Okay, so maybe flying *is* dangerous...

On 30 Aug 2005 08:45:23 -0700, wrote:

"Jim" wrote in message
.. .

It isn't dangerous to go skydiving (1-in-10000 chance of dying) once.
But "being a regular skydiver" where one jumps 100 or perhaps 1000
times in a lifetime gives you a much less trivial chance of being killed

Curious.

Not really.


Maybe I should have written "Curious to me."?


If each jump carries a 1-in-10000 chance of dying, wouldn't
the 1000th jump also carry a 1-in-10000 chance of dying?

Yes, assuming the jumper survived the first 9999 jumps.


Yes.

The probability of a person having successfully made 9999
jumps surviving his 10000th jump is very different (and less)
than the probability of a person who has made no jumps,
successfully making 10000 safe jumps.


Well, I just don't see this. I'll have to think on it some more.
I've been inclined to see each event as independent of and
not influenced by any preceding events.

The difference is that the first jumper's probability of the
first 9999 jumps are all 100% successfully, having been
made in the past.

If one flips a fair coin, over the long run there is a 1-in-2
chance of either side coming up.

Yes, but there isn't a 1-in-2 chance of flipping ten heads (say)
in a row. Except, of course, if one is improbable enough to
flip nine heads in a row, then the tenth head is 1-in-2.


I'm missing this one too. I may not have a very good grasp
of probability theory.

If one flips a fair coin 1 million times do the odds of
either side coming up change?

One is not just loooking at the last flip, one is looking at
the accumulation of *all* the flips. For instance, it's no
good surviving the 10000th jump, if you didn't survive
the 7359th. :-) :-( :-S

"Don Tuite" wrote:

Actuary's numbers relate to populations, not individuals.

Does the parachute know whether 10,000 jumpers made one
jump each, or whether one jumper made 10,000 jumps?

"Jim" wrote in message
...

The probability of a person having successfully made 9999
jumps surviving his 10000th jump is very different (and less)
than the probability of a person who has made no jumps,
successfully making 10000 safe jumps.

I've been inclined to see each event as independent
of and not influenced by any preceding events.

Well, in a case like this, it's not true. While each jump
taken in isolation has the same probability, the odds
of a successful 10,000th jump certainly *are* dependent
on having 9,999 successful jumps before hand.

If any of those previous 9,999 jumps are unsuccessful,
then the probability of a successful 10,000 jump is 0%.

Jeff Shirton jshirton at cogeco
dot ca

Keep thy airspeed up, lest the earth come from below
and smite thee. - William Kershner
Challenge me (Theophilus) for a game of chess at Chessworld.net!

On Tue, 30 Aug 2005 09:22:45 -0700, Jim wrote:

On 30 Aug 2005 08:45:23 -0700, wrote:

"Jim" wrote in message
. ..

It isn't dangerous to go skydiving (1-in-10000 chance of dying) once.
But "being a regular skydiver" where one jumps 100 or perhaps 1000
times in a lifetime gives you a much less trivial chance of being killed

Curious.

Not really.


Maybe I should have written "Curious to me."?


If each jump carries a 1-in-10000 chance of dying, wouldn't
the 1000th jump also carry a 1-in-10000 chance of dying?

Yes, assuming the jumper survived the first 9999 jumps.


Yes.

The probability of a person having successfully made 9999
jumps surviving his 10000th jump is very different (and less)
than the probability of a person who has made no jumps,
successfully making 10000 safe jumps.


Well, I just don't see this. I'll have to think on it some more.
I've been inclined to see each event as independent of and
not influenced by any preceding events.

The difference is that the first jumper's probability of the
first 9999 jumps are all 100% successfully, having been
made in the past.

If one flips a fair coin, over the long run there is a 1-in-2
chance of either side coming up.

Yes, but there isn't a 1-in-2 chance of flipping ten heads (say)
in a row. Except, of course, if one is improbable enough to
flip nine heads in a row, then the tenth head is 1-in-2.


I'm missing this one too. I may not have a very good grasp
of probability theory.

If one flips a fair coin 1 million times do the odds of
either side coming up change?

One is not just loooking at the last flip, one is looking at
the accumulation of *all* the flips. For instance, it's no
good surviving the 10000th jump, if you didn't survive
the 7359th. :-) :-( :-S

"Don Tuite" wrote:

Actuary's numbers relate to populations, not individuals.

