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ThatDonGuy
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August 14th, 2025 at 5:18:50 PM permalink

Again, choose some positive integer N, and then declare distances of D / N, 2D / N, 3D / N, and so on.
If the treasure is <= D/N away, the distance is 2 D/N
If it is between D/N and 2D/N away, the distance is 2 D/N + 2 x 2 D/N = 2 D/N (1 + 2)
If it is between 2D/N and 3D/N away, the distance is 2 D/N + 2 x 2 D/N + 2 x 3 D/N = 2 D/N (1 + 2 + 3)
...
If it is between (N-1) D/N and D away, the distance is 2 D/N + 2 x 2 D/N + 2 x 3 D/N + ... + 2 x ND/N = 2 D/N (1 + 2 + 3 + ... + N)
Since each of these is equally possible, the expected distance is
1/N x 2D / N x (1 + 3 + 6 + ... + N (N+1) /2)
= 2/3 D x (N + 1) (N + 2) / N
This is a minimum where N = sqrt(2)
It turns out that N = 1 and N = 2 both have an expected distance of 2D, so the two equally efficient answers are:
(a) Start by saying, "I will walk to the North Pole"
(b) Start by saying, "I will walk halfway to the North Pole"; if that is unsuccessful, say, "I will walk to the North Pole"

I will have to see if something like, "I will walk sqrt(2) / 2 of the distance to the North Pole" followed by "I will walk to the North Pole" is a better answer.

ChesterDog
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August 14th, 2025 at 5:50:12 PM permalink

Call the maximum distance from the start 1.

Suppose I decide I will make three trips at most. Let a = the distance from the start for the first attempt, and b = the distance from the start for the second attempt.

The probability that I find the treasure on the first attempt is a and the distance walked would be 2a.

The probability that I find the treasure on the second attempt is b - a and the total distance walked would be 2a + 2b.

The probability that I find the treasure on the final attempt is 1 - b and the total distance walked would be 2a + 2b + 2.

The mean distance I would walk would be: D = a (2a) + (b - a)(2a + 2b) + (1-b)(2a + 2b + 2).

Simplifying yields D = 2 + 2a(1 - b).

Since b is less than or equal to 1, D cannot be less than 2. If a = 0, then D = 2. Or if b = 1, D = 2. So, D = 2 is the best that can be done with a max of 3 trips.

D = 2 is also the distance for one trip max or two trips max. And I found the same for four trips max, so I D = 2 is the answer.
Wizard
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Wizard
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August 14th, 2025 at 6:28:22 PM permalink
Let me emphasize you don't know the maximum distance the treasure could be away. I could be 1 inch and it could be a googleplex miles. So, I prefer to not see any assumptions about a maximum distance. Whatever you assume, the treasure could be further away.

What would be good for purposes of discussion is an expected ratio of necessary travel to total travel.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
unJon
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August 14th, 2025 at 8:06:35 PM permalink
I have a feeling the answer to this question will contain the term:

1/e
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard
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August 14th, 2025 at 9:39:18 PM permalink
Quote: unJon

I have a feeling the answer to this question will contain the term:

1/e

link to original post



Mine doesn't.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Ace2
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August 14th, 2025 at 10:30:10 PM permalink
Quote: Wizard

Let me emphasize you don't know the maximum distance the treasure could be away. I could be 1 inch and it could be a googleplex miles. So, I prefer to not see any assumptions about a maximum distance. Whatever you assume, the treasure could be further away.

What would be good for purposes of discussion is an expected ratio of necessary travel to total travel.
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If the distance to the treasure is a random number between zero and infinity (if possible) then the expected distance is infinity. So you'd never get there regardless of method used
It’s all about making that GTA
AutomaticMonkey
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August 14th, 2025 at 11:14:04 PM permalink
Quote: Ace2

Quote: Wizard

Let me emphasize you don't know the maximum distance the treasure could be away. I could be 1 inch and it could be a googleplex miles. So, I prefer to not see any assumptions about a maximum distance. Whatever you assume, the treasure could be further away.

What would be good for purposes of discussion is an expected ratio of necessary travel to total travel.
link to original post

If the distance to the treasure is a random number between zero and infinity (if possible) then the expected distance is infinity. So you'd never get there regardless of method used
link to original post



It's specified you're in the middle of a desert, so it can't be infinite if there is a middle. You also have a compass to find north which implies a spherical surface with a magnetic field and that excludes infinity too.

Maybe there's a missing specification, that the treasure is in the desert, and if that's so can we use r as the distance from the middle of the desert to its northern limit?
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