==> pickover/pickover.03.s <==
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You wrote (in article <1992Sep14.141704.26532@watson.ibm.com>):
>Title: Cliff Puzzle 3: Too many 3's
>1. How can it be that almost all of the numbers have a 3 in them?
Because as the numbers get larger, they contain more digits,
increasing the probability that one of the digits in them might be a
3. In fact, the probability that a 3 will _not_ appear in a very long
number is very low.
I like this puzzle. Simple, but it made me think for a moment.
A three in every number? Preposterous! ;)
As for the other information you requested from responders: You have
my name and email address now, I don't give out my home address unless
it's necessary, and what sort of 'affiliation' are you seeking --
religious, business, or what?
<< Brian >>
--
_/_/_/ Brian Kendig Macintosh Jedi Live never to be ashamed
_/_/ Starfleet Captain Oracle Employee if anything you do or say
_/ Intrepid Adventurer Saturn SL2 Owner is published around the world
bskendig@netcom.com Wizard of Frobozz -- even if what is published
Princeton '92! BSE/CS Writer/Actor/Singer is not true.
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In article <1992Sep14.141704.26532@watson.ibm.com> you write:
>Title: Cliff Puzzle 3: Too many 3's
>From: cliff@watson.ibm.com
>
>If you respond to this puzzle, if possible please include your name,
>address, affiliation, e-mail address. If you like, tell me a little bit
>about yourself. You might 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
>
>* * *
>
>How many numbers have at least one digit -- a three?
>
>In the first 10 numbers, 1,2,3,4,5,6,7,8,9,10 there is only one number
>which contains the digit 3. This means that 1/10 or 10% of the numbers
>have the number 1 in the first 10 numbers. In the first 100 numbers the
>occurrence of numbers with at least one three seems to be growing. In
>fact there are 19 numbers: 3,13,23,33,43,53,63,73,83,93,
>30,31,32,34,35,36,37,38,39. This means that about 19% of the digits
>contain the number 3 in the first 100 numbers.
>
>We can make a table showing the percentage of numbers with
>at least one 3-digit for the first N numbers.
>N %
>10 1
>100 19
>1000 27
>10000 34
>
>The percentages rapidly increase to 100% indicating that almost all of
>the numbers have a 3 in them! In fact, a formula describing the
>proportion of 3's can be written: 1-(9/10)**N. The proportion gets
>very close to 1 as N increases.
>
>Stop And Think
>
>1. How can it be that almost all of the numbers have a 3 in them?
>
Thaddeus Crews, 509 Windsor Green Blvd., Goodlettsville, TN, 37072
Graduate Student (Ph.D.) @ Vanderbilt University, Computer Science
thaddeus@vuse.vanderbilt.edu
This problem seems a little bit simple to me, but I was never that
great at math problems so I am not betting the farm on this answer.
The percentages you show for # of the first N numbers with at least
one 3-digit is also true (about) for the # of the first N numbers
with at least one 4-digit, at least one 5-digit, etc...
Basically, as N increases, so does the number of digits in N, and
therefore so does the number of chances for the digit 3 to appear
(as well as all other digits). Given a number N with enough (?)
digits, there is a 100% chance of all digits 0-9 appearing in that
number (of course, 1.0E10000000000) does not have a 3 in it, but
if you take the next 1.0E10000000000 numbers the percent that has
a 3 will be (I suspect) 100%.
My proof is clearly weak, but the claim is this: as N increases,
the number of digits in N also increases. As N approaches
infinity, the number of digits in N approaches infinity (at a
slower rate, however). As the number of digits approaches infinity,
the likelyhood of any specific digit appearing at least once
approaches 100%.
I think the real question (to be answered by someone with a better
math training) would be "At what number N does the statistical
likelyhood become 100% of at least one 3-digit appearing in the
first N numbers."
Hope this helps....
--
-- Thad Crews (email thaddeus@vuse.vanderbilt.edu)
--------------------------------------------------------------------------
"Some people have a way with words, and some people, ... oh ... *not* have
a way, I suppose..." -- Steve Martin
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Heh. As the numbers get larger, they have more digits. Assuming a random occu
various digits in the larger numbers (not unreasonable when n-> infinity) the pr
number NOT having a 3 is very low.
-john 'I know it's not a proof...' karakash-
-------------------------
>Title: Cliff Puzzle 3: Too many 3's
Seth Breidbart
PO Box 5157
New York, NY 10185
Morgan Stanley & Co.
>1. How can it be that almost all of the numbers have a 3 in them?
