View Full Version : Unusual birthday puzzle
pzhon
09-02-2004, 04:53 AM
This was posted on usenet a few years ago. Which of the following is more likely?
A) 365 random people have 365 different birthdays.
B) 365^2 random people don't have all 365 birthadys.
The probabilities are close. Ignore February 29th.
random
09-02-2004, 06:50 AM
Post deleted by random
--deleted since what was once a question will now look like an unwarranted attack--
random
09-02-2004, 07:03 AM
aradsfa;lsk i'm retarded, 6am, going to sleep
FrankLu99
09-02-2004, 09:32 PM
[ QUOTE ]
This was posted on usenet a few years ago. Which of the following is more likely?
A) 365 random people have 365 different birthdays.
B) 365^2 random people don't have all 365 birthadys.
The probabilities are close. Ignore February 29th.
[/ QUOTE ]
ok i am foolish enough to take a shot ( just like how i bluff the river with 8 high)
P(A) == 1/(365!)
P(B) >= 365*((364/365)^(365^2))
Am i close?
I will guess that B is more likely
random
09-03-2004, 12:21 AM
[ QUOTE ]
..will now look like an unwarranted attack
[/ QUOTE ] Certainly not. I said something stupid and it needed to be pointed out. I appreciate you deleting it, though. See ya.
bigpooch
09-03-2004, 02:57 AM
Well, they are close in order of magnitude. If anyone is
so inclined, they can check that the first probability is
roughly bigger by a factor of sqrt(2xpix365). (For those
who are mathematically inclined, it's almost immediate from
Stirling's approximation).
RiverTheNuts
09-03-2004, 04:34 AM
im just curious, but how the hell did 2 pi get into a formula with 365 days?? /images/graemlins/smile.gif
I should really learn this crap
[ QUOTE ]
how the hell did 2 pi get into a formula with 365 days?
[/ QUOTE ]
Pi has nothing to do with the day count. Call it N days and then relate the 2 functions/series/whatever. That relation is where the guy is saying that pi comes from.
I don't know if he's right since I didn't investigate further, but that's what he means. I think.
~D
Gordon Scott
09-03-2004, 09:42 AM
Not a math guy but B
My guess would be that A isn't close.
[ QUOTE ]
A) 365 random people have 365 different birthdays.
[/ QUOTE ]
365! / (365^365) = 1.45 * 10^-157 (appr.)
[ QUOTE ]
B) 365^2 random people don't have all 365 birthadys.
[/ QUOTE ]
The probability of someone having been born on day n (of the year) given all previous n-1 days are "filled" is:
1-(364/365)^(265^2-(n-1))
So the probability of not all days being "filled" is now
1 - Product[1-(364/365)^(365^2-(n-1)),{n,1,365}] = 1.15 * 10^-156 (appr.)
[ QUOTE ]
Which [...] is more likely?
[/ QUOTE ]
B.
bigpooch
09-04-2004, 12:29 AM
A) The first probability is exactly
P(A) = 365! /( 365^365)
By Stirling's approximation formula, for large n,
n! is roughly sqrt(2 x pi x n) x (n/e)^n where pi is about
3.14159265 and e is about 2.71828183 and so
365! is approximately
sqrt(2 x pi x 365)x(365^365)/(e^365)
and hence,
P(A) is about
sqrt(2 x pi x 365)/(e^365).
By using a better Stirling approximation (asymptotic
expansion of the Gamma function, which is just a function
that extends factorial to all the real numbers), n! is
actually bigger by a factor of
(1 + 1/(12n) + 1/(288 x n^2) - 139/(51840 x n^3) +...)
and so P(A) is really closer to
1.0002283365 x sqrt(2 x pi x 365)/(e^365)
or about
47.9/(e^365)
B) Well, I see where I committed a common error in my
back of the envelope calculation for my previous post!
The second probability p(B) is just
365 x p(everyone is not born on Jan. 1)
+C(365,2) x p(everyone is not born on Jan. 1/2)
+C(365,3) x p(everyone is not born on Jan. 1..3)
+...
