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The Sky Really Did Fall 100 Years Ago Today
Today is the 100th anniversary of the Tunguska event which blasted hundreds of square miles of Siberia.
Don Yeomans, manager of the Near-Earth Object Office at the Jet Propulsion Laboratory tells Science@NASA:
"A century later some still debate the cause and come up with different scenarios that could have caused the explosion," said Yeomans. "But the generally agreed upon theory is that on the morning of June 30, 1908, a large space rock, about 120 feet across, entered the atmosphere of Siberia and then detonated in the sky."
It is estimated the asteroid entered Earth's atmosphere traveling at a speed of about 33,500 miles per hour. During its quick plunge, the 220-million-pound space rock heated the air surrounding it to 44,500 degrees Fahrenheit. At 7:17 a.m. (local Siberia time), at a height of about 28,000 feet, the combination of pressure and heat caused the asteroid to fragment and annihilate itself, producing a fireball and releasing energy equivalent to about 185 Hiroshima bombs.
"That is why there is no impact crater," said Yeomans. "The great majority of the asteroid is consumed in the explosion."
In my 2005 column, "Earth Killers from Outer Space," I noted:
...the probability of such a catastrophic asteroid strike is very small. Researchers at NASA's Spaceguard Survey estimate that a hit similar in size to the 1908 Tunguska event in Siberia occurs once every 300 years and would take out an area of about 5,000 square kilometers (1/100,000th of the Earth's surface). Huge portions of the planet are uninhabited (ocean) or sparsely inhabited, so Spaceguard calculates that another Tunguska is apt to hit a large urban area about once every 100,000 years. Ultimately the researchers calculate that the annual probability of an individual's death from a Tunguska-type impact is 1 in 30 million.
Whole NASA story on Tunguska here.
Guy Montag|6.30.08 @ 4:10PM|#
"A century later some still debate the cause and come up
with different scenarios that could have caused the explosion,"
said Yeomans."
PAAAALEEZE! It was the Ancient Astronauts. This has been settled
science since the 1970s when Dr. Erich Anton Paul von Däniken
examined the evidence and explained it to the world.
|6.30.08 @ 4:20PM|#
the annual probability of an individual's death from a
Tunguska-type impact is 1 in 30 million
So, with a population of six billion, the statistical average
global death rate from such impacts is 200. How many people die
each year from shark attacks, again?
Neu Mejican|6.30.08 @ 4:25PM|#
Thomas Pynchon's latest, Against the Day has a nice
fictional account of the incident...
Read that section just a couple of days ago...
Taktix®|6.30.08 @ 4:27PM|#
I seriously doubt this thing was caused by Asteroids (or Jesus tears, as I like to call them). Clearly, someone was trying to teach evolution for the first time, and, well... we've been warned...
|6.30.08 @ 4:31PM|#
"Ultimately the researchers calculate that the annual
probability of an individual's death from a Tunguska-type impact is
1 in 30 million..."
The California Lottery says that the odds of winning the big
jackpot are 1 in 41,416,353. So you are more likely to be struck
from the sky by a killer asteroid than to be the big lottery
winner. Lots of people seem encouraged by the thought that they
will win the big prize, but by the same token, they should be
deathly afraid that a big space rock will never let them spend
it.
|6.30.08 @ 4:31PM|#
Um, it would be nice if we could, I dunno, prevent an asteroid or comet from hitting the Earth and, you know, killing people and stuff. Rather than leave it to chance.
Nigel Watt|6.30.08 @ 4:32PM|#
The California Lottery says that the odds of winning the big jackpot are 1 in 41,416,353...
Yeah, but that's weekly/monthly, so you multiply it by 52 or 12 to
get the annual probability.
Taktix®|6.30.08 @ 4:40PM|#
The California Lottery says that the odds of winning the big
jackpot are 1 in 41,416,353...
Yeah, but that's weekly/monthly, so you multiply it by 52 or 12 to
get the annual probability.
Wrong, that's the gambler's fallacy. No many how many chances you
have, the probability is still the same.
Every given coin flip has a 50% chance of landing on heads, no
matter how many times you flip it.
