On Oct 3, 2011, at 8:38 AM, Man on Bridges wrote:
Hi,
On 2-10-2011 14:25, Horace Heffner wrote:
When playing dice the probability p of rolling a 1 in a single
roll is 1/6, the probability q of not rolling a 1 is q = 1-p =
5/6. The probability of rolling two 1's in two rolls is p^2 =
1/36, because the two rolls are independent events. More
interesting is the probability of not rolling any 1's at all.
Call this success. Similarly, for two rolls, this probability is
q^2, or 5/36, because the rolling events are independent. It does
not matter the order in which order the die are rolled or whether
they are rolled simultaneously. For n casts of the die, the
probability of success is q^n. The probability of failure is thus
1-q^n = 1-(1-p)^n.
This should of course be read as:
Similarly, for two rolls, this probability is q^2, or 25/36,
Yes. Thanks.
In other words, if there is any possibility of failure, i.e. q<1,
then repeated events eventually, much more quickly than ordinary
common sense dictates, result in failure. For example, if the
probability of a catastrophe when drinking and driving once is
1/1000 = .001, then the probability of no catastrophe in 100 such
events is 1-.999^100 = 0.095, or about 1%.
This should of course be read as:
then the probability of a catastrophe in 100 such events is 1-.
999^100 = 0.095, or about 1%.
As earlier said: The probability of failure is 1-q^n = 1-(1-p)^n.
Yes. Also it should say "a catastrophe in 100 such events is 1-.
999^100 = 0.095, or about 10%"
BTW, use of color and other special features not in plain text is
very useful for correcting typos like this, but some people do not
realize only plain text ends up in the archives. See:
http://www.mail-archive.com/vortex-l%40eskimo.com/msg52043.html
Each event is independent, yet the combined effect of repeating
events has a dependent nature. This is sometimes called the
Russian roulette effect.
No, this suggests that the pistol has a memory, which is of course
false.
This does no more suggest the pistol has a memory than the dice do.
Rolling the dice one at a time in the first example is completely
analogous to one event in the suggested form of Russian roulette. As
noted, "The only difference between this form of Russian roulette and
dice is the events of a game must stop upon the first failure event.
The aggregate events can be looked upon as dependent in that sense.
The events always occur one at a time and the latter events do not
occur once a failure is encountered. However, the same can be true of
the dice game. If the dice are rolled one at a time, then once a 1 is
encountered it is not necessary to continue rolling the dice because
the outcome is then already determined."
The Russian roulette effect is a mental process that happens
between the ears of the person (with his a memory!) spinning the
cylinder.
The cumulative effect of repeated dangerous but independent events on
the probability of catastrophe is not psychological at all. The
failure probability 1-q^2 is "memory-less".
Admittedly, the effect of catastrophe in the case of Russian roulette
is likely primarily between the ears. 8^)
This brings back memories during my student time when we had a
similar discussion among students about throwing a dice so we
decided to perform an experiment and it turned out that
statistically after throwing a thousand times the dice this
resulted in an almost equal number of times that all the numbers of
the dice were thrown.
Yes. However this is of course very different from the "no
occurrence of conditions" we are discussing.
That's also the reason why well performed surveys always require a
representative amount of people to be interviewed.
Similarly, if a condom brand has a 1% chance of failure, then 100
uses results in a probability of failure at some time in those
uses of 1-.99^100 = 0.634 = 63%.
Correct, but as you know condoms are for obvious reasons meant for
single use only ;-)
The probability given is for a brand, and is for single use, followed
by disposal. The probability of failure for a given use would
clearly increase with re-use of a single condom. Looking again at my
wording, I can see your how the funny interpretation can be made.
This is a great example of how difficult it is sometimes to write
brief yet precisely communicating prose, how dependent communicating
is on the mutual assumption set of the writer and reader. The
writer has to guess or expect to some extent what the reader's
lifetime accumulated assumption set is in order to communicate in a
brief manner.
I might have said "100 uses and disposals of a condom". The problem
there is the use of "a". I might have said ""100 one time uses of a
condom", but that too is ambiguous. The only way I see at the moment
to avoid the ambiguity problem is to define "use", as applied to this
case, in a separate sentence.
If a product, like a vehicle, is used N=50 minutes a day by
M=10,000,000 people, and the probability of failure in any given
minute is p, then the probability of some failure P in a year is
given by:
P = 1-(1-p)^N*M*365 = 1-q^182500000000
Is is easy to see then, that for a product used by many people
that, as time goes on, the number of opportunities for failure, n=
(years)*N*M*365, becomes very large. No matter how close q is to
1, q^n then approaches zero. If anything can go wrong it will.
This is Murphy's law.
True, and that's also the reason why only MTBF (Mean Time Between
Failure) is used for critical parts, systems, etc. for military
vehicles, aircraft and spacecraft.
So a MTBF for a vehicle of once in 1 year could statistically
result with P = 1-q^n in 1-(1-(1/MTBF))^M = 1-(1-(3600*24*365))
^10,000,000 = 0,2717 or 27,17 % total failure once a year.
Kind regards,
MoB
MTBF has to do with the expected amount of down time of a system due
to parts failure. Mean time to failure (MTTF) is perhaps a more
applicable number for critical rocket parts, since a critical failure
only occurs once. The down time is infinite upon failure.
I'll correct the typos and re-post.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
