Thank you for the comment, this conceptual difficulty is largely the reason I wanted to make a video on the topic. I would say the easiest way to convince yourself of the result of this problem is to take two coins, flip them, and mark how many times one is heads and one is tails vs how many times both are heads (and just ignore the result when they're both tails). The fact that TH and HT are two unique cases will become apparent and demonstrate that HH has only a 1/3 probability of occurring. In terms of revealing information, the fact that the premise of the problem states that one coin must be heads "reveals" that the TT permutation must be discarded, but it isn't necessary to cover one/both coins at the start as that will have no impact assuming that you discard TT even when they're covered. The Monty hall problem can similarly be broken down into permutations (C = car G = Goat, CGG GCG GGC) which I could elaborate more on if desired

That's actually a different concept than what is presented here, you should try it out for yourself to see the result! Just follow the premise that 1 of the two coins must be heads

Hi, because the premise of the problem is that at least one of the coins is heads, the possibility of "tails, tails" is omitted, so there are 3 possibilities

Not to be rude but I just wanted to clarify for you, this is similar to the Monty hall problem. He said “at least” one of them is heads, so if you look at the 4 possibilities, one is removed due to there being 2 tails, then there is a 33% chance of it being heads as he is not saying that the first one is heads, only at least one of them kinda like the riddle “Someone has 15 cents in 2 coins, one coin is not a dime how is this possible” this problem relies on you believing that the *first* one is heads, hope this helps you and anyone in the future who sees this comment

Good, I’m not the only one.

I'm glad you think so! I encourage you to try out the experiment for yourself, it's really an interesting problem

You're definitely right about the issue of floating point numbers, however the existence of irrational numbers means that there is no system that can finitely represent any number (particularly, the result of any calculation) without rounding. For the purposes of this demo however a double precision float is probably more than enough precision.

The precision can also be greatly increased by adjusting the algorithm used to calculate the next step. Especially in a system like this, it’s important to use an algorithm that is energy conserving so that the system doesn’t magically add or remove energy from the system that will cause it to collapse or go haywire when it shouldn’t. There’s also the issue of adjusting the initial system to something “normalized” so that total system momentum is 0 (the center of mass of the system stays centered around zero and doesn’t drift), the total system rotation rotation is planar (so that the system of equations is simplified down to only the X-Y plane, avoiding extra error added in each step), and even ensuring all three points begin on the same plane (though this is only really evident in systems where the velocities are also somewhat coplanar, it does avoid making systems that have the same symmetries from looking wildly different), there could also be something said about ensuring some other initial ratios of distance and velocity/positions/masses of particles with certain relative properties are fixed and normalized.

Possibly... Start with body 1 and 2 in such a tight energetic orbital around each other, that they don't even affect body 3 as two separate bodies.

which one?

@ the KZbin one.

@@parallelgaming8424 oh

That's too kind of you, I hope both our channels can continue to grow. Happy holidays :]

@@Wunoumenal Happy holidays! :D i promoted your channel in my community tab kzbin.infoUgkxQeKD5jXw89FI7R6A9PJKh7xdJoz2uq6Y

Special Relativity relies heavily on the definition of simultaneity. Two events are defined to be simultaneous if light from those events reaches the midpoint at the same time. Clock synchronisation, that is, the definition of the clock time at which an event happens, is based on that definition as well. Imagine a long train, and the lightning strikes making burn marks on the roof of the train. The observer on the train is at the midpoint between those two burn marks, but the light from the front lightning strike reaches him before the light from the back lightning strike. Therefore, given the definition of simultaneity, they were not simultaneous for him.

@@renedekker9806 Interesting, I would describe Simultaneity as a relative measurement, of all events that occurred at a given time. This would not take into account the time light would take to reach the observer, so the events would appear at a time relative to their distance from the observer. As such, each observer would have their own line/plane/volume of simultaneity. (Depending on the number of spatial dimensions) By extension, I would not describe events that lie on your past light-cone as simultaneous, even if you would observe them happening simultaneously. With the train example, I'll correct my statement and agree that the observer on the train will witness the front lightning bolt before the point when they observe the lightning bolts is not the midpoint between them. However I don't think this experiment is the best for describing the effects of relativity. Instead, let's place the observer on the train slightly behind the middle such that they'll reach the midpoint between the lightning bolts at the same time as the light reaches them. Unfortunately I don't have the time to work this out at the moment, I need to include length contractions into this properly for it to make sense.

@@tachrayonic2982 _"relative measurement, of all events that occurred at a given time"_ - that is correct, The measurement itself does not take into account the time it takes for light to travel to any observer. But the question is: how do you determine whether two events occur at the same given time? For that, you could put a clock at the location of each event, that is stopped at the time the event occurs. But for it to be a valid measurement those clocks need to be synchronised beforehand. How do you determine whether two clocks are in sync? When the light of the clocks reaching 12 o'clock reaches the MIDPOINT between the clocks at the same time, then they are in sync. That is, the definition of simultaneity depends on the time it takes for light to travel to the midpoint. _"...is not the midpoint between them"_ - in the reference frame of the train, the observer on the train IS at the midpoint between the two events. So he can shortcut the whole two-synchronised-clocks thing, and simply check whether the light reaches him at the same time. After all, that is the definition of simultaneity.

@@renedekker9806 Alright, I've slept on it and I've got it sorted out. But first to answer your question, you can determine when an event happened in your reference frame by measuring how far an event is away from you when you observe it. From there you can calculate where+when it occurred in spacetime, and simultaneous events occurred at the same time in spacetime in your reference frame. Events being simultaneous will depend on the reference frame from which they're observed, although the events will always sit beyond the past and future light cones for all observers. However, all observers with the same velocity will be able to agree on their reference frame, and by extension whether or not events are simultaneous. In the typical experiment of special relativity, the precise location the person is standing on the train is irrelevant as we are not taking into account the time it takes for light to reach them. From there, an outside observer might witness the front of a train exit a tunnel at the same same time as the rear of the train enters it. From this outside reference frame, the train appears to the the same length as the tunnel. From the observer on the train however, the tunnel seems contracted. The front of the train exits the tunnel before the rear of the train enters it. The two observer's lines/planes/volumes of simultaneity do not align, the observer off of the train sees expect both events to be simultaneous and the observer on the train expects them not to be. And, to be clear, this does not depend on where each observer is standing, only that the observers have the same velocity as their objects. (The Train or the Ground) Once you take into account the speed of light from the event to the observer, your results will vary depending on where they're standing. Someone on the ground at the entrance to the tunnel will witness the rear of the train enter the tunnel before the front exits the tunnel, merely because the exit of the tunnel is further away. If they calculate how long it took the light from the end of the tunnel to reach them, they'll find the events were simultaneous. Similarly, it doesn't matter where on the train the observer sits. They will always be able to calculate the the event at the front occurred before the event at the rear, even if they're standing at the rear and witness it first.

Added a pinned comment with a link! If you're more interested in the code/project files, feel free to send an email to [email protected]

Thank you very much

😂