Nigel Watt|6.30.08 @ 4:43PM|#
Every given coin flip has a 50% chance of landing on heads, no matter how many times you flip it.
This helps explain why I did poorly in statistics.
|6.30.08 @ 4:44PM|#
And if you flip a coin 52 times, Taktix, your chance of getting
one or more heads is much greater than 50/50.
The Gambler's Fallacy is about thinking that past events change the
odds of this particular roll, not over the course of several
events.
If you play the lottery every week, your chances of winning once
that year are 52x those of someone who plays just once.
|6.30.08 @ 4:48PM|#
The California Lottery says that the odds of winning the big
jackpot are 1 in 41,416,353...
Yeah, but that's weekly/monthly, so you multiply it by 52 or 12
to get the annual probability.
Wrong, that's the gambler's fallacy. No many how many chances
you have, the probability is still the same.
Every given coin flip has a 50% chance of landing on heads, no
matter how many times you flip it.
But you have a 3/4 chance to get heads once in 2 flips, a 7/8
chance to get heads once in 3 flips and so on. So while the
relationship isn't strictly multiplicative, your odds do improve
the more times you play.
That said, because of the payout ratio, you'll most likely lose
money in the process.
kinnath|6.30.08 @ 4:48PM|#
Each independent event has a given probability. Depending on
"or" versus "and" you add or multiple when combining
probabilities.
If you can win the lottery this week or the next week or the next
week or etc, the you add the probabilities together. However, 52 in
40 million ain't that much better that 1 in 40 million.
If you wanted the probability of holding the wining ticket that day
that an asteroid hit, then you would multiply the probabilities: 1
in 40 million times 1 in 30 million would be 1 in 120 million
million (british for trillion)
Salvius|6.30.08 @ 4:48PM|#
Wrong, that's the gambler's fallacy.
No, what Nigel is saying is that you're comparing a weekly
probability with an annual probability. If you play the lottery
every week, that's 52 iterations per year, and multiple iterations
increases the likelihood that at least one of them will be
a win.
Every individual coin flip has a 50% chance of landing on heads,
but if you flip 52 times in a row, the chances are much higher than
50% that it will come up heads at least one of those times.
|6.30.08 @ 4:49PM|#
What is the Bush Administration doing to protect us from this threat? Haliburton and Cheney will get rich off the fallout from a meteor strike. In fact, they are probably planting explosives all over the world so when it hits, they can make it even more devastating. This would never have happened under the Clinton administration. By the way, what does the LP's platform say?
|6.30.08 @ 4:50PM|#
Joe is right, Taktix is wrong: if you play the lottery weekly,
and your odds of winning each week are 1:41,416,353, then if you
play every week of the year, your odds of winning at least once in
the course of the year are 0.0001% in the year (which is a bit less
than 52 * 1/41,416,353).
However, playing the lottery weekly costs your $52 instead of $1.
The cost/benefits do NOT change when you repeat, since you pay more
for your higher chance of winning.
|6.30.08 @ 4:50PM|#
Lots of people seem encouraged by the thought that they will win the big prize, but by the same token, they should be deathly afraid that a big space rock will never let them spend it.
I know you're being facetious, but calculate for me, if you will, the odds of winning the California State Lottery and then being killed by an asteroid.
kinnath|6.30.08 @ 4:52PM|#
By the way, the gambler's fallacy would be if you have already flipped heads 49 straight times, what are the odds the next one would be heads -- answer 50/50; history is not relative to future odds.
kinnath|6.30.08 @ 4:53PM|#
but calculate for me, if you will, the odds of winning the
California State Lottery and then being killed by an
asteroid.
Already done above.
kinnath|6.30.08 @ 4:56PM|#
. . . history is not relative
relavent to future odds.
so close, but so far . . . .
Taktix®|6.30.08 @ 4:57PM|#
Yeah, I never took statistics, so I am quite wrong.
Sorry, it even happens to me sometimes.
However, I ain't gettin' hit by no damn Jesus-Rock today!
robc|6.30.08 @ 5:00PM|#
kinnath,
If you flip tails 49 straight times, Im checking your coin before I
assume 50/50 on the next flip.
