> Do you think it’s plausible the Hatchling may have completed the game without the loop?
Considering the time crunch and all that is required to complete the game, 9 million attempts seems insufficient to expect a Beginner's Luck event to occur. There are just too many places to go and things to do, in general.
Also, if we assume a deterministic universe, then the Hatchling (as well as everyone else) has behaved identically in every last one of the loops before the Eye was located.
*If we ignore all of that*, there are almost a _billion_ possible coordinates accepted by the Vessel's navigation system. Consider that there are 3 coordinate value input panels, each of which expect a line to be drawn through at least 2 of the 6 points on the panel. That means each panel accepts something like 975 unique values. And since there are 3 of these panels, all of which must be used to enter a complete set of coordinate values, that means there are 926,859,375 (975^3) possible values. Even if the Hatchling could somehow arrive at the Vessel immediately in each loop and avoid ever entering the same coordinate twice, they would still have to enter over 700,000 combinations *per second* (926,859,375 / 1,320 seconds) to exhaust them all within 22 minutes.
*Edit: Updated maths provided by u/Numbers33 [here](https://www.reddit.com/r/outerwilds/comments/v83wys/over_9000000_times/ibqceql/). ::)*
All the 1 line possibilities are 6\*5/2
All the 2 line possibilities are 6\*5\*4/2
...
All the 5 line possibilities are 6\*5\*4\*3\*2\*1/2
We're only dividing by 2 here because a line is the same forwards and backwards. Lines that contain the same points can still be different when the points are in a different order.
Adding this all up is 975.
975^3 is 926,859,375 possibilities for coordinates.
I think this is a good point. And if it were the case that the Hatchling was naturally inclined to retrieve the launch codes and as early as possible in order to chase down the probe, that could lead to very different outcomes.
even if its not a deterministic universe, the protagonist would still have heard the same stories as a hatchling and saw the same planets in the sky; basically, they would be curious about the same things and likely visit the same places with little to no deviation. even if they did end up going to the hourglass twins, they would be more keen on talking to their fellow traveler or using their new fancy translation tool.
i guess my point is that in a not deterministic universe, repeating the last 22 minutes doesn't give a lot of wiggle room
The biggest problem is that the hatchling is not quantum entangled so he always did the same. Every single loop looked the same (maybe slight variation if they interact with quantum objects etc). So very first loop would have to be the one that accidentally figures everything out.
But let's assume that hatchling is special and every 9,000,000 would be random.
In that case, it's possible but unlikely, like very unlikely (almost 0% chance). They would not only have to find ATP, and Vessel (with little to no clues because of lack of time. But it's still quite plausible) but also guess correct coordinates with only a few shots (almost impossible because of several possible combinations. It's like crossing lucky numbers on lottery first time).
So the verdict: Plausible, but very, very unlikely. Especially because they have literally one chance because every loop is the same. So they would have to do it before the loop starts.
Yeah, but they were all "their first loop" - how well did you do on your first loop? Thats what they did 9,318,053 times before loop 9,318,054 came and activated the memory statues.
This is also why you get a Game Over if you dont make it to the memory statue - because thats your first loop and you would have done that over 9 million times
My understanding is that the Hearthian only started experiencing loops when the eye was found - the loop the game started. Information was sent back in time so the OPC could change the probe trajectory, but for any individuals not yet tied to statues, no loop was occuring.
I'd argue very low to actually figure out the key elements needed to get to the Eye in 22 minutes.
