| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1886.1 | You got it | VMSDEV::HALLYB | Fish have no concept of fire | Fri Aug 05 1994 17:08 | 9 |
| I believe your calculations are correct. The 50% mark is between
22 and 23, exactly as presented.
There's an old (true) story about this, where 22 people are having a
lunch. They go around the table with no duplications. Some doubts are
raised about the calculations. The waitress (person #23) chimes in,
saying her birthday is the same as "the general's over there".
John
|
| 1886.2 | Coincidences and Pepsi Diet Cola | MOVIES::GREEN | Use Tabasco as Gravy | Wed Aug 10 1994 10:10 | 44 |
| About a year ago I went to a public lecture on coincidences given in
Edinburgh given by Persi Diaconis, a well known and outspoken
statistician.
The talk covered a number of "coincidences" including discussion around
the birthday paradox (.0), "psychic" tricks and multiple winners of
national/state lotteries. Diaconis analysed the problems debunking
their coincidental appearance.
He gave a few interesting "rules of thumb" around general category
selection problems (of which the birthday paradox is one). The problem
is for a given set of randomly chosen events, the category size (C = #
of events - e.g. 365 for birthday paradox), number of event matches (n
= 2 for b.p.), determine a sample set size (S) to achieve a specified
probability of match (p = 0.50 for b.p.)
He gave a few approximate formulae, two of which are:
S = 1.2 sqrt(C) defines S for p = 0.50 and n = 2
S = 1.6 sqrt(C) 0.75 n = 2
[both of these fit the birthday paradox quite well]
Interestingly the analysis as in .0 (and these approximations) give
*worst case* set sizes. Very few category sets are randomly chosen -
e.g. depending on latitude, power cuts etc. birth dates are skewed,
another example (how many people do you need in a room to have 50%
chance of having 2 of them with the same PIN code # for a card). The
120 from formula above is conservative - there's a large set of PINs
unused - all the "obvious" passwords such as dddd (e.g. 3333), simple
sequences 1234 etc.
He also gave a formula for determining S from all 3 variables.
(I'll add this if I find it ...)
The single variable case seems simple enough to curve fit over a
reasonable category range. The more general case possibly done
similarly with larger sets.
Anyone seen similar treatment of this problem?
Russ
|
| 1886.3 | Capitol Hill birthday problem | EVMS::HALLYB | Fish have no concept of fire | Tue Nov 29 1994 10:19 | 13 |
| The incoming U.S. Congress has a total of 100+435 = 535 voting members.
Assuming uniform distribution and ignoring February 29th (for now),
what is the probability every day of the year is the birthday of one
or more Senators or Congressmen?�
What is the expected number of dates that are not the birthday of a
Senator or Congressman?
John
______________________
�Please, no responses of the form "nobody's birthday is every day"
|
| 1886.4 | a naive answer | CSSE::NEILSEN | Wally Neilsen-Steinhardt | Tue Nov 29 1994 12:36 | 45 |
| Consider a given day and a given congressperson, defined as a member of the US
Seante or the House of Representatives. The probability that this day is their
birthday is
1/365
The probability that this day is not their birthday is
1 - 1/365 = 0.9972...
To get the probability that this is no congressperson's birthday, multiply
the factor above by itself 535 times
( 1 - 1/365 )^535 = 0.2276...
The probability that this date is one or more congressperson's birthday,
is given when you subtract this from 1, to get .7723...
The probability that all 365 days are the birthday of one or more
congresspersons is given by multiplying this by itself 365 times, to get
0. forty zeroes 11...
pretty close to zero.
The expected number of days which are the birthday of no congressperson is
given by
0.2276... * 365 or about 83 days.
Why is this a naive answer?
It assumes, in the last two steps, that the probabilities for given days are
independent. This is probably a good approximation, since the numbers
involved are greater than 30 (old statistical rule of thumb), but it cannot
be exact. If I check a hundred days at random, and find that they are all
the birthday of no congressperson, then all 535 birthdays must be somewhere
in the remaining 265 days, so the probability I would compute for the
remaining 265 days would be larger.
I have seen the correct answer worked out for a related problem in statistical
sampling. It involves the hypergeometric distribution, as I remember. Almost
everyone solves the approximate problem, as above.
|
| 1886.5 | Kind of an inverse of the Birthday Problem | SSAG::LARY | Laughter & hope & a sock in the eye | Wed Nov 30 1994 00:38 | 11 |
| In fact, it would take close to 2300 senators and congressmen (no thanks!) to
raise the probability of total birthday coverage to 50%; a general formula for
approximating this is that given M pigeonholes the number of uniformly
distributed pigeons it would take to yield a probability P of covering them
all is approximately
M (ln M - ln (-ln P))
The approximation is pretty good for largish M and non-infinitesimal P, it
yields 2287 politicians in my above example which worked back through Wally's
formula yields a 50.2% chance of total coverage, not bad...
|
| 1886.6 | | RUSURE::EDP | Always mount a scratch monkey. | Thu Dec 01 1994 15:38 | 103 |
| Re .4:
> doit:=proc(m,n)
> local prob;
> prob:=array(-1..m);
> prob[-1]:=0;
> prob[0]:=1;
> for j from 1 to m do
> prob[j]:=0;
> od;
> for i from 1 to n do
> for j from min(m,i) to 0 by -1 do
> prob[j]:=j*prob[j]+(m-j+1)*prob[j-1];
> od;
> od;
> expected:=0;
> for j from 0 to m do\
> prob[j]:=prob[j]/m^n;
> expected:=expected+(m-j)*prob[j];
> od;
> RETURN(prob[m], expected)\
> end:
> doit(365,535); evalf(");
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0984270529175705683256520792547683552789943610769408
--------------------------------------------------------------------------------
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56459678825922310352325439453125
,
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6810376359245683534042497024
--------------------------------------------------------------------------------
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00232507288455963134765625
-59
.6028365965*10 , 84.11058092
>
|
| 1886.7 | | EVMS::HALLYB | Fish have no concept of fire | Fri Dec 02 1994 08:59 | 12 |
| > -59
> .6028365965*10 , 84.11058092
[1] How many times does ^^^^^ go into Avogadro's number?
[2] Was this done on a Pentium? :-)
I'm quite certain these numbers are correct (thanks, Eric) but I find
them highly nonintuitive, especially in light of the birthday problem.
John
|