Does the parachute know whether 10,000 jumpers made one
jump each, or whether one jumper made 10,000 jumps?

Well, maybe I see the difference between the probability of the
outcome of an individual event and the probability of the SEQUENCE of
outcomes of a SEQUENCE of the event. Does the following make
sense?

If one has a chance of surviving an event of 9-in-10 (to simplify a
little), and the outcome of one such event has no bearing on the
outcome of a following identical event, then each such event is
independent of others that precede it and each such individual event
carries a survival chance of 9-in-10. The odds for each independent
occurrence do not change.

But, the probability of a given outcome occurring in each of a
SEQUENCE of events, taken as a SEQUENCE, changes with each
repetition of the event.

Since we are stipulating that the events are independent of each
other, the probability of a given sequence of outcomes is calculated
as the product of the probabilities of each individual outcome.

If I haven't completely mangled this then, the probability of
surviving through TWO sequential occurrences of an event, each
occurrence of which carries a 9-in-10 probability, is:

.9 * .9 = .81

If one were to survive through these two trials and try a third
the odds of surviving all three would be:

.9 * .9 * .9 = .729

Doesn't look good for an event with a 1-in-10 chance of dying!

Have I got this sorted out?

If I haven't completely mangled this then, the probability of
surviving through TWO sequential occurrences of an event, each
occurrence of which carries a 9-in-10 probability, is:

.9 * .9 = .81

If one were to survive through these two trials and try a third
the odds of surviving all three would be:

.9 * .9 * .9 = .729

Doesn't look good for an event with a 1-in-10 chance of dying!

Have I got this sorted out?

Yep.

And you even got the part about multplying survival rates rather than
death rates (a sometimes subtle point - the sequence depends on multple
survivals, not multple deaths)

Jose
--
Quantum Mechanics is like this: God =does= play dice with the universe,
except there's no God, and there's no dice. And maybe there's no universe.
for Email, make the obvious change in the address.

Jim wrote:

Have I got this sorted out?

Yep. I knew this but couldn't figure out how to put it clearly. You did it very
well.

George Patterson
Give a person a fish and you feed him for a day; teach a person to
use the Internet and he won't bother you for weeks.

The probability of a person having successfully made 9999
jumps surviving his 10000th jump is very different (and less)
than the probability of a person who has made no jumps,
successfully making 10000 safe jumps.



Well, I just don't see this.

Think of it this way - the =reason= the probability "adds up" (so to
speak) is that the more jumps you make, the greater the chance that you
have already died from a jump in the past. (to imagine this as a
non-silly example, suppose you did make 10,000 jumps, either
self-propelled (while alive) or tossed out of the plane on a static line
(if already dead). They toss you out (dead) as many times as necessary
to bring the total jumps to 10,000 (because the pilot is paid by the
jump and he needs the money)

Jose
--
Quantum Mechanics is like this: God =does= play dice with the universe,
except there's no God, and there's no dice. And maybe there's no universe.
for Email, make the obvious change in the address.

wrote in message
oups.com...
"Jim" wrote in message
...

It isn't dangerous to go skydiving (1-in-10000 chance of dying) once.
But "being a regular skydiver" where one jumps 100 or perhaps 1000
times in a lifetime gives you a much less trivial chance of being killed

Curious.

Not really.

If each jump carries a 1-in-10000 chance of dying, wouldn't
the 1000th jump also carry a 1-in-10000 chance of dying?

Yes, assuming the jumper survived the first 9999 jumps.

The probability of a person having successfully made 9999
jumps surviving his 10000th jump is very different (and less)
than the probability of a person who has made no jumps,
successfully making 10000 safe jumps.



No, if the abolute odds of not surviving A jump are 1:10,000. The odds of
death are 1:10,000 on jump #1,#2,...#10000...#20000. The dice don't have a
memory. Which is exatly what you say below so I think you just mis-spoke
above.




The difference is that the first jumper's probability of the
first 9999 jumps are all 100% successfully, having been
made in the past.

If one flips a fair coin, over the long run there is a 1-in-2
chance of either side coming up.

Yes, but there isn't a 1-in-2 chance of flipping ten heads (say)
in a row. Except, of course, if one is improbable enough to
flip nine heads in a row, then the tenth head is 1-in-2.

If one flips a fair coin 1 million times do the odds of
either side coming up change?

