Alien Abduction of Farmer Brown

[BMAD]

Farmer Brown is standing in the middle of his perfectly circular field feeling very content. It is midnight and there is no moon and unknown to the farmer, Martian zoologists are landing randomly at points on the circumference of his field. They land at one minute intervals, starting at midnight. As soon as there are martians at points A,B,C such that triangle ABC contains the center of the field, Farmer Brown will be teleported to the waiting space-ship and transported to spend the rest of his life as an exhibit in a Martian zoo. What is the expected time until he is abducted?

[gavinksong]

At any given time, if farthest Martians are X radians apart,

 

There is a X/2π chance that Farmer Brown will be captured in one minute.

There is a X/2π chance that the farthest Martians will still be X radians apart in one minute.

There is a 1-X/π chance that the farthest Martians will be Y radians apart in one minute, where π ≥ Y > X.

 

E(X) = X / 2π + (X / 2π) (1 + E(X)) + (1-X/π) (1/(π-X)) ∫ 1 + E(y) dy

[bonanova]

we determined the probability of 3 random circumferential points creating a triangle that included the center. It turned out that to be 1/4. That means that if this puzzle were changed to say that every minute three new Martians replaced the previous three, it would take on average four sets of Martians to kidnap the old farmer, which would happen at 12:03.

But this puzzle adds Martians to the already existing ones. The probability that the center is included, vs n was found to be n / 2^n-1: This gives, for 3 <= n <= 20, the following probabilities for the farmer to be taken by n Martians. OK how do we convert that to expected wait time?

 

0.25

0.5

0.6875

0.8125

0.890625

0.9375

0.964844

0.980469

0.989258

0.994141

0.996826

0.998291

0.999084

0.999512

0.999741

0.999863

0.999928

0.999962

[gavinksong]

we determined the probability of 3 random circumferential points creating a triangle that included the center. It turned out that to be 1/4. That means that if this puzzle were changed to say that every minute three new Martians replaced the previous three, it would take on average four sets of Martians to kidnap the old farmer, which would happen at 12:03.

But this puzzle adds Martians to the already existing ones. The probability that the center is included, vs n was found to be n / 2^n-1: This gives, for 3 <= n <= 20, the following probabilities for the farmer to be taken by n Martians. OK how do we convert that to expected wait time?

 

0.25

0.5

0.6875

0.8125

0.890625

0.9375

0.964844

0.980469

0.989258

0.994141

0.996826

0.998291

0.999084

0.999512

0.999741

0.999863

0.999928

0.999962

 

Actually, I just did the math. The probability of three random points on a circumference forming a triangle that contains the center is 3/4. You got the general formula right though. It was probably just a slip of the finger.

 

That said, it still doesn't answer the question because, like you said, it's not like Martians who already landed randomize their positions every minute. We need a formula that tells us the probability that n+1 Martians capture farmer Brown given that the first n Martians failed. Or something like that.

[gavinksong]

we determined the probability of 3 random circumferential points creating a triangle that included the center. It turned out that to be 1/4. That means that if this puzzle were changed to say that every minute three new Martians replaced the previous three, it would take on average four sets of Martians to kidnap the old farmer, which would happen at 12:03.

But this puzzle adds Martians to the already existing ones. The probability that the center is included, vs n was found to be n / 2^n-1: This gives, for 3 <= n <= 20, the following probabilities for the farmer to be taken by n Martians. OK how do we convert that to expected wait time?

 

0.25

0.5

0.6875

0.8125

0.890625

0.9375

0.964844

0.980469

0.989258

0.994141

0.996826

0.998291

0.999084

0.999512

0.999741

0.999863

0.999928

0.999962

 

I made the same mistake I accused you of. For some reason, I had solved for 1-P. The n=3 case is in fact 1/4, but you made the same mistake with your general formula. That is, it's really 1 - n/2

^n-1, which you already know.

 

Anyways, E(t) = ∑ t P(t), but P(t) has to be the probability that the martians catch poor farmer Brown at exactly t minutes, so not 1 - n/2^n-1. Could it be as simple as P(t) = (1 - n/2^n-1) - (1 - (n-1)/2^n-2) = (n-2)/2^n-1? I... My brain isn't functioning.

[BMAD]

I would first focus on the expected number of Martians that would land.

[bonanova]

Let's say that after k Martians have landed p(covering center) = .5.

Would that mean expect number is k and the time is 12:(k-1)?

[BMAD]

Let's say that after k Martians have landed p(covering center) = .5.

Would that mean expect number is k and the time is 12:(k-1)?

Are you asking me or Gavinksong or is this Rhetorical?

[DejMar]

"As soon as there are martians at points A,B,C such that triangle ABC contains the center of the field..."

Are the three points, A,B,C any set of three random points on the circumference of the field from an increasing permutation of triangles that would enclose the field's center? That is, if Farmer Brown has not been transported after the appearance of three alien zoologists, a,b,c, but, a minute later, a fourth martian, d, landed, there are four permutations of triangles, such that zero, one or two of the four permutations that involve the four landed martians form triangles that enclose the field's center point. Would each of the triangles that enclosed the field's center be considered triangle ABC (I.e., more than one triangular transportation zone might exist)?

Is it possible that a martian may land on an already occupied point [a Dorothy-Gingema event], and are martian's considered point-like beings for calculation of the probability?

 

Edited March 16, 2015 by DejMar

[k-man]

Not sure why this hasn't been answered yet, but I'll try to wrap up some loose ends here....

 

Given the general formula for the probability

p(n) that n martians will capture the poor farmer is p(n)=1-n/2^n-1 we can convert it into a recursive function P(n) for the probability that exactly the nth martian captures the farmer.

 

P(n) = p(n) * (1 - CP(n-1) ), where CP(n-1) is a cumulative probability that n-1 martians have captured the farmer (i.e. CP(n-1) = Sum of P(k) for k in 1...n-1)

 

For the first 10 martians the P(n) looks like this:

n  | P(n)

1   |  0

2   |  0

3   |  0.25

4   |  0.375

5   |  0.2578125

6   |  0.09521484375

7   |  0.01956939697

8   |  0.00225305557

9   |  0.00014492311

10 |  0.00000517746

 

Beyond 10 the probability drops off very quickly and is negligible for the expected value calculation. The expected value for the martian that captures the farmer is the sum of all n*P(n), which comes out to about 4.266, so the expected time of abduction is around 00:03:16.

 

 

Edit: formatting

Edited March 16, 2015 by k-man