Occam's razor says fraud.
kinnath|6.30.08 @ 5:02PM|#
Yeah, I never took statistics,
Me neither, but you might want to learn some basic probabilities if
you play any poker ;-)
|6.30.08 @ 5:06PM|#
2^49 may be improbable, but it's not impossible.
But a one-in-a-quadrillion event is far, far, far, ..., far more
suggestive of an unfair coin than of a fair one.
|6.30.08 @ 5:07PM|#
If you flip tails 49 straight times, Im checking your coin before I assume 50/50 on the next flip.
Occam's razor says fraud.
Someone, somewhere flipped heads 49 straight times.*
*sez intuition.
|6.30.08 @ 5:09PM|#
1/2^49 is not impossible, no, but it's considerably less likely
than winning the lottery. (1.8*10^-15)
So, you know, checking that coin would definitely be the way to go.
That said, kinnath correctly stated the Gambler's Fallacy, plus or
minus some typos and an extreme example.
|6.30.08 @ 5:18PM|#
You'll have to forgive my typos. I'm tired as fuck, and I'm still sticking by my power of 15 slur.
|6.30.08 @ 5:19PM|#
>>2^49 may be improbable, but it's not
impossible.
>Someone, somewhere flipped heads 49 straight times.*
*sez intuition.
Heh. No way. Impossible in all but theory. If you flipped 49 coins
every day it ought to come up heads every time once in about
1.54*10^12 years, or more than 100 times the current estimated of
the age of the universe.
|6.30.08 @ 5:23PM|#
But, again, one is per year and one is per week. The odds that
both happen in one year is 1 in 24 trillion. The odds that both
happen in one week is 1 in 1.2 quadrillion.
So the probability of winning the lottery and being hit by an
asteroid in the same fortnight are about the same as the
probability of getting tails on a fair coin 49 straight times.
kinnath|6.30.08 @ 5:24PM|#
But, again, one is per year and one is per week.
Sorry, wasn't playing that close attention.
|6.30.08 @ 5:26PM|#
Heh. No way. Impossible in all but theory. If you flipped 49 coins every day it ought to come up heads every time once in about 1.54*10^12 years, or more than 100 times the current estimated of the age of the universe.
One person or every person flippin'? Still, yeah I've never observed anything even close to that happening before.
|6.30.08 @ 5:28PM|#
Are the odds of an asteroid/comet hitting a particular spot on the Earth actually evenly distributed? In other words, is it actually true that an object is equally likely to hit any given location on the planet as any other?
|6.30.08 @ 5:34PM|#
In other words, is it actually true that an object is
equally likely to hit any given location on the planet as any
other?
Trailer parks are more likely to be hit.
|6.30.08 @ 5:35PM|#
The odds that both happen in one week is 1 in 1.2
quadrillion.
This is not right. It's actually 1 in 60 quadrillion. So getting 49
tails in a row (in one attempt) is about as likely as winning the
lottery and getting hit by an asteroid in the same calendar
quarter.
Neu Mejican|6.30.08 @ 5:36PM|#
Let me think about this....in a truly random process, over time,
isn't the likelihood of the sequence
xxxxxxxxxx
xxxxxxxxxx
xxxxxxxxxx
xxxxxxxxxx
xxxxxxxxxY
the same as it is for any other 50 iteration long string?
Why would you feel the need to check your coin?
It would be more troublesome if after a very large number of coin
tosses this sequence never came up.
No?
Neu Mejican|6.30.08 @ 5:41PM|#
In other words, is it actually true that an object is
equally likely to hit any given location on the planet as any
other?
Trailer parks are more likely to be hit.
And the safest place is in bed...under the covers.
|6.30.08 @ 5:42PM|#
MikeP,
So if I do a Tom Stoppard and flip that fiftieth consecutive heads,
I can expect to simultaneously win the lottery and die from an
asteroid strike? Looks like I picked a good week to stop flipping
coins.
As for my unanswered question, I was just thinking that asteroids
and comets might have a tendency to strike some parts of the planet
more than others, given the Earth's axial tilt, etc.
|6.30.08 @ 5:45PM|#
Well, each coin toss is an individual event and the "meta-event" of 50 tosses is a mental construct which can be represented by a permutation but is indeed 50 seperate events with only two possible outcomes.
