==> pickover/pickover.12.s <==
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In article <1992Oct23.160130.166012@watson.ibm.com> you write:
: Consider a metallic slide with 10 large holes in it equally spaced from
: top to bottom. If you attempt to slide down the slide you have a 50%
: probability of sliding through each hole in the slide into an oleaginous
: substance beneath the slide during each encounter with a hole.
:
: 1. If you were a gambling person, which hole would you bet a person
: would fall through?
The chance of falling thru the first hole is 50%. For the second hole, it
is (.5)(.5) = 25%, the thrid is (.5)^3 = .125. The chance by the tenth
hole is about .0097 %. Obviously, since I am limited to one hole, I would
place my money on hole #1 (best chance).
: 2. If you were a gambling person, how many attempts would it require
: for a person to slide from the top of the slide to the bottom without
: falling through a single hole.
The sum of the prob for falling thru a hole is .5 + .5^2 + .5^3 +...+.5^10.
This is about 99.902% = .99902. So about 98 times out of 100000, someone
will make it through without falling. This is about 1 time out of 1020.
So give or take about 1020 tries....
:
: 3. If all the people on earth lined up to go down the slide, and they
: slid down a more horrifying slide with 100 holes at a rate of 1 person
: per second, when would you expect the first person to arrive at the
: bottom of the slide without falling through.
: An hour? A day? A decade? ...
The prob for falling thru the last hole is .5^100 = 7.88x10^-31. There must
be some chance less than this that one WILL make it thru the slide. The MIN
number of tries that it must take is 1/.5^100 = 1.26x10^30. At the given rate
this is about 9.647 x 10^23 years, much older than the universe if I remeber
correctly.
Also, the chance of making it must be GREATER than .5^101. or with
all the math, the MAX amount of time is 1.929x10^24 years. So give or
take about 1.5x10^24 years....
--
Michael Neylon aka Masem the Great and Almighty Thermodynamics GOD!
// | Senior, Chemical Engineering, Univ. of Toledo
\\ // Only the | Summer Intern, NASA Lewis Research Center
\ \X/ AMIGA! | mneylon@jupiter.cse.utoledo.edu /
--------+ How do YOU spell 'potato'? How 'bout 'lousy'? +----------
"Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L
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In rec.puzzles you write:
>Title: Cliff Puzzle 12: Slides in Hell
>From: cliff@watson.ibm.com
>
>If you respond to this puzzle, if possible please send me your name,
>address, affiliation, e-mail address, so I can properly credit you if
Jeff Rogers
Rensselaer Polytechnic institute
rogerj@rpi.edu
>Consider a metallic slide with 10 large holes in it equally spaced from
>top to bottom. If you attempt to slide down the slide you have a 50%
>probability of sliding through each hole in the slide into an oleaginous
>substance beneath the slide during each encounter with a hole.
>
>1. If you were a gambling person, which hole would you bet a person
>would fall through?
The first one. There's only a 50% chance of them getting past it, and a
small chance of them falling into each succeeding hole.
hole # percent chance of reaching and falling into
1 50
2 25
3 12.5
4 6.25
5 3.125
6 1.5625
7 0.78125
8 0.390625
9 0.1953125
10 0.09765625
>
>2. If you were a gambling person, how many attempts would it require
>for a person to slide from the top of the slide to the bottom without
>falling through a single hole.
The chances for reaching each succeeding hole are the same as reaching and
falling into the previous one. Therefore, the chances of passing all the
holes are the same as reaching and falling into the last hole (see previous
answer for stats), which makes the probability .0009765625, so
statistically, 1024 slides would be required to guarantee reaching the
bottom. If I was a gambling person, I'd probably bet about half this,
because the actual events can happen in any order, and on average, I'd guess
that he'd get down in about 512 slides.
>
>3. If all the people on earth lined up to go down the slide, and they
>slid down a more horrifying slide with 100 holes at a rate of 1 person
>per second, when would you expect the first person to arrive at the
>bottom of the slide without falling through.
>An hour? A day? A decade? ...
This is solved similarly; it is represented by powers of 2. To successfully
get past the last hole, it would require (statistically, at least) 2^100
or (by my trusty pocket calculator) 1.2676506 *10^30 slides.
More significant figures? dc! Which gives 1267650600228229401496703205376.
In similar logic as the last problem, I'd expect about half that, or
633825300114114700748351602688 slides. How much time would this be? Excluding
leap years, I calculate 20098468420665737593491 years. That's 20 sextillion
years, significantly more than the age of the universe, by about 11 orders
of magnitude. So I'd guess that no one will ever reach the bottom, they'll
all try and fail (assuming everyone only gets to go once), or die waiting in
line.
Diversion
--
"I can see 'em | "Want me to create a diversion?"
