Is there any simple way to solve $366^n(366-n)!geq 2times 366!$?
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I am teaching probability for junior high school students. On the last page there is a "challenge yourself" section asking
What is the least number of students in a classroom for the probability that at least two of them have their birthday falling on the same day of the year to be greater than $1/2$?
It leads to an inequality $366^n(366-n)!geq 2times 366!$. I have solved it with program as follows.
The answer is 23 students.
Question
I wonder whether there is another simpler way to solve it for junior high school students with pencil and paper only.
inequality
asked Nov 26 at 17:28
Artificial Stupidity
346110
add a comment |
up vote
2
down vote
favorite
I am teaching probability for junior high school students. On the last page there is a "challenge yourself" section asking
What is the least number of students in a classroom for the probability that at least two of them have their birthday falling on the same day of the year to be greater than $1/2$?
It leads to an inequality $366^n(366-n)!geq 2times 366!$. I have solved it with program as follows.
The answer is 23 students.
Question
I wonder whether there is another simpler way to solve it for junior high school students with pencil and paper only.
inequality
asked Nov 26 at 17:28
Artificial Stupidity
346110
en.m.wikipedia.org/wiki/Birthday_problem
– Sorfosh
Nov 26 at 17:33
This inequality is hard to solve explicitly, logarithms don't help much. Ways to go around the problem is to use an approximative model which yields good approximate results (fine enough since n is an integer). All presentations of this material that I've seen reference a table similar to yours.
– Olivier Moschetta
Nov 26 at 17:37
If I had to attack this by hand, I'd convert the factorial on the left to a Gamma function and try to see if calculus would help to find roots. I wouldn't be too optimistic though, and besides, I doubt any of your junior high students know calculus.
– Ben W
Nov 26 at 17:42
I think I have to reduce the number by asking "What is the least number of students in a classroom for the probability that at least two of them have the same blood type to be greater than 1/2?"
– Artificial Stupidity
Nov 26 at 17:56
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am teaching probability for junior high school students. On the last page there is a "challenge yourself" section asking
What is the least number of students in a classroom for the probability that at least two of them have their birthday falling on the same day of the year to be greater than $1/2$?
It leads to an inequality $366^n(366-n)!geq 2times 366!$. I have solved it with program as follows.
The answer is 23 students.
Question
I wonder whether there is another simpler way to solve it for junior high school students with pencil and paper only.
inequality
asked Nov 26 at 17:28
Artificial Stupidity
346110
I am teaching probability for junior high school students. On the last page there is a "challenge yourself" section asking
What is the least number of students in a classroom for the probability that at least two of them have their birthday falling on the same day of the year to be greater than $1/2$?
It leads to an inequality $366^n(366-n)!geq 2times 366!$. I have solved it with program as follows.
The answer is 23 students.
Question
I wonder whether there is another simpler way to solve it for junior high school students with pencil and paper only.
inequality
inequality
asked Nov 26 at 17:28
Artificial Stupidity
346110
asked Nov 26 at 17:28
Artificial Stupidity
346110
asked Nov 26 at 17:28
Artificial Stupidity
346110
asked Nov 26 at 17:28
Artificial Stupidity
346110
asked Nov 26 at 17:28
Artificial Stupidity
346110
346110
en.m.wikipedia.org/wiki/Birthday_problem
– Sorfosh
Nov 26 at 17:33
This inequality is hard to solve explicitly, logarithms don't help much. Ways to go around the problem is to use an approximative model which yields good approximate results (fine enough since n is an integer). All presentations of this material that I've seen reference a table similar to yours.
– Olivier Moschetta
Nov 26 at 17:37
If I had to attack this by hand, I'd convert the factorial on the left to a Gamma function and try to see if calculus would help to find roots. I wouldn't be too optimistic though, and besides, I doubt any of your junior high students know calculus.
– Ben W
Nov 26 at 17:42
I think I have to reduce the number by asking "What is the least number of students in a classroom for the probability that at least two of them have the same blood type to be greater than 1/2?"
