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Buffon's needle problem
By Yatir Halevi on Tuesday, July 30, 2002 - 06:51 pm:

I read once that the probability that a pin dropped from a certain height from a plain that consists of parallel lines will not fall on any line has in it p.
Does any body know anything about this?


Yatir

By Andre Rzym on Wednesday, July 31, 2002 - 08:30 pm:

Yatir,

Lets take the case when the length of the needle=separation of the lines, d. Hopefully you can then solve the general case.

Suppose the lines are at y=0, y=+/-d, y=+/2d etc.

Consider a square of side d centred on the origin. If we can figure out the probability of the pin touching a line given that the centre of the pin falls in the square, then can you see that this is the result for the entire plane?

What does it mean for the pin to fall 'randomly'? The centre of the pin can lie anywhere in the square with equal probability and at any angle to the x axis with equal probability. Therefore the probability of the pin being in [x,x+dx;y,y+dy;q,q+dq] is dxdydq/2d2p.

As a check,

òx=-d/2d/2òy=-d/2d/2òq=02pdxdydq/2d2p=1

which is a good start.

Now define 1(x,y,q) to be a function that is 1 if the pin touches a line and 0 otherwise. Then the probability of the pin touching is

òx=-d/2d/2òy=-d/2d/2òq=02p1(x,y,q)dxdydq/2d2p

Now 1 is not actually dependent on y, so

òx=-d/2d/2òy=-d/2d/2òq=02p1(x,y,q)dxdydq/2d2p=òx=-d/2d/2òq=02p1(x,q)dxdq/2dp

Symmetry for positive and negative x reduces this to

òx=0d/2òq=02p1(x,q)dxdq/dp=òq=02pòx=0d/21(x, q)dxdq/dp

Now the pin touches the line (i.e. 1=1) if |d/2.cos(q)|>x, so 1 is a function of |cos(q)| so we simplify further to

2òq=-p/2p/2òx=0d/21(x, q)dxdq/dp

and using the actual condition for touching:

2òq=-p/2p/2òx=0d/21(x, q)dxdq/dp=2òq=-p/2p/2òx=0d/2cos(q)dxdq/dp=2òq=-p/2p/2d/2cos(q)dq/dp

=òq=-p/2p/2cos(q)dq/p=2/p

Andre

By Yatir Halevi on Wednesday, July 31, 2002 - 09:16 pm:

I don't know much probability, so can you elaborate on how exactly you reached the integral. I guess it has to do with the fact you added an infinite number of small probabilities. (definition of an integral)...

Yatir

By Andre Rzym on Thursday, August 01, 2002 - 05:12 pm:

Yatir, you are correct, it is a question of sums converging to an integral.

First, a few bits of basic probability. If there are exactly 4 possible outcomes (a,b,c,d), with probability Pa,…Pd then we know

Pa+ Pb+ Pc+ Pd=1

If we have a function f which has values depending on the outcomes, then the expected value of f is

f(a).Pa+ f(b). Pb+ f(c). Pc+ f(d). Pd

Now suppose our trial yields not 4 discrete outcomes but x, a continuous variable for example in the range [0,1].

We could denote the prob. that x is in [0,0.5], [0.5,1.0] by p(0) & p(0.5). Similarly if we divided the range into 4 parts the probabilities would be q(0/4), q(1/4), q(2/4), q(3/4). Now our p’s will sum up to 1, just as our q’s will. However [ignoring pathogenic distributions] the more we divide the interval, the smaller the probabilities will become (think, for example, of x being uniformly distributed in [0,1]. Then each p=1/2, each q=1/4).

Instead, therefore, we denote the probability of being in an interval as being a function multiplied by the width of the interval. Therefore the analogue to p above would be a function P, such that the probability of being in [0,0.5], [0.5,1] is P(0).Dx, P(0.5).Dx respectively, where Dx=0.5. Ditto the analogue of q is Q, the probabilities being Q(0/4). Dx, Q(1/4). Dx, Q(2/4). Dx, Q(3/4). Dx where Dx=0.5

As you might guess, in the limit as Dx -> 0 we get the probability of x being in the range [x,x+dx]=Pr(x)dx. We call Pr(x) the probability density.

Since the sum of probabilities add up to 1, we must have

òPr(x)dx=1

Similarly the expected value of some function, f(x), is just the sum of the product of f(x) and Pr(x)dx, i.e.

òf(x)Pr(x)dx=1

Andre

By Yatir Halevi on Friday, August 09, 2002 - 01:30 pm:

why is the probability:
1/2d2p?

Yatir

By Andre Rzym on Friday, August 09, 2002 - 07:41 pm:

I take it you are referring to the probability density 1/2pd2 in paragraph 4.

Consider, first, the x coordinate of the pin centre. Since we assume that the distribution is uniform, the probability of it being in {x,x+D} is proportional to D, i.e.

pr[pin centre in {x,x+D}]=kD

Hence for a small interval, dx,

pr[pin centre in {x,x+dx}]=k.dx

Since we are assuming that the centre of the pin is in [-d/2,d/2] we require

1=ò-d/2d/2k.dx

=> k=1/d => probability density=1/d

The same argument applied to the y coordinate gives

pr[pin centre in {y,y+dy}]=1/d.dy However since we assume that there is no correlation between the x and y coord,

pr[pin centre in {x,x+dx;y,y+dy}]=pr[pin centre in {x,x+dx}].pr[pin centre in {y,y+dy}]
=[1/d.dx].[1/d.dy]=dxdy/d2

Now the angle of the pin to the x axis is also assumed to be uniformly distributed, so

pr[pin centre in {q,q+D}]=k.D

where q is in radians. Since the range of angles is [0,2p] we must have

1=ò02pk.dq => k=1/2p

Using the independence argument again,

pr[pin centre in {x,x+dx;y,y+dy;q,q+dq}]=pr[pin centre in {x,x+dx;y,y+dy}].pr[pin centre in {q,q+dq}]=[dxdy/d2].[dq/2p]

Andre

By Yatir Halevi on Saturday, August 10, 2002 - 10:32 pm:

Thanks a lot Andre, I get it now!


Yatir