Neu Mejican|6.30.08 @ 5:48PM|#
A study looking at the estimation of Tunguska-type events.
http://impact.arc.nasa.gov/news_detail.cfm?ID=179
...If this revision (down to an estimated energy of 3-5
megatons, and a corresponding diameter perhaps as low as 40 m) is
correct, the expected frequency of such impacts changes, from once
in a couple of millennia to once in a few hundred years. If smaller
impactors can do the damage previously associated with larger ones,
of course, the total hazard from such impacts is
increased....
Art-POG:
Meaning...?
|6.30.08 @ 5:50PM|#
Let me think about this....in a truly random process, over
time, isn't the likelihood of the sequence
xxxxxxxxxx
[ . . .]
the same as it is for any other 50 iteration long
string?
Yes.
Why would you feel the need to check your coin?
Because you're comparing and actual outcome with a given
single pattern (all head, in this case). The odds that those two
things match are, as noted, beyond infinitesimal. If you had
specified any other pattern and someone then flipped exactly that
pattern for 49 flips and wanted to bet you on the next flip, I'd
check the coin then too.
Neu Mejican|6.30.08 @ 5:51PM|#
Risk of impact by country...includes a weighting for population,
it seems.
http://www.dailygalaxy.com/my_weblog/2007/07/the-neo-code--c.html
LLStone|6.30.08 @ 5:51PM|#
But what if the asteroids don't like lottery winners? What are your precious odds then, Fancy Brain?
|6.30.08 @ 5:56PM|#
I did a little googling and found only a passing reference to a
Q and A about whether meteors have a tendency to strike from the
plane of the ecliptic. The answer was no, but I don't know how
authoritative that answer was.
I do believe that meteors prefer to hit sinners.
Mike M.|6.30.08 @ 5:56PM|#
Um, it would be nice if we could, I dunno, prevent an
asteroid or comet from hitting the Earth and, you know, killing
people and stuff. Rather than leave it to chance.
There is no doubt in my mind that eventually we'll be able to do
just that, but it will be a while.
Before that happens, it's far more likely that we'll have
space-based kinetic projectile weapons operated via the "mass
driver". The former capability will quite likely grow out of the
latter technology.
|6.30.08 @ 5:56PM|#
Why would you feel the need to check your coin?
Because exactly 4 of the ways you can toss a coin 50 times have the
property of the first 49 tosses being heads or tails. On the other
hand there are 1,125,899,906,842,620 ways to toss a coin 50 times
that don't start with 49 straight heads or tails.
Neu Mejican|6.30.08 @ 5:58PM|#
I'd check the coin then too.
How would you rig a coin to create a particular pattern of 50
results?
I can see ways to make it always come up heads or tails, but not
how to make it come up
TTThThTThTTTThThhhhThThhhTTThhThThh...
|6.30.08 @ 5:58PM|#
Art-POG:
Meaning...?
What Brian Courts said, I think.
But what if the asteroids don't like lottery winners? What are your precious odds then, Fancy Brain?
Speaking of which, what are the odds that a sentient asteroid (that can change its own trajectory and knows the locations of the lottery winners on our planet) enters our atmosphere are? Not good, I'm guessing.
|6.30.08 @ 6:07PM|#
I was just thinking that asteroids and comets might have a
tendency to strike some parts of the planet more than others, given
the Earth's axial tilt, etc.
I was wondering if gravitational lensing might have an effect, but
I figure that will just make a larger circular target into a
smaller one. More interplanetary crap is in the ecliptic than
outside it, so I would think that nearer the poles the impacts
would be more tangential. But that probably has little relevance to
what happens on the ground.
It is true that, due to the earth's motion around the sun, you are
more likely to get hit by an asteroid in the predawn hours. So
being in bed might not be the safest place since you are usually
there in the predawn hours.
Neu Mejican|6.30.08 @ 6:09PM|#
MikeP,
You forgot to account for the protective factor of the covers,
which I believe more than offsets the problem.
Unless your bed is in a mobile home and the mobile home is in a
trailer park.
