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?