– Artificial Stupidity
Nov 26 at 17:56
add a comment |
en.m.wikipedia.org/wiki/Birthday_problem
– Sorfosh
Nov 26 at 17:33
This inequality is hard to solve explicitly, logarithms don't help much. Ways to go around the problem is to use an approximative model which yields good approximate results (fine enough since n is an integer). All presentations of this material that I've seen reference a table similar to yours.
– Olivier Moschetta
Nov 26 at 17:37
If I had to attack this by hand, I'd convert the factorial on the left to a Gamma function and try to see if calculus would help to find roots. I wouldn't be too optimistic though, and besides, I doubt any of your junior high students know calculus.
– Ben W
Nov 26 at 17:42
I think I have to reduce the number by asking "What is the least number of students in a classroom for the probability that at least two of them have the same blood type to be greater than 1/2?"
– Artificial Stupidity
Nov 26 at 17:56
en.m.wikipedia.org/wiki/Birthday_problem
– Sorfosh
Nov 26 at 17:33
en.m.wikipedia.org/wiki/Birthday_problem
– Sorfosh
Nov 26 at 17:33
This inequality is hard to solve explicitly, logarithms don't help much. Ways to go around the problem is to use an approximative model which yields good approximate results (fine enough since n is an integer). All presentations of this material that I've seen reference a table similar to yours.
– Olivier Moschetta
Nov 26 at 17:37
This inequality is hard to solve explicitly, logarithms don't help much. Ways to go around the problem is to use an approximative model which yields good approximate results (fine enough since n is an integer). All presentations of this material that I've seen reference a table similar to yours.
– Olivier Moschetta
Nov 26 at 17:37
If I had to attack this by hand, I'd convert the factorial on the left to a Gamma function and try to see if calculus would help to find roots. I wouldn't be too optimistic though, and besides, I doubt any of your junior high students know calculus.
– Ben W
Nov 26 at 17:42
If I had to attack this by hand, I'd convert the factorial on the left to a Gamma function and try to see if calculus would help to find roots. I wouldn't be too optimistic though, and besides, I doubt any of your junior high students know calculus.
– Ben W
Nov 26 at 17:42
I think I have to reduce the number by asking "What is the least number of students in a classroom for the probability that at least two of them have the same blood type to be greater than 1/2?"
– Artificial Stupidity
Nov 26 at 17:56
I think I have to reduce the number by asking "What is the least number of students in a classroom for the probability that at least two of them have the same blood type to be greater than 1/2?"
– Artificial Stupidity
Nov 26 at 17:56
add a comment |
1 Answer
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The arithmetic-geometric inequality tells us
$$frac{N!}{(N-n)!}leleft(N-frac{n-1}{2}right)^n $$
and is quite sharp when $nll N$. Thus a good approximation for $n$ might be the solution of
$$N^n=2cdot left(N-frac{n-1}{2}right)^n,$$
or:
$$left(1-frac{n-1}{2N}right)^n approx frac 12$$
With $c:=n^2/N$ and for $ngg 1$,
$$left(1-frac{n-1}{2N}right)^napproxleft(1-frac c{2n}right)^napprox e^{-frac c2}.$$
This suggests $capprox 2ln 2$ and so $napprox sqrt{2Nln 2}$. With $N=366$, this crude approximation gives us $napprox 22.53$.
answered Nov 26 at 17:59
Hagen von Eitzen
275k21268495
add a comment |
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1 Answer
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1 Answer
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up vote
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The arithmetic-geometric inequality tells us
$$frac{N!}{(N-n)!}leleft(N-frac{n-1}{2}right)^n $$
and is quite sharp when $nll N$. Thus a good approximation for $n$ might be the solution of
$$N^n=2cdot left(N-frac{n-1}{2}right)^n,$$
or:
$$left(1-frac{n-1}{2N}right)^n approx frac 12$$
With $c:=n^2/N$ and for $ngg 1$,
$$left(1-frac{n-1}{2N}right)^napproxleft(1-frac c{2n}right)^napprox e^{-frac c2}.$$
This suggests $capprox 2ln 2$ and so $napprox sqrt{2Nln 2}$. With $N=366$, this crude approximation gives us $napprox 22.53$.
