Need for needles

[JasonH]

http://xkcd.com/356/

:D This site sucked up almost all my reading time last night/this morning :|

Thanks!

 

[foges]

Can anyone recommend some reading material to understand these type of probability questions. I am an electrical engineering student, I have had basic probability and statistics, but nothing really similar to this..

thanks

 

[Andy Nguyen]

One of the most popular books in our Master reading list is 50 Challenging Probability problems. Other quant interview books have Probability questions as well.

 

[albertino]

Dropping the needles so that they all fall within the stick is equivalent to randomly picking \(n\) numbers (the left endpoints) each uniformly and independently drawn from \([0,1-h]\). Under this interpretation, the volume of the sample space is \((1-h)^n\). The volume of the region where the first needle is to the left of the second is to the left of the third ... is to the left of the \(n\)-th, is given by \(\int_0^{1-nh}\int_{x_1+h}^{1-(n-1)h}\cdots\int_{x_{n-1}+h}^{1-h}1dx_ndx_{n-1}\cdots dx_1\), which comes out to \(\frac{(1-nh)^n}{n!}\). To account for the ordering we need to multiply this by \(n!\). The probability we want is then \(\frac{(1-nh)^n}{(1-h)^n}\)

peterruse, you are right but I think that your solution will be more understandable if you simply use the definition of the uniform random variable instead of introducing sample space.
The joint probability density function of X1, X2, ... ,Xn will be fX1 X2 .. Xn(x1,.. , xn) = fX1(x1)^n = \(\frac{1}{(1-h)^n}\) since each marginal density function fXi = \(\frac{1}{1-h}\) if 0<xi< 1-h and 0 otherwise. Thus, the probability will be
\(n!\int_0^{1-nh}\int_{x_1+h}^{1-(n-1)h}\cdots\int_{x_{n-1}+h}^{1-h}\frac{1}{(1-h)^n}dx_ndx_{n-1}\cdots dx_1 = \frac{n!}{(1-h)^n} \int_0^{1-nh}\int_{x_1+h}^{1-(n-1)h}\cdots\int_{x_{n-1}+h}^{1-h}1dx_ndx_{n-1}\cdots dx_1\) = \(\frac{(1-nh)^n}{(1-h)^n}\)

 

[pruse]

peterruse, you are right but I think that your solution will be more understandable if you simply use the definition of the uniform random variable instead of introducing sample space.
The joint probability density function of X1, X2, ... ,Xn will be fX1 X2 .. Xn(x1,.. , xn) = fX1(x1)^n = \(\frac{1}{(1-h)^n}\) since each marginal density function fXi = \(\frac{1}{1-h}\) if 0<xi< 1-h and 0 otherwise. Thus, the probability will be
\(n!\int_0^{1-nh}\int_{x_1+h}^{1-(n-1)h}\cdots\int_{x_{n-1}+h}^{1-h}\frac{1}{(1-h)^n}dx_ndx_{n-1}\cdots dx_1 = \frac{n!}{(1-h)^n} \int_0^{1-nh}\int_{x_1+h}^{1-(n-1)h}\cdots\int_{x_{n-1}+h}^{1-h}1dx_ndx_{n-1}\cdots dx_1\) = \(\frac{(1-nh)^n}{(1-h)^n}\)

Okay? And? You essentially repeated my solution...

 

[albertino]

Okay? And? You essentially repeated my solution...

Yes, I actually repeated your solution only changing its explanation. If possible explain why do you use 1 instead of \(\frac{1}{(1-h)^n}\) in the expression
\(\int_0^{1-nh}\int_{x_1+h}^{1-(n-1)h}\cdots\int_{x_{n-1}+h}^{1-h}1dx_ndx_{n-1}\cdots dx_1\)?

 

[pruse]

Yes, I actually repeated your solution only changing its explanation. If possible explain why do you use 1 instead of \(\frac{1}{(1-h)^n}\) in the expression
\(\int_0^{1-nh}\int_{x_1+h}^{1-(n-1)h}\cdots\int_{x_{n-1}+h}^{1-h}1dx_ndx_{n-1}\cdots dx_1\)?

Because I am talking about a volume (specifically, of the part of the sample space where the needles fall properly onto the stick); when you find volume you integrate 1 over the region. Then I divide this volume by the volume of the sample space, \((1-h)^n\). This is essentially equivalent to what you did with density functions.

I chose not to go the more "mathematical" route of talking about density functions for a more intuitive feel...

 

[albertino]

Thanks, didn't know about it.