I can see 'em | Diversion
Someone wake me when it's over" | rogerj@rpi.edu
-------------------------
In article <1992Oct23.160130.166012@watson.ibm.com> you write:
Title: Cliff Puzzle 12: Slides in Hell
>Consider a metallic slide with 10 large holes in it equally spaced from
>top to bottom. If you attempt to slide down the slide you have a 50%
>probability of sliding through each hole in the slide into an
>oleaginous substance beneath the slide during each encounter with a
>hole.
>
>1. If you were a gambling person, which hole would you bet a person
>would fall through?
None. The best chance is the first hole but I got a 50-50 chance. Why
bother? (2nd hole is 1/4, 3rd 2**-3, ...)
>2. If you were a gambling person, how many attempts would it require
>for a person to slide from the top of the slide to the bottom without
>falling through a single hole.
No gurantee. Each slide is an independent event. Now, if you are
talking mere probability, on the average, one in 1024 slides may make
it through all 10 holes.
>3. If all the people on earth lined up to go down the slide, and they
>slid down a more horrifying slide with 100 holes at a rate of 1 person
>per second, when would you expect the first person to arrive at the
>bottom of the slide without falling through. An hour? A day? A decade?
Again, can't tell. It could be the first one, it could be none. Probablity
can not foretell actual events. But if you have infinite number of people
sliding down till eternity, on the average, you may see 1 person slide over
all holes every (2**100)/(365*24*69*6) years. This number is many times
bigger than the world population for now.
-------------------------
Some answers to your questions:
1. As the puzzle states there is a 50% chance of falling into each
hole, I would bet a person would fall into the first hole -- in a large
enough sample, 1/2 of the people will fall through the first hole, 1/4
through the second, 1/8 through the third, etc.
2. In a large sample, 1/(2^10) people would make it all the way down
the slide without falling through any of the holes (1/1024). This means
that 1023 out of 1024 people would fall through a hole. Using the
formula (1023/1024)^x=1/2, we can determine out of the first x people
to go down the slide, there is a 50% chance that one person will make
it down without falling through a hole. The answer to this equation is
x=709.4 Thus I would bet that a person would make it all the way down
on one of the first 710 attempts.
3. As 2^100=1.2676*10^30 (roughly), and (including leaps years under
the Gregorian calendar) there are 31556952 seconds in the average year,
then statistically one person should make it down the slide every
4.017*10^22 YEARS. However, and this is a very rough estimate, I figure
the log of (1-1/(1.2676*10^30)) to be about -5.5*10^(-29). [I'm doing
the calculations on a scientific calculator which only has 10 places.]
Thus, using the formula xlog(1-1/2^100)=log(1/2), I get x=5.5*10^27.
Thus, there's about a 50% chance that after 5.5*10^27 seconds, someone
will have made it down the slide. To be on the safe side, I'd bet only
if I were given at least 6*10^27 seconds, a value which equals
1.901*10^20 YEARS.
I hope this answers the questions.
Ted Schuerzinger
email: J.Theodore.Schuerzinger@Dartmouth.EDU
snailmail: HB 3819
Dartmouth College
Hanover, NH 03755
USA
In case you're wondering, I'm just a junior at Dartmouth who's
interested in puzzles like these. I'm not even a math major -- I'm a
double major in government and Russian.
-------------------------
In article <1992Oct23.160130.166012@watson.ibm.com> you write:
>Title: Cliff Puzzle 12: Slides in Hell
>From: cliff@watson.ibm.com
>Consider a metallic slide with 10 large holes in it equally spaced from
>top to bottom. If you attempt to slide down the slide you have a 50%
>probability of sliding through each hole in the slide into an oleaginous
>substance beneath the slide during each encounter with a hole.
>
>1. If you were a gambling person, which hole would you bet a person
>would fall through?
There's a 50% chance of falling through the first hole, 25% the
second, 2^-n the n'th. If the odds offered were the same, I'd go for
the first hole.
>2. If you were a gambling person, how many attempts would it require
>for a person to slide from the top of the slide to the bottom without
>falling through a single hole.
You expect to make it 1 out of 1024 times; after 710 tries, the chance
of someone succeeding exceeds 1/2. (Log base (1023/1024) of 1/2 is
709.4).
>3. If all the people on earth lined up to go down the slide, and they
>slid down a more horrifying slide with 100 holes at a rate of 1 person
>per second, when would you expect the first person to arrive at the
>bottom of the slide without falling through.
>An hour? A day? A decade? ...
Never. OK, 1/2^100 will make it. There being under 2^33 people on
the planet, ...
After 4.2e22 years, the expected number of people who succeeded is 1;
after about 2.9e22 years, the chance of someone having succeeded is
about 1/2.
Like I said, never.
Seth sethb@fid.morgan.com
-------------------------
In rec.puzzles you write:
>1. If you were a gambling person, which hole would you bet a person
>would fall through?
If the pay-back odds were the same regardless of the hole, then obviously,
I'd bet on the first hole! There's a 1:2 chance the person falls through
the first hole, a 1:4 combined chance of the person falling though the
second hole, etc...