It will be clear that the second and subsequent terms are
much smaller than the first term.
The first term is just 365 x (364/365)^(365^2)
=365 x ((364/365)^365)^365.
Now, (364/365)^365 is roughly 1/e; actually, it's about
0.99862857x(1/e). Thus, the first term is close to
365 x (0.99862857/e)^365
or about 221.181/(e^365).
Also, the second term is
(365x364/2) x {((363/365)^365)} ^365
where (363/365)^365 is close to 1/(e^2) and so this term is
in the order of e^-(2x365) which is much smaller than the
first term. One can see that each of the following terms
is then roughly smaller by a factor of e^365 (not quite
because of the combinatorial factor) compared with the
previous term.
In any case, P(B)/P(A) is about 4.61756 and P(B) is really
bigger and these probabilities are much closer than I had
originally thought!
LAZY MAN's APPROXIMATION:
This is the idea that gives one an immediate guess as to
which of these probabilities is bigger. Without caring too
much about how closely 1/e approximates (364/365)^365, one
can see that
P(A) looks like sqrt(2 x pi x 365)/(e^365)
and P(B) is (very!) roughly
365/(e^365).
(I had forgotten the numerator when typing my original
post!). So someone who is really careful and astute can
readily deduce that roughly
P(A)/P(B) = sqrt(2 x pi/365) or about 0.1312. (Actually,
you only need to know that 365 > 2 x pi for guessing that
B is more probable!)
Of course, unfortunately, the 1/e approximation is extremely
crude and hence one has to be careful since raising a crude
approximation to the power of 365 really distorts the
result!
pzhon
09-04-2004, 03:36 AM
[ QUOTE ]
This was posted on usenet a few years ago. Which of the following is more likely?
A) 365 random people have 365 different birthdays.
B) 365^2 random people don't have all 365 birthadys.
The probabilities are close. Ignore February 29th.
[/ QUOTE ]
Instead of just giving the answers, I'll give ways to estimate the values by hand.
A) The probability is (365/365)(364/365)(363/365)...(1/365) = 365!/(365^365).
A good way to estimate n! is Stirling's formula: n! ~ sqrt(2 pi n ) (n/e)^n. Stirling's formula is low by a factor of roughly (1+1/(12n))= 1.00023 when n=365. 365!/(365^365) ~ sqrt(2 pi 365)/(e^365) = 1.454623 x 10^-157.
The exact value is 1.454955 x 10^-157.
B) The probability that no one was born on January 1st is (364/365)^(365^2). A good approximation to the probability that 365^2 people don't have all 365 birthdays is 365 times (364/365)^(365^2). The exact value (by inclusion-exclusion) is 365 (364/365)^(365^2) - (365 choose 2) (363/365)^(365^2) + (365 choose 3) (362/365)^(365^2) - ... but the higher order terms are smaller than the first term by a factor of more than 10^100.
How can we estimate 365 (364/365)^(365^2)? (1-1/n)^n ~ 1/e, so one estimate is that this is 365((364/365)^365)^365 = 365(1/e)^365 = 365/(e^365). This estimate is off by a lot, a factor of 1.6. (1-1/n)^n is not close enough to 1/e.
1/e = 1 - 1 + 1/2 - 1/6 + 1/24 - 1/120 + ...
(1-1/n)^n = 1 - n/n + (n choose 2)/n^2 - (n choose 3)/n^3 + ...
Compare the third terms: In 1/e, this is 1/2, but in (1-1/n)^n, the term is ((n^2-n)/2) /n^2 = 1/2 - 1/(2n), lower by 1/(2n). Compare the fourth terms: In 1/e, this is -1/6, but in (1-1/n)^n, the term is -((n^3 - 3n^2 + 2n)/6)/n^3, higher by about 3/(6n). The coefficients of 1/n in the difference are 0 + 0 + 1/2 - 3/6 + 6/24 - 10/120 + 15/720 - ... = 1/(2e).