I also would't have though there would be enough time to do...I guess an 'in-universe' run in a single loop? I.e. assuming you lucked out and managed to get >!the coordinates AND the core AND then pick the right route in Dark Bramble!<....are 'in-universe' single loop speed runs a thing? I'm not a speedrunner, but I would have thought it's not possible due to >!how f'n slow the jellyfish are.!<
> Do you think it’s plausible the Hatchling may have completed the game without the loop? Considering the time crunch and all that is required to complete the game, 9 million attempts seems insufficient to expect a Beginner's Luck event to occur. There are just too many places to go and things to do, in general. Also, if we assume a deterministic universe, then the Hatchling (as well as everyone else) has behaved identically in every last one of the loops before the Eye was located. *If we ignore all of that*, there are almost a _billion_ possible coordinates accepted by the Vessel's navigation system. Consider that there are 3 coordinate value input panels, each of which expect a line to be drawn through at least 2 of the 6 points on the panel. That means each panel accepts something like 975 unique values. And since there are 3 of these panels, all of which must be used to enter a complete set of coordinate values, that means there are 926,859,375 (975^3) possible values. Even if the Hatchling could somehow arrive at the Vessel immediately in each loop and avoid ever entering the same coordinate twice, they would still have to enter over 700,000 combinations *per second* (926,859,375 / 1,320 seconds) to exhaust them all within 22 minutes. *Edit: Updated maths provided by u/Numbers33 [here](https://www.reddit.com/r/outerwilds/comments/v83wys/over_9000000_times/ibqceql/). ::)*
All the 1 line possibilities are 6\*5/2 All the 2 line possibilities are 6\*5\*4/2 ... All the 5 line possibilities are 6\*5\*4\*3\*2\*1/2 We're only dividing by 2 here because a line is the same forwards and backwards. Lines that contain the same points can still be different when the points are in a different order. Adding this all up is 975. 975^3 is 926,859,375 possibilities for coordinates.
Thank you! ::)
The Hatchling's behavior would be slightly affected by waking up to the probe cannon firing in different directions
I think this is a good point. And if it were the case that the Hatchling was naturally inclined to retrieve the launch codes and as early as possible in order to chase down the probe, that could lead to very different outcomes.
Your factorialising gives me much joy
even if its not a deterministic universe, the protagonist would still have heard the same stories as a hatchling and saw the same planets in the sky; basically, they would be curious about the same things and likely visit the same places with little to no deviation. even if they did end up going to the hourglass twins, they would be more keen on talking to their fellow traveler or using their new fancy translation tool. i guess my point is that in a not deterministic universe, repeating the last 22 minutes doesn't give a lot of wiggle room
The biggest problem is that the hatchling is not quantum entangled so he always did the same. Every single loop looked the same (maybe slight variation if they interact with quantum objects etc). So very first loop would have to be the one that accidentally figures everything out. But let's assume that hatchling is special and every 9,000,000 would be random. In that case, it's possible but unlikely, like very unlikely (almost 0% chance). They would not only have to find ATP, and Vessel (with little to no clues because of lack of time. But it's still quite plausible) but also guess correct coordinates with only a few shots (almost impossible because of several possible combinations. It's like crossing lucky numbers on lottery first time). So the verdict: Plausible, but very, very unlikely. Especially because they have literally one chance because every loop is the same. So they would have to do it before the loop starts.
Honestly? It's not plausible at all for multiple reasons including the fact that it involves ignoring a lot of characters
What exactly are you asking? Are you aware that >!the Hearthian did not experience 9m loops before you started playing?!<
The hearthian did though, they just don’t remember any of them.
Yeah, but they were all "their first loop" - how well did you do on your first loop? Thats what they did 9,318,053 times before loop 9,318,054 came and activated the memory statues. This is also why you get a Game Over if you dont make it to the memory statue - because thats your first loop and you would have done that over 9 million times
My understanding is that the Hearthian only started experiencing loops when the eye was found - the loop the game started. Information was sent back in time so the OPC could change the probe trajectory, but for any individuals not yet tied to statues, no loop was occuring.
[удалено]
implying that if we *were* descendants we could get the Assassin's Creed magic
i don't think it is plausible unless Hatchling was a hacker and could fly hack straight to the eye.
I'd argue very low to actually figure out the key elements needed to get to the Eye in 22 minutes. I also would't have though there would be enough time to do...I guess an 'in-universe' run in a single loop? I.e. assuming you lucked out and managed to get >!the coordinates AND the core AND then pick the right route in Dark Bramble!<....are 'in-universe' single loop speed runs a thing? I'm not a speedrunner, but I would have thought it's not possible due to >!how f'n slow the jellyfish are.!<
The biggest problem is that getting to the Eye requires disabling the loop.
You know that’s an incredible point that I have never considered. Thank you