36 am, Ian wrote:
How old are you? A 1 in 80 chance of dying today means that you
have a

79 in 80 chance of making it to tomorrow, which is a (79/80)^365 = 1%
chance of making it through a year. I'll bet even 100 year olds have a
better survival rate than that ...

Ian

Well under 100. Ian (although not as far under as I'd like). It's a
Not exactly. The odds ratio applies to any point in time; it's
neither cumulative or additive. I have a 98.75% chance of being here
tomorrow; on a day a year from now I'll have roughly the same odds of
being here a day after that.

I do wear a parachute, BTW, but again, given my age and condition,
it's not certain that I could extract myself successfully from a lawn
dart. I wear it because I'd rather be busy trying to use it, than
sitting on my hands waiting to hit the ground.

As you know, FLARM isn't usable here (US), but I'm planning to add a
transponder and an ELT this year. I take aspirin as well. Not sure
why I post to Usinet, perhaps the riskiest activity of all.

I control what I can and accept the rest and, yes, I think I'll go
flying.

Ray Warshaw
1LK

multifactorial analysis and some of the factors that go into it are
personal, but, given the assumptions, it's probably a decent
characterization. The risk isn't additive, BTW; it remains about the
same day to day as long as the factors used to calculate it are
stable.

So you really do only have a 1% chance of being here next year? What
are you doing posting to Usenet - go flying. It's probably not worth
buying a flarm, by the way. Or a parachute.

Ian- Hide quoted text -

- Show quoted text -

On 31 Oct, 19:11, 1LK wrote:

Not exactly. The odds ratio applies to any point in time; it's
neither cumulative or additive. I have a 98.75% chance of being here
tomorrow; on a day a year from now I'll have roughly the same odds of
being here a day after that.

If you have a 98.75% chance of being here tomorrow, then you have a
98.75% x 98.75% chance of being here a day after that, a 98.75% x
98.75% x 98.75% chance of seeing Sunday and so on.

Sure, /if/ you make it to 1st November 2008 you have a 98.75% chance
of making it to 2nd ... but there's only a 1% chance, on these
figures, that you'll put it to the test.

Ian

The calculation which yields the 80/1 is only true for the single
instance; my odds of being here next year are another thing entirely.
For that you need to do mortality computations.

Ray

On 1 Nov, 14:07, 1LK wrote:
The calculation which yields the 80/1 is only true for the single
instance; my odds of being here next year are another thing entirely.
For that you need to do mortality computations.

I'm having real difficulty understanding this. Are you saying that ...

1) There is a 1/80 chance you'll be dead by this time tomorrow and
that

2) There is a 1/80 chance you'll be dead by this time the day after?

Ian

On Oct 31, 7:11 pm, 1LK wrote:
Not exactly. The odds ratio applies to any point in time; it's
neither cumulative or additive. I have a 98.75% chance of being here
tomorrow; on a day a year from now I'll have roughly the same odds of
being here a day after that.

Think of it like tossing a coin to see whether you live. The
probability
of each toss being "heads" is 0.5, and is independent of the previous
results.

To stay alive for a week, you have to toss "heads" 7 times in a row,
and the probability of that is 0.5 ^ 7 = 0.078125 = 1 in 128

Personally, I hope you have more than a 0.9875 ^ 365 chance of
being alive in a year. (i.e. 0.01014 or 1 in 98).

Of course, if you are still alive in a year, there would be the same
chance (0.9875) that you would be alive in a year and a day.

results.

To stay alive for a week, you have to toss "heads" 7 times in a row,
and the probability of that is 0.5 ^ 7 = 0.078125 = 1 in 128

It's not binary, it's multifactorial.

Ray

On Nov 1, 2:11 pm, 1LK wrote:
The calculation which yields the 80/1 is only true for the single
instance;

Single instance of what? If it is the "single instance of a day",
then the calculations are correct.

my odds of being here next year are another thing entirely.
For that you need to do mortality computations.

To stay alive for a week, you have to toss "heads" 7 times in a row,
and the probability of that is 0.5 ^ 7 = 0.078125 = 1 in 128

It's not binary, it's multifactorial.

I don't understand: what do you mean by "it" and "multifactorial"?
Binary? Well yes, flipping a coin is binary; that's why I
subsequently
used your figures (that you didn't bother to include).

It might help if you could explain the reasons (based on an
equivalent
example, if you prefer) why you believe that the calculations are
wrong.
Examples I can think of are
- it is not a 1.25% chance of dying on every day, only on some days
- each day shouldn't be treated as independent from the preceding
days (but that doesn't fit with your original statement)

Anyway, I am glad that your mortality isn't as imminent as it
at first appeared.

Tom's calculations indicate how hard it is for the human mind to grasp
probability, and thus why we cannot calculate risk properly.

Tom's coin analogy fails because he is looking for an unbroken sequence
of survival, which therefore takes into account the past in predicting
the future. His calculations are cumulative. Even with coin tosses, we
can see that once we ignore the past and stop cumulating results, the
calculation changes.

Thus, at the start of the week, the chance of survival for a week at
coin toss levels is 1 in 128. The chance of surviving for 8 days is
worse, at 1 in 256. However, if our subject survives day 1, his chance
of making day 8 increases to 1 in 128, and by the end of day 7 it has
risen to 50:50. The older he gets, the longer his chances of living
forever! I think (but as a European writing after what UK
government-sponsored has recently described as a "hazardous" level of
wine consumption I cannot be sure) this may be related to Zeno's paradox
(in Tom Stoppard's words, "... thus proving that the arrow never reaches
it's target and Saint Sebastian died of fright").

If we ignore the past, however, each day's chance is the same at 0.5.
Thus Ray (may he live forever) is able to state that next year his
chances will be pretty much the same, if he makes it that far.