So, (1-1/n)^n ~ 1/e - 1/(2en) = 1/e (1-1/(2n)).
365 (364/365)^(365^2)
~ 365 (1/e (1 - 1/(2*365)))^365
~ 365/(e^365) * sqrt(1/e)
= 6.724496 x 10^-157
The actual value is 6.718345 x 10^-157.
B is larger, roughly by a factor of sqrt(365/(2 pi e)). Because 365 > 2 pi e, it is more likely that 365^2 people don't have all 365 birthdays than that 365 people have all 365 birthdays. Since 7 < 2 pi e, it is more likely that 7 people were born on different days of the week (.61%) than that 49 people were not born on all 7 days of the week (.37%).
FrankLu99
09-04-2004, 04:06 AM
thx 4 confirming that i still suck at math
BarronVangorToth
09-06-2004, 09:19 AM
[ QUOTE ]
A) The first probability is exactly
P(A) = 365! /( 365^365)
By Stirling's approximation formula, for large n,
n! is roughly sqrt(2 x pi x n) x (n/e)^n where pi is about
3.14159265 and e is about 2.71828183 and so
365! is approximately
sqrt(2 x pi x 365)x(365^365)/(e^365)
and hence,
P(A) is about
sqrt(2 x pi x 365)/(e^365).
By using a better Stirling approximation (asymptotic
expansion of the Gamma function, which is just a function
that extends factorial to all the real numbers), n! is
actually bigger by a factor of
(1 + 1/(12n) + 1/(288 x n^2) - 139/(51840 x n^3) +...)
and so P(A) is really closer to
1.0002283365 x sqrt(2 x pi x 365)/(e^365)
or about
47.9/(e^365)
B) Well, I see where I committed a common error in my
back of the envelope calculation for my previous post!
The second probability p(B) is just
365 x p(everyone is not born on Jan. 1)
+C(365,2) x p(everyone is not born on Jan. 1/2)
+C(365,3) x p(everyone is not born on Jan. 1..3)
+...
It will be clear that the second and subsequent terms are
much smaller than the first term.
The first term is just 365 x (364/365)^(365^2)
=365 x ((364/365)^365)^365.
Now, (364/365)^365 is roughly 1/e; actually, it's about
0.99862857x(1/e). Thus, the first term is close to
365 x (0.99862857/e)^365
or about 221.181/(e^365).
Also, the second term is
(365x364/2) x {((363/365)^365)} ^365
where (363/365)^365 is close to 1/(e^2) and so this term is
in the order of e^-(2x365) which is much smaller than the
first term. One can see that each of the following terms
is then roughly smaller by a factor of e^365 (not quite
because of the combinatorial factor) compared with the
previous term.
In any case, P(B)/P(A) is about 4.61756 and P(B) is really
bigger and these probabilities are much closer than I had
originally thought!
LAZY MAN's APPROXIMATION:
This is the idea that gives one an immediate guess as to
which of these probabilities is bigger. Without caring too
much about how closely 1/e approximates (364/365)^365, one
can see that
P(A) looks like sqrt(2 x pi x 365)/(e^365)
and P(B) is (very!) roughly
365/(e^365).
(I had forgotten the numerator when typing my original
post!). So someone who is really careful and astute can
readily deduce that roughly
P(A)/P(B) = sqrt(2 x pi/365) or about 0.1312. (Actually,
you only need to know that 365 > 2 x pi for guessing that
B is more probable!)
Of course, unfortunately, the 1/e approximation is extremely
crude and hence one has to be careful since raising a crude
approximation to the power of 365 really distorts the
result!
[/ QUOTE ]
I just envisioned a new person to this site going to this thread first to pick up some basic odds for, I don't know, a draw or something and coming to the above post.
Math overload, alert, alert, his brain thinks, as he believes that his two soooooted cards should make him a flush, you know, "most of the time."
Barron Vangor Toth
www.BarronVangorToth.com (http://www.BarronVangorToth.com)
Couldn't you at least say if you came up with the same answer?
And what you think of my approach...
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