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Dieter
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Dieter
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May 8th, 2026 at 6:42:03 PM permalink
Quote: Wizard


p.s. Why am I advised to capitalize Googleplex?
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I think "googolplex" is the number, and "Googleplex" is Alphabet City.
May the cards fall in your favor.
SkinnyTony
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May 8th, 2026 at 9:19:01 PM permalink
I think that the answer depends on how the location of the portal is chosen. I assume that either path is equally likely but the probability distribution of the distance along the correct path is important (and without a rigorous definition of this we can't calculate the expected distance traveled, so we can't minimize it).

The probability distribution can't be uniform because there can't be a uniform probability distribution over an unbounded set (you can have one over an infinite set, like all the real numbers between 0 and 1, but it needs to be bounded -- there's no uniform probability distribution over all the positive reals)
Ace2
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May 8th, 2026 at 10:46:55 PM permalink
Quote: Wizard

An Evil Wizard transports you to a distant planet, where you are placed at the beginning of two different paths. As in the Led Zeppelin song, there are two paths you may go down and there's always time to go back and switch to the other path any time you wish as many times as you wish.

Somewhere, on one of the paths, is a portal taking you back home.

The path with the portal has a finite but unknown distance (for example, an inch or a Googleplex light years). The other is infinitely long.

What strategy should you take to minimize the expected walking time to find the portal?

p.s. Why am I advised to capitalize Googleplex?
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This problem has already been posted twice before and never answered. Even the “googleplex” term was used. Formerly it was framed as an evil wizard in an infinite “but not completely infinite” desert. Don’t waste your time on it
It’s all about making that GTA
charliepatrick
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May 9th, 2026 at 12:49:32 AM permalink
It sounds like an interesting problem however I'm quite busy this weekend. So these are my initial thoughts.

I had assumed for simplicitiy that there was an East-West path and with one road going West and the other road going East. There were mile markers and for simplicity the prize is at one of those markers. I had assumed that initially the prize had equal probability of being East or West of the starting position (MM0).

It became apparent that the probabilities of where the prize is couldn't be uniform but must be something like the setter picks a path at random and then works their way along the path. At each Mile Marker they roll a multi-sided die and (say) if it's a 1 drops the prize otherwise moves on. This means that at MM0 the probability distribution along a path is similar to being at MM1.

As has been said you travel to a mile point and have to ascertain whether it's best to look at the next one or turn round. The upside is finding the prize at the next MM quicker, and the downside is if the prize is at all the other MMs, then you've lost 2 miles.

For instance if the setter tossed a coin, at the beginning the chance of [the prize] being at MM1E is 1/4 (MM2E=1/8 etc.). At the beginning the chances of being West is 1/2 and of being East beyond MM2E=1/8. So at MM1E the chances are now W=4/6, MM2E=1/6, MM2+E=1/6. The additional distance saved for MM2E is at least 4, and probably 6. But there is a 5/6 chance you add 2 miles to the journey. So for it's best to tun round at MM1E.

I was then going to look at different probabilities and spot a pattern, but it does look as if you turn round when the above calculation says the chances of (the downside) being West [x 2 miles] outwieghs the chances of (the upside) being further East (x >4 distance from staring position). Given you know the shape of the probaibility curve, you can now look at the differential as you change the distances involved.

This would then give an answer such you turn round when the chances of being further East is lower than p.

The problem I see is you don't yet know the turning point for the West leg, so can't tell the "upside" accurately. It seems obvious you go further along the West leg than when you first turned on the East leg. However you do know the chances of being East or West are identical when you get to the same distance.
SkinnyTony
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May 9th, 2026 at 3:10:50 AM permalink
Quote: Ace2

This problem has already been posted twice before and never answered. Even the “googleplex” term was used. Formerly it was framed as an evil wizard in an infinite “but not completely infinite” desert. Don’t waste your time on it



I'm fairly sure that the point I made above is the reason it hasn't been answered (the question is not well defined) but I'd like to get your feedback (particularly if I'm wrong)
gordonm888
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May 9th, 2026 at 6:15:40 AM permalink
Quote: Wizard

Quote: ThatDonGuy

Label the two paths A and B, and let A(x) and B(x) be the points on each path that are distance x from the starting point.

Start by walking on path A, then the turnaround points are A(1), B(2), A(3), B(4), A(5), B(6), and so on.
The exact distance from, say, 0 to A(1) is irrelevant.

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You are right about establishing turnaround points. I would say the distances between them are very relevant. That is the point of the problem. For example, if you walked 100 miles down path A and turned around and eventually returned to path A, after not finding the portal on path B, then you wouldn't turn around again at 101 miles from the junction.
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If every mile of each path is essentially equally probable to contain the portal then after taking a hundred years to walk 1,000,000 miles down an infinite path why would you turn around and spend the next 100 years walking back to your point of origin? It would have profited you more to spend that next 100 year increment walking another 1,000,000 miles down the first path because you would have a chance to find the portal during that time.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
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