Neu Mejican|6.30.08 @ 6:20PM|#
So if Shatner wins the lottery will it be
7307 Takei
4659 Roddenberry
or
68410 Nichols
That defies physics to come gunning?
http://www.usatoday.com/life/people/2007-10-02-george-takei_N.htm
|6.30.08 @ 6:25PM|#
How would you rig a coin to create a particular pattern of
50 results?
Oh, I have no idea. I wasn't really taking the coin-toss example
that literally. Nonetheless, if someone really did flip a specified
pattern for 49 flips I'd assume he had found a way to do it!
Anyway, your example is interesting when applied to lotteries. In
the pick six numbers type of game, the pattern 1,2,3,4,5,6 is just
as likely to win as any other, but most people would never play
those numbers. Which makes me think that if you do play
the lottery and it's one of those mega-prizes that happen from time
to time and attracts massive numbers of players often resulting in
split pots, you'd be wise to play 1-6. It's just as likely to win
as any other pattern, and if it does happen to win it's
less likely that you'll have to share the pot with anyone else. Of
course, what it really emphasizes is how absurdly unlikely you are
to win so the much better advice is not to play.
|6.30.08 @ 6:27PM|#
Wait. Does this mean that Uhura and Sulu got asteroids, but The
Shat did not? Weird.
I think it will be extra horrific if we take a major hit while
we're sitting here twiddling our thumbs. Or flipping coins. Even a
much smaller strike than Tunguska could be devastating if it hits
just wrong.
Neu Mejican|6.30.08 @ 6:30PM|#
Brian,
Regarding the lottery advice.
Of course, the odds of winning the lottery are not substantially
better if you buy 52 or 520 tickets a year.
However, the odds of winning are infinitely better for someone who
buys even one ticket than they are for someone who never buys a
ticket.
So, if you've got a dollar and find the moment of fantasizing about
what you would do with 50 million more worth the cost of a ticket,
then the best advice is to buy it...because you've just made
yourself infinitely more likely to win that 50 million you were
thinking about than you were before you bought that ticket.
5^)
Skoal|6.30.08 @ 6:30PM|#
Dr Ray Stantz: You have been a participant in the biggest
interdimensional cross rip since the Tunguska blast of 1909!
Louis: Felt great.
Dr. Egon Spengler: We'd like to get a sample of your brain
tissue.
Louis: Okay.
|6.30.08 @ 6:33PM|#
I think people missed the point of my comment. If you are going to mentally magnify the nearly infinitesimal probability of winning the lottery to the point of actually being hopeful for a win, it is only fair that you magnify into a proportionately strong fear the much larger probability (whether or not you adjust for weekly/monthly play) that Tunguska II will vaporize you. Whether it does so before you win, as you win, or sometime after you win (but probably well before you spent the whole fortune) doesn't seem to make a whole lot of difference. But as the wag once probably said, "I've been vaporized poor and I've been vaporized rich, and believe me, rich is better."
White Trash Neu Mejican|6.30.08 @ 6:37PM|#
Wait. Does this mean that Uhura and Sulu got asteroids, but
The Shat did not? Weird.
Makes you wonder if, perhaps, blow jobs are involved in the process
somehow....
Not that there's anything wrong with that.
jtuf|6.30.08 @ 6:58PM|#
JAM, looking forward to winning the lottery is like looking forward to dating a movie star. Deep down you know it won't happen, but imagining that it will can be entertaining for an hour.
Neu Mejican|6.30.08 @ 7:04PM|#
Because exactly 4 of the ways you can toss a coin 50 times
have the property of the first 49 tosses being heads or tails. On
the other hand there are 1,125,899,906,842,620 ways to toss a coin
50 times that don't start with 49 straight heads or
tails.
The odds of the 50th toss being heads or tails, however, doesn't
change no matter which one of the possible patterns of 49 tosses
preceded it. Aren't you just restating the gambler's fallacy
mentioned above?
8^)
|6.30.08 @ 7:16PM|#
However, the odds of winning are infinitely better for
someone who buys even one ticket than they are for someone who
never buys a ticket.
True. Infinitely greater, yet almost the same. :)
|6.30.08 @ 7:33PM|#
Kinnath, Probability has NO BEARING on poker. Shame on you for
trying to convince people otherwise.
Now everyone repeat after me: Any two will do, any two will do . .