>2. If you were a gambling person, how many attempts would it require
>for a person to slide from the top of the slide to the bottom without
>falling through a single hole.
1024 is the median value for this case... There's a 1:2**n chance of
a person falling through the nth hole, having missed all of the holes
before n. Since the probability of falling through = the probability
passing over the hole safely (vs not ever getting there), the
probability that a person makes it to the end is also 1:1024.
>3. If all the people on earth lined up to go down the slide, and they
>slid down a more horrifying slide with 100 holes at a rate of 1 person
>per second, when would you expect the first person to arrive at the
>bottom of the slide without falling through.
>An hour? A day? A decade? ...
There is a 1:2**(100-Log2(5 billion people)) chance that somebody makes
it through... Given a finite # of people on the planet (approx 5 bil.)
I think we'll run out first...
--Joseph Zbiciak im14u2c@camelot.bradley.edu
-------------------------
Subject: Re: Cliff Puzzle 12: Slides in Hell (SPOILER)
Newsgroups: rec.puzzles
References: <1992Oct23.160130.166012@watson.ibm.com>
In article <1992Oct23.160130.166012@watson.ibm.com>, Cliff Pickover writes:
> Consider a metallic slide with 10 large holes in it equally spaced from
> top to bottom. If you attempt to slide down the slide you have a 50%
> probability of sliding through each hole in the slide into an oleaginous
> substance beneath the slide during each encounter with a hole.
> 1. If you were a gambling person, which hole would you bet a person
> would fall through?
The probability of falling into hole i is (1/2)^i, so your best bet
would be hole 1.
> 2. If you were a gambling person, how many attempts would it require
> for a person to slide from the top of the slide to the bottom without
> falling through a single hole.
The probability of success is p = (1/2)^10, and as each trial is
independant the expected number of trials before success is 1/p or
2^10.
> 3. If all the people on earth lined up to go down the slide, and they
> slid down a more horrifying slide with 100 holes at a rate of 1 person
> per second, when would you expect the first person to arrive at the
> bottom of the slide without falling through.
In this case the number of expected trials is 2^100, which is much
larger than the total number of people.
> An hour? A day? A decade? ...
Try about 10^24 years. As another problem, assuming a large enough
supply of sliders estimate when the slide will wear through from
friction.
-------------------------
In article <1992Oct23.160130.166012@watson.ibm.com> you write:
>Title: Cliff Puzzle 12: Slides in Hell
>From: cliff@watson.ibm.com
>
>If you respond to this puzzle, if possible please send me your name,
>address, affiliation, e-mail address, so I can properly credit you if
>you provide unique information. PLEASE ALSO directly mail me a copy of
>your response in addition to any responding you do in the newsgroup. I
>will assume it is OK to describe your answer in any article or
>publication I may write in the future, with attribution to you, unless
>you state otherwise. Thanks, Cliff Pickover
>
> * * *
>
>Consider a metallic slide with 10 large holes in it equally spaced from
>top to bottom. If you attempt to slide down the slide you have a 50%
>probability of sliding through each hole in the slide into an oleaginous
>substance beneath the slide during each encounter with a hole.
>
>1. If you were a gambling person, which hole would you bet a person
>would fall through?
I'd bet that they fell through the first hole. The probability of that
happening is 50%. The probability of them falling through the second
hole is:
P(didn't fall through the first)*P(fell through the second) = 50%*50% = 25%
In general, P(falls through hole n)=
P(no fall through 1)*P(no fall through 2)*...*P(no fall through n-1)
*P(fell through hole n).
For this problem, P(falls through hole n) is (50%)^n, where n is the hole #
from the top.
>2. If you were a gambling person, how many attempts would it require
>for a person to slide from the top of the slide to the bottom without
>falling through a single hole.
(Hey, after the first failed attempt, they're screwed, no?)
P(success)=P(no fail)=P(no fall 1)P(no fall 2)...P(no fall 10)
=50%^10
=1/1024
They should make it at least one time in 1024.
>3. If all the people on earth lined up to go down the slide, and they
>slid down a more horrifying slide with 100 holes at a rate of 1 person
>per second, when would you expect the first person to arrive at the
>bottom of the slide without falling through.
>An hour? A day? A decade? ...
Oh, one in about 4.02*10^22 years... I wouldn't hold my breath.
-Richard
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1. I would bet on the first hole, as there is a 0.5 probability of a person's
falling into it, which is the highest such probability.
2. The probability of reaching the end of the slide on a particular try is
1/2^10 = 1/1024. In 709 tries, there is an approximately 0.5 probability of
3. Beats me - the even money bet is for a number of tries (approximately) equal
((2^100 - 1)/(2^100))
calculate it.
--
_______________________________________________________________________
Dan Blum Institute for the Learning Sciences Room 327
blum@ils.nwu.edu 1890 Maple Ave., Evanston, IL 60201 708-467-2306
"Let it be granted that a controversy may be raised about any question,
and at any distance from that question."
Lewis Carroll
_______________________________________________________________________