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Edited: Mar 30, 2020

Fun Probability Problem

in Math Problems

Hey everyone! I came up with a nice problem and I was really surprised with the result when I solved it! Here it is:

11 comments

11 Comments

Ian Fowler

Apr 16, 2020

Check this out.

Ian Fowler

Apr 02, 2020

I stand corrected. Thank You. That's what happens when you are in too big of a hurry. You are also correct in that there is more to this problem than meets the eye. Thanks for the update and I will soldier on - Ian

Ayrat Valiullin

Mar 30, 2020

I also want to suggest harder problem which I am still trying to solve. Go one dimension higher and now consider two concentric circles instead of two pairs of points. Find the probability for a random section inside the big circle to cross the smaller circle.

Ayrat Valiullin

Mar 31, 2020

Replying to

@Ian Fowler Exactly. I want the section that connects my two random picked points to include one or both of x=+-r.

I got your explanation of why it approaches infinity. Sounds reasonable, thanks!

Ian Fowler

Mar 31, 2020

Replying to

Circle problem

I shall use k for "pi"

Area of large circle = kR^2

Area of small circle = kr^2

Area of ring = kR^2 - kr^2 = k(R^2 - r^2)

Choose 2 points P1 and P2 at random inside the large circle. What is the probability that the line joining P1 and P2 will cross the smaller circle. I will show 2 methods.

1) P1 is in the small circle. P = kr^2 / kR^2 = r^2 / R^2

P2 must lie in the ring. P = k(R^2-r^2) / kR^2 = (R^2 - r^2)/ R^2

Mulitply since both must happen

P(crossing) = r^2(R^2 - r^2) / R^4

We repeat the process for P2 in the small circle and P1 in the ring and get the same answer so we double.

P(crossing)

= [r^2/R^2] [(R^2-r^2)/R^2] *2

= 2r^2(R^2 - r^2) / R^4

2) P(crossing)

= 1 - both lie in the ring - both lie in the small circle

= 1 - [k(R^2-r^2)/kR^2][k(R^2-r^2)/kR^2] - [kr^2/kR^2][kr^2/kR^2]

cancelling out the "k"s

= 1 - [(R^2 - r^2)^2/R^4] - [r^4/R^4]

R^4 is the common denominator

= [ R^4 - (R^4 - 2r^2R^2 + r^4) - r^4 ] / R^4

= [R^4 - R^4 + 2r^2R^2 - r^4 - r^4] / R^4

= [ 2r^2R^2 - 2r^4] / R^4

factoring the numerator

= 2r^2(R^2 - r^2) / R^4

BTW you can still apply your analysis when r is fixed and R ---> inf

Now the chances are that both P1 and P2 will be in the ring and P(crossing) ---> 0

Ayrat Valiullin

Apr 02, 2020

Replying to

@Ian Fowler I understand your analysis, but it's not correct. If P1 and P2 both lie in the ring, they still can cross the smaller circle (they can also touch it):

Let φ=r/R. Then the line joining P1(x1,y1) and P2(x2,y2)

1) Definately never crosses if they both lie in the small circle

Probability of that happening = φ^4

2) Definately crosses if they lie in different areas (you considered it would be enough)

P = 2φ^2(1-φ^2)

3) We don't know/depend on the angle/distance from the centre/a lot of parameters

if they both lie in the ring

You may look at what I did in Desmos:

https://www.desmos.com/calculator/agcwvhxo2f

I tried to simulate "the line crossing the circle" as the problem of the system of equations having roots

Ian Fowler

Mar 30, 2020

Nice analysis. Makes sense.

Consider this:

P(1) = 2r(R-r) / R^2 and keeping r constant

Differentiate P(1) wrt R and set equal to 0 yields R = 2r as a stationary point.

Differentiating again to get the second derivative and substituting R = 2r yields the constant -2 which proves R = 2r gives the maximum value of P(1).

Substituting R = 2r back into P(1) gives the maximum value of P(1) = 1/2r^2

I think this is worth a video on bprp.

Ayrat Valiullin

Mar 30, 2020

Replying to

Indeed. I understand why P(R/r=3)=2/3 have to be a maximum value. But how can you say it without using differentiation and P(R) function? I know there should be simple explanation. And the question about the probability approaching 1/2 at infinity is still up in the air.

Ian Fowler

Mar 30, 2020

Note: (a,b) is the open interval from a to b, a>b 1) Exactly 2: P1 lies in either (r,R) or (-R,r). P = 2(R-r)/2R P2 must lie in the other interval. P = (R-r)/2R Multiplying and reducing gives p(2) = (R-r)^2 / 2R^2

2) Exactly 1: P1 lies in either (r,R) or (-r,R), P = 2(R-r)/2R

P2 lies in (-r,r). P = 2r/2R

doubling since we can switch P1 and P2

Multiplying and reducing gives P(1) = 2r(R-r) / R^2

3) Exactly 0 : There are 3 cases:

a) P1 and P2 both lie in (-R,-r). P = (R-r)^2/4R^2

b) P2 and P1 both lie in (-r,r). P =( 2r/2R)^2 = r^2/R^2

c) P1 and P2 lie both on (r,R). P = (R-r)^2/4R^2

Getting a common denominator and adding gives P(0) = [2(R-r)^2 + 4r^2] / 4R^2

Check:

P(2) + P(1) + P(0)

= (R-r)^2/2R^2 + 2r(R-r)/R^2 +[ 2(R-r)^2 + 4R^2 ] / 4R^2

Getting a common denominator and simplifying gives:

4R^2/4R^2 = 1

Thanks for the great problem. I enjoyed it. - Ian

Ayrat Valiullin

Mar 30, 2020

Replying to

Thanks, Ian. Glad you liked it. Did you notice that at R=3 (when r=1) probability function P(R) reaches its maximum P(3)=2/3? And the fact that P(R) approaches 1/2 as R goes to infinity also surprises me.

It it quite obvious as you look at the function but isn't it interesting from the "probability point of view"?

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Fun Probability Problem