answered Nov 26 at 17:59
Hagen von Eitzen
275k21268495
add a comment |
up vote
2
down vote
The arithmetic-geometric inequality tells us
$$frac{N!}{(N-n)!}leleft(N-frac{n-1}{2}right)^n $$
and is quite sharp when $nll N$. Thus a good approximation for $n$ might be the solution of
$$N^n=2cdot left(N-frac{n-1}{2}right)^n,$$
or:
$$left(1-frac{n-1}{2N}right)^n approx frac 12$$
With $c:=n^2/N$ and for $ngg 1$,
$$left(1-frac{n-1}{2N}right)^napproxleft(1-frac c{2n}right)^napprox e^{-frac c2}.$$
This suggests $capprox 2ln 2$ and so $napprox sqrt{2Nln 2}$. With $N=366$, this crude approximation gives us $napprox 22.53$.
answered Nov 26 at 17:59
Hagen von Eitzen
275k21268495
add a comment |
up vote
2
down vote
up vote
2
down vote
The arithmetic-geometric inequality tells us
$$frac{N!}{(N-n)!}leleft(N-frac{n-1}{2}right)^n $$
and is quite sharp when $nll N$. Thus a good approximation for $n$ might be the solution of
$$N^n=2cdot left(N-frac{n-1}{2}right)^n,$$
or:
$$left(1-frac{n-1}{2N}right)^n approx frac 12$$
With $c:=n^2/N$ and for $ngg 1$,
$$left(1-frac{n-1}{2N}right)^napproxleft(1-frac c{2n}right)^napprox e^{-frac c2}.$$
This suggests $capprox 2ln 2$ and so $napprox sqrt{2Nln 2}$. With $N=366$, this crude approximation gives us $napprox 22.53$.
answered Nov 26 at 17:59
Hagen von Eitzen
275k21268495
The arithmetic-geometric inequality tells us
$$frac{N!}{(N-n)!}leleft(N-frac{n-1}{2}right)^n $$
and is quite sharp when $nll N$. Thus a good approximation for $n$ might be the solution of
$$N^n=2cdot left(N-frac{n-1}{2}right)^n,$$
or:
$$left(1-frac{n-1}{2N}right)^n approx frac 12$$
With $c:=n^2/N$ and for $ngg 1$,
$$left(1-frac{n-1}{2N}right)^napproxleft(1-frac c{2n}right)^napprox e^{-frac c2}.$$
This suggests $capprox 2ln 2$ and so $napprox sqrt{2Nln 2}$. With $N=366$, this crude approximation gives us $napprox 22.53$.
answered Nov 26 at 17:59
Hagen von Eitzen
275k21268495
answered Nov 26 at 17:59
Hagen von Eitzen
275k21268495
answered Nov 26 at 17:59
Hagen von Eitzen
275k21268495
answered Nov 26 at 17:59
Hagen von Eitzen
275k21268495
275k21268495
add a comment |
add a comment |
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en.m.wikipedia.org/wiki/Birthday_problem
– Sorfosh
Nov 26 at 17:33
This inequality is hard to solve explicitly, logarithms don't help much. Ways to go around the problem is to use an approximative model which yields good approximate results (fine enough since n is an integer). All presentations of this material that I've seen reference a table similar to yours.
– Olivier Moschetta
Nov 26 at 17:37
If I had to attack this by hand, I'd convert the factorial on the left to a Gamma function and try to see if calculus would help to find roots. I wouldn't be too optimistic though, and besides, I doubt any of your junior high students know calculus.
– Ben W
Nov 26 at 17:42
I think I have to reduce the number by asking "What is the least number of students in a classroom for the probability that at least two of them have the same blood type to be greater than 1/2?"
– Artificial Stupidity
Nov 26 at 17:56