Cumulation of probabilities is what the human brain does automatically.
Suppose the chance of being killed on a glider flight is 1 in 1,000. The
mind (without extensive training) deals with this in a number of ways:

1. I can fly safely 999 times, then have to give up or I will certainly
die on the 1,000th. If I'm already dead, I was "statistically" unlucky.

2. I've had 500 flights, so my risk level has risen to 50:50.

3. At my club we fly 1,000 flights a year between us, so one of us is
sure to die flying.

Unless I'm badly mistaken, none of these are true statements.

I try to think as follows:

a. In the UK where I fly, gliding fatalities are on average around 2.5
per annum out of 5,000 pilots, so my "statistical" risk is around 1 in
2,000 of dying through gliding each year.

b. I can do a number of things to reduce my personal risk to less than 1
in 2,000, so I'll try to do those things.

c. This is, to me, an acceptable level of risk for the pleasure I get
from gliding.

The good thing is that these probabilities are not cumulative. I've been
flying for 11 years, so if they were cumulative my "statistical" risk
might be down to under 1 in 20. It ain't.

What can be cumulative are personal mistakes - careless rigging, no
positive check, lack of sleep, etc. etc. These are the things I worry about.


Tom Gardner wrote:
On Nov 1, 2:11 pm, 1LK wrote:
The calculation which yields the 80/1 is only true for the single
instance;

Single instance of what? If it is the "single instance of a day",
then the calculations are correct.

my odds of being here next year are another thing entirely.
For that you need to do mortality computations.

To stay alive for a week, you have to toss "heads" 7 times in a row,
and the probability of that is 0.5 ^ 7 = 0.078125 = 1 in 128
It's not binary, it's multifactorial.

I don't understand: what do you mean by "it" and "multifactorial"?
Binary? Well yes, flipping a coin is binary; that's why I
subsequently
used your figures (that you didn't bother to include).

It might help if you could explain the reasons (based on an
equivalent
example, if you prefer) why you believe that the calculations are
wrong.
Examples I can think of are
- it is not a 1.25% chance of dying on every day, only on some days
- each day shouldn't be treated as independent from the preceding
days (but that doesn't fit with your original statement)

Anyway, I am glad that your mortality isn't as imminent as it
at first appeared.

On Nov 1, 10:46 pm, Chris Reed wrote:
Tom's calculations indicate how hard it is for the human mind to grasp
probability, and thus why we cannot calculate risk properly.

Tom's coin analogy fails because he is looking for an unbroken sequence
of survival,

Well, yes. If I die on Tuesday, it would seem somewhat
optimistic to assume that I'll be alive on Wednesday!
Am I missing something?

which therefore takes into account the past in predicting
the future.

Er, no. Conditional probability and all that!

His calculations are cumulative. Even with coin tosses, we
can see that once we ignore the past and stop cumulating results, the
calculation changes.

Er, for the conditions I stated, no.

Thus, at the start of the week, the chance of survival for a week at
coin toss levels is 1 in 128. The chance of surviving for 8 days is
worse, at 1 in 256. However, if our subject survives day 1, his chance
of making day 8 increases to 1 in 128, and by the end of day 7 it has
risen to 50:50.

True, but missing the point.

The chance of getting to day 7 from day 1 is 1 in 128, so
the chance of getting to day 8 from day 1 is still 1 in 256.
No change.

No one is disputing if you've reached day 7 then the
chance of getting to day 8 is 1 in 2. Conditional probability, etc.

But on day N the chance of getting to day N+356 is
vanishingly small.

The older he gets, the longer his chances of living
forever! I think (but as a European writing after what UK
government-sponsored has recently described as a "hazardous" level of
wine consumption I cannot be sure) this may be related to Zeno's paradox
(in Tom Stoppard's words, "... thus proving that the arrow never reaches
it's target and Saint Sebastian died of fright").

Rats. You took the words right out my mouth! Leibnitz
and Newton also had a few things to say in this area

If we ignore the past, however, each day's chance is the same at 0.5.
Thus Ray (may he live forever) is able to state that next year his
chances will be pretty much the same, if he makes it that far.

Quite correct. But of course we are actually talking about the
chance of him getting there (which would seem to be unfortunately
small based on his statements).

Cumulation of probabilities is what the human brain does automatically.
Suppose the chance of being killed on a glider flight is 1 in 1,000. The
mind (without extensive training) deals with this in a number of ways:

1. I can fly safely 999 times, then have to give up or I will certainly
die on the 1,000th. If I'm already dead, I was "statistically" unlucky.

2. I've had 500 flights, so my risk level has risen to 50:50.

3. At my club we fly 1,000 flights a year between us, so one of us is
sure to die flying.

Unless I'm badly mistaken, none of these are true statements.

Correct (except under pathologically perverse circumstances

I try to think as follows:

a. In the UK where I fly, gliding fatalities are on average around 2.5
per annum out of 5,000 pilots, so my "statistical" risk is around 1 in
2,000 of dying through gliding each year.

b. I can do a number of things to reduce my personal risk to less than 1
in 2,000, so I'll try to do those things.

c. This is, to me, an acceptable level of risk for the pleasure I get
from gliding.

The good thing is that these probabilities are not cumulative. I've been
flying for 11 years, so if they were cumulative my "statistical" risk
might be down to under 1 in 20. It ain't.

I think the concept of "cumulative" is seriously misleading
in this context.

I think what you're really trying to say is that the probability
of dying on day X from cause Y is *not independent* of the
probability of dying on day X+1 from cause Y.

Under such conditions the "1 in P^N" calculation is clearly
and simply invalid.

In the absence of other information, I chose to presume
"independent" and you have chosen "not independent".