.
Game starts at 8:00 p.m. Friday night . . .
|6.30.08 @ 7:46PM|#
I hate to be the first one to bring this up, but isn't it just possible that the rock brought...a special child named Obama?
|6.30.08 @ 7:49PM|#
Ultimately the researchers calculate that the annual
probability of an individual's death from a Tunguska-type impact is
1 in 30 million.
So if we assume for the sake of argument that my life span from
here on out is 100 years, and nothing will possibly kill me except
a meteor during that time, my probability of successfully surviving
for that 100 years is about:
(29,999,999/30,000,000) ^ 100 = 99.99966666721680%
My probability of dying from one is:
100% - 99.99966666721680% =
about 0.0003333327832%
I've been freely switching between numbers and percentages (i.e.
100% = "1").
I haven't looked through all the comments so I don't know if
anyone's made this exact point yet.
Untermensch|6.30.08 @ 8:23PM|#
That is why there is no impact crater
Actually, in a recent Scientific American some guys reported pretty
compelling (although not yet definitive) evidence that there
is an impact crater in the form of a lake in the
region.
|6.30.08 @ 9:55PM|#
Math challenge:
If your odds of winning a lottery jackpot are 1 in X (where X is a
really big number approaching infinity), then what would be your
probability of winning at least one jackpot if you were to play the
lottery X consecutive times (assuming each lottery ticket was
independent of the others)?
MikeP|6.30.08 @ 10:05PM|#
Which makes me think that if you do play the lottery and
it's one of those mega-prizes that happen from time to time and
attracts massive numbers of players often resulting in split pots,
you'd be wise to play 1-6.
I thought exactly that for a time too. But then I figured that, if
I came to that conclusion, it is likely that someone else would as
well. You have just proven me correct.
You are best off picking the most random numbers you can. Any
pattern you use is more likely to coincide with another's pattern
than a random selection is.
|6.30.08 @ 10:11PM|#
Incidentally, I once saw one of those little grocery check line
books called "Winning the Lottery". They said that, for a 1-56
lottery, your picked numbers should average 28, because most
winning lottery numbers average near 28.
Absolutely wrong. Each collection of numbers is as likely as each
other -- including that that averages 3.5 and that that averages
53.5. There are so, so many more collections of numbers that
average 28, but that doesn't mean that each individual collection
with that property is any more likely.
|6.30.08 @ 10:44PM|#
If your odds of winning a lottery jackpot are 1 in X (where
X is a really big number approaching infinity), then what would be
your probability of winning at least one jackpot if you were to
play the lottery X consecutive times (assuming each lottery ticket
was independent of the others)?
It's 1-((X-1)^X/X^X)
|6.30.08 @ 11:22PM|#
If your odds of winning a lottery jackpot are 1 in X (where X is
a really big number approaching infinity), then what would be your
probability of winning at least one jackpot if you were to play the
lottery X consecutive times (assuming each lottery ticket was
independent of the others)?
Well, the probability that you win the lottery at least
once is one minus the probability that you lose every
lottery. The probability of losing any individual lottery in your
example is 1-1/X. The probability of losing X lotteries in a row
then is (1-1/X)^X. So the probability of winning at least one
lottery is given by P=1-(1-1/X)^X.
Now the question is what happens as X approaches infinity. If we
let say Q = (1-1/X)^X, then P=1-Q. The limit as X->oo of Q can
be found with a bit of work which involves taking the natural log
of both sides and then getting it into a form with a 0/0 type limit
and applying l'Hoptial's rule. Ultimately you get that lim X->oo
of ln(Q) = -1. This means the lim X->oo of Q= e^-1.
So… lim X->oo P = 1-e^-1 ~= 0.63212. So the chance of winning at
least once is about 63%
TallDave|7.1.08 @ 12:25AM|#
On the plus side, our worthless wasteful pie-in-the-sky "it will never work" missile defense system successfully hit a detached warhead the other day. So our ability to hit bullet with bullets is improving, maybe enough to deflect something like this headed to, say, the Eastern seaboard rather than a desolate wilderness.
|7.1.08 @ 4:31AM|#
Earlier I was on my way out so I didn't have time to specify all
the steps to solve the challenge, but just in case anyone is
interested (though I figure anyone who is all that interested would
be able to figure out the intermediate steps anyway) I'll give the
details below. Of course I'm not going to list every justification
for each step such as limit rules etc since those can be looked up
easily enough.