The probability that a random sequence of n digits does not contain a
3 is .9^n; as n->infinity, this probability -> 0. Since almost all
numbers have a lot of digits (there are only a finite number of
integers with n), the limiting
probability is 0.
-------------------------
In article <1992Sep14.141704.26532@watson.ibm.com> you write:
>Title: Cliff Puzzle 3: Too many 3's
>From: cliff@watson.ibm.com
>
>If you respond to this puzzle, if possible please include your name,
>address, affiliation, e-mail address. If you like, tell me a little bit
>about yourself. You might 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
>
>* * *
>
>How many numbers have at least one digit -- a three?
>
>In the first 10 numbers, 1,2,3,4,5,6,7,8,9,10 there is only one number
>which contains the digit 3. This means that 1/10 or 10% of the numbers
>have the number 1 in the first 10 numbers. In the first 100 numbers the
>occurrence of numbers with at least one three seems to be growing. In
>fact there are 19 numbers: 3,13,23,33,43,53,63,73,83,93,
>30,31,32,34,35,36,37,38,39. This means that about 19% of the digits
>contain the number 3 in the first 100 numbers.
>
>We can make a table showing the percentage of numbers with
>at least one 3-digit for the first N numbers.
>N %
>10 1
>100 19
>1000 27
>10000 34
>
>The percentages rapidly increase to 100% indicating that almost all of
>the numbers have a 3 in them! In fact, a formula describing the
>proportion of 3's can be written: 1-(9/10)**N. The proportion gets
>very close to 1 as N increases.
>
>Stop And Think
>
>1. How can it be that almost all of the numbers have a 3 in them?
>
No problem. In fact almost all very large numbers have all digits in them.
It is rather hard for a number with zillions of digits to avoid "3"s (or any
other digit).
It fact, the sequences "15", "172", and "666" (and any other finite sequence)
are also contained (in order) within almost all numbers.
Dan Shoham
shoham@ll.mit.edu
-------------------------
Before I forget:
Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618
clong@remus.rutgers.edu
--
Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618
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>Title: Cliff Puzzle 3: Too many 3's
>From: cliff@watson.ibm.com
>If you respond to this puzzle, if possible please include your name,
>address, affiliation, e-mail address. If you like, tell me a little bit
>about yourself. You might 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
>* * *
>How many numbers have at least one digit -- a three?
>In the first 10 numbers, 1,2,3,4,5,6,7,8,9,10 there is only one number
>which contains the digit 3. This means that 1/10 or 10% of the numbers
>have the number 1 in the first 10 numbers. In the first 100 numbers the
>occurrence of numbers with at least one three seems to be growing. In
>fact there are 19 numbers: 3,13,23,33,43,53,63,73,83,93,
>30,31,32,34,35,36,37,38,39. This means that about 19% of the digits
>contain the number 3 in the first 100 numbers.
>
>We can make a table showing the percentage of numbers with
>at least one 3-digit for the first N numbers.
>N %
>10 1
>100 19
>1000 27
>10000 34
>The percentages rapidly increase to 100% indicating that almost all of
>the numbers have a 3 in them! In fact, a formula describing the
>proportion of 3's can be written: 1-(9/10)**N. The proportion gets
>very close to 1 as N increases.
>Stop And Think
>1. How can it be that almost all of the numbers have a 3 in them?
I'm not sure this is the answer you are looking for, but:
9 = 9
9*9 = 81
9*9*9 = 729
9*9*9*9 = 6561
etc.
The probability of having 3 as the digit in a one-digit number is 1/10.
" of not having 3 " is 9/10.
In a two-digit number, the prob. of NOT having 3 as the first digit or
the second digit, ie. not having 3 in the two-digit number, is simply
the product of (NOT having 3 in first digit) times (NOT having 3 in second):
(9/10)*(9/10) = 81/100
= 0.81
For a three-digit number: (9/10)*(9/10)*(9/10) = 729/1000
= 0.729
For an n-digit number: (9/10)**n = probability.
We can see that as "n" becomes larger and larger, the
probability of NOT having a three at all in the number
becomes smaller and smaller. Indeed, as "n" approaches
infinity, this probability approaches zero. In other
words, it is very rare for a large number NOT to have 3
as one of its digits. In fact, it is very rare for a
large number NOT to have any of the ten possible integers
represented at least once.
[Aside, N %
10 1 = (10 - 9)/1 times 100
100 19 = (100 - 81)/100 times 100
1000 27 = (1000 - 729)/1000 times 100
10000 34 = (10000 - 6561)/10000 times 100
etc. ]
Kumar
kumar@ug.cs.dal.ca
ps: I'll leave it as a small exercise to tie up the loose ends.