One is not just loooking at the last flip, one is looking at
the accumulation of *all* the flips. For instance, it's no
good surviving the 10000th jump, if you didn't survive
the 7359th. :-) :-( :-S

"Don Tuite" wrote:

Actuary's numbers relate to populations, not individuals.

Does the parachute know whether 10,000 jumpers made one
jump each, or whether one jumper made 10,000 jumps?

--
Jeff Shirton jshirton at cogeco
dot ca

Keep thy airspeed up, lest the earth come from below
and smite thee. - William Kershner
Challenge me (Theophilus) for a game of chess at Chessworld.net!

"Gig 601XL Builder" wr.giacona@coxDOTnet wrote in message
news:ed0Re.6670$7f5.4709@okepread01...

The probability of a person having successfully made 9999
jumps surviving his 10000th jump is very different (and less)
than the probability of a person who has made no jumps,
successfully making 10000 safe jumps.

No, if the abolute odds of not surviving A jump are 1:10,000.
The odds of death are 1:10,000 on jump #1,#2,...#10000...
#20000. The dice don't have a memory.

Yes, but a jumper *does* have a memory.
A jumper cannot have a second jump *unless* the first jump was
successful, correct?

Above, I was comparing two jumpers, one who had 9999
jumps under his belt, and another who had 0 jumps under his
belt.

For the new jumper, his odds are 1:10,000 (if that is accurate)
for his first jump.

For the experienced jumper, his odds of surviving his *first*
jump are 100%, since he already survived his first jump. It
is no longer in the realm of "probability", it is now in the realm
of certainty, since it is in the unchangeable past.

To give another example that might make things more clear,
suppose we have two people:

1) One person is going to take a revolver, put one bullet in
the gun, and play "Russian Roulette" 1000 times.

2) A second person has already played (and survived) a
game of Russian Roulette 999 times, and only has to
play it for one more time.

The second person has a 5/6 chance of survival.

Do you honestly give the first person 5/6 chance of survival?
I would give him (without calculating precisely) somwhere
around 0.005 % chance of survival.

There is a difference.

--
Jeff Shirton jshirton at cogeco
dot ca

Keep thy airspeed up, lest the earth come from below
and smite thee. - William Kershner
Challenge me (Theophilus) for a game of chess at Chessworld.net!

wrote in message
oups.com...
"Gig 601XL Builder" wr.giacona@coxDOTnet wrote in message
news:ed0Re.6670$7f5.4709@okepread01...

The probability of a person having successfully made 9999
jumps surviving his 10000th jump is very different (and less)
than the probability of a person who has made no jumps,
successfully making 10000 safe jumps.

No, if the abolute odds of not surviving A jump are 1:10,000.
The odds of death are 1:10,000 on jump #1,#2,...#10000...
#20000. The dice don't have a memory.

Yes, but a jumper *does* have a memory.
A jumper cannot have a second jump *unless* the first jump was
successful, correct?

Above, I was comparing two jumpers, one who had 9999
jumps under his belt, and another who had 0 jumps under his
belt.

For the new jumper, his odds are 1:10,000 (if that is accurate)
for his first jump.

For the experienced jumper, his odds of surviving his *first*
jump are 100%, since he already survived his first jump. It
is no longer in the realm of "probability", it is now in the realm
of certainty, since it is in the unchangeable past.

To give another example that might make things more clear,
suppose we have two people:

1) One person is going to take a revolver, put one bullet in
the gun, and play "Russian Roulette" 1000 times.

2) A second person has already played (and survived) a
game of Russian Roulette 999 times, and only has to
play it for one more time.

The second person has a 5/6 chance of survival.

Do you honestly give the first person 5/6 chance of survival?
I would give him (without calculating precisely) somwhere
around 0.005 % chance of survival.

There is a difference.



The odds for an something to happen on any given roll,trigger pull or jump
don't change

"Jim" wrote in message
...
On 30 Aug 2005 06:01:00 -0700, wrote:


It isn't dangerous to go skydiving (1-in-10000 chance of dying) once.
But "being a regular skydiver" where one jumps 100 or perhaps 1000
times in a lifetime gives you a much less trivial chance of being
killed.

Curious.

If each jump carries a 1-in-10000 chance of dying, wouldn't the 1000th
jump also carry a 1-in-10000 chance of dying?

If one flips a fair coin, over the long run there is a 1-in-2 chance
of either side coming up. If one flips a fair coin 1 million times do
the odds of either side coming up change?

No. Gambler's Fallacy.

moo