So starting with P = 1-(1-1/x)^x which can be written as P=1-Q
(where Q is the probability of losing x consecutive lotteries),
with Q=(1-1/x)^x we want to find what happens as x gets very large
and approaches infinity.
lim x->oo Q = lim x->oo (1-1/x)^x
lim x->oo ln(Q) = lim x->oo x*ln(1-1/x) (this gives an
indeterminate limit of oo*0 form and can be rearranged to apply
l'Hopital's rule)
lim x->oo ln(Q) = lim x->oo (ln(1-1/x)/(1/x)) (this is now in
the 0/0 indeterminate form so to apply l'Hoptial's rule just
differentiate both top and bottom)
Top: d/dx ln(1-x^-1) = x^-2 / (1-x^-1); bottom: d/dx x^-1 = -x^-2
(note the derivatives are found separately for top and bottom
function)
Since from l'Hopital's rule we want the ratio of the derivatives of
the top and bottom functions, the x^-2 terms conveniently cancel...
always makes life easier when that happens.
so putting the fraction back together:
lim x->oo ln(Q) = lim x->oo -1/(1-1/x) (by inspection this
limit is clearly -1)
lim x->oo ln(Q) = -1
lim x->oo Q = e^-1
And finally, since P=1-Q, that gives:
lim x->oo P=1-e^-1 ~= 0.63212
Interestingly, x doesn't have to get very close to infinity at all
for that to be a very good estimate of the odds of winning at least
once in x tries. At x=10, it's about 65.1%, at x=100 it's about
63.4% and by x=1000 it's essentially reached it's final value of
about 63.2%
Kolohe|7.1.08 @ 4:36AM|#
Absolutely wrong. Each collection of numbers is as likely as
each other -- including that that averages 3.5 and that that
averages 53.5. There are so, so many more collections of numbers
that average 28, but that doesn't mean that each individual
collection with that property is any more likely.
What is interesting is that you should *always* use the random
number generator to pick numbers. All combo's are equally likely,
but not all combo's have equal payout, because some have an
increased likelyhood of multiple people playing the same numbers.
E.g. numbers 1-12 and to a lesser extent 1-31 are overrepresented
from people playing birthdates. And I believe a few lotteries have
had their winning numbers make a pattern on the form, causing
several dozen winners of the jackpot or sub-jackpot.
|7.1.08 @ 7:11AM|#
Actually, isn't there an opportunity for improvement here? If you know of a bias for people's choosing certain numbers, are you better off striking those out of selection by the random number generator? For example, should you only select random numbers from 32 and higher? Or is being too clever by half going to cost you by making it more likely to walk into some other pattern?
Kolohe|7.1.08 @ 12:27PM|#
or example, should you only select random numbers from 32
and higher? Or is being too clever by half going to cost you by
making it more likely to walk into some other pattern?
My guess would be 'too clever by half'. I think any pattern is more
likely to be picked then a random collection of numbers. There is
also the related notion that it is impossible to pick numbers that
appear random, trying to do so will always result in some kind a
pattern when done repeatedly. (this is why, no kidding, to
implement random anti-terrorism measures in the military, your
supposed to roll a D&D die to pick which additional measure to
use)
Paul|7.1.08 @ 6:15PM|#
Wrong, that's the gambler's fallacy. No many how many
chances you have, the probability is still the same.
Not entirely.
For instance, I don't play the lottery, therefore my chances are 0.
See, increasing the number of instances increases my chances.
;)
|7.2.08 @ 11:07AM|#
Brian Courts, MikeP and Mo,
Congratulations. I've asked that exact question to dozens of rather
clever people and almost all of them have got it wrong... miserably
wrong.
Very few people clue in that "the probability of winning at least
one jackpot" is the same as "the probability of NOT losing every
jackpot" (which is much easier to calculate).
Fewer still realize that as X approaches infinity, the whole
equation simplifies into [1-(1/e)] or approximately 63.2%.
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