RotMG WORLD RECORD using Gambler's Fate (O3 White)
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- Опубликовано: 6 ноя 2024
- I used Python to figure out how long this might take! If you're interested in coding, ask me about it in the comments-I'd love to chat and can point you toward resources :)
Are you willing to risk it all on a coin?
The world record for longest documented Gambler's Fate prism is now 10 seconds... Can you top that? If you're not looking for a little luck in your life, you could also just use the T7 prism to get get 10 seconds-but it will cost you a whopping 5 more MP!
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Basically, in simple terms, there's a 50% chance that the coin will change from a gold coin to a silver coin, then a 50% chance it will change from a silver coin to a gold coin. We can just find the average time for one iteration and then multiply by two to cover the other iteration. The probability you get a gold coin to land when you press spacebar is 100%, or 1. After that, you multiply the probability of getting a gold coin by 50%, or 1/2, since there's a 50% chance that the gold coin continues to exist. It then becomes 1/2, then 1/4, then 1/8, so on forever. The average time of the coin is the sum of this infinite series, where the first term is 1 and the common ratio is 1/2. 1(1-0.5) = 1/0.5 = 2. Therefore, the average lifetime is expected to be 4 seconds, which is what you got through simulation.
I don't think the sum of that series gives the average time in general. Those are just chances, not multiplied by the actual times they are weights for. It gave the right answer here, but that seems coincidental.
@@hecate_uwu This still gives the total average time. Think about it as both probability and time. Since the new decision of whether a coin will change color/disappear changes every second, the 0.5 means that 50% of the time, the coin will continue to exist or it will disappear; in other words, it will either last 1 second or 0 seconds, which averages to 0.5 seconds. 0.25 now means that the coin will exist for 1 additional second 1 out of 4 times and 0 additional seconds 3 out of 4 times, which averages to 0.25. This goes on for each additional second the coin exists. Therefore, instead of 1 + 0.5 + 0.25... being the probabilities, it's the average lifetime of the coin each additional second; therefore, when we add up the infinite series, we get the average time of the entire lifetime of the coin.
@@ankitroy1366 ah yeah good point
Once it’s silver it has a chance to disappear not turn gold, your reasoning says it would last forever
@@Lucas-tn7xq it's clearly just a typo
Why does the outro give nostalgia
Good old mothrship days
Brings me back to the early days :')
so basically to find the average time the coin takes to turn silver, you're looking at a 1/2 for the first second, and if it doesn't turn silver (the other half of that 1/2) there's a 1/2 it will next second, for a 1/4, giving us a 1/(2^n) it turns silver each second. If we then multiply by the time that would have passed at that point, we get n/(2^n). Summing this over all positive n gives us the expected time it will be gold. The same can be done for silver, leading to the sum of 2n/(2^n) over all positive n. The sum of x^n over all positive n converges to 1/(1-x) if |x| < 1. If we take the derivative of both sides of this equation, we get the sum of nx^(n-1) for all positive n converges to 1/(1-x)^2. if we pick x = 1/2, this shows the sum of n/2^(n-1) (which is equivalent to what we found for the coin above) = 1/(1-1/2)^2 = 4 exactly, as your simulation suggested.
ok so this is pretty unreadable but i did my best for a youtube comment
@Ryan Puska Combinatorics, I'd say. If you're interested in using power series like this, you should look into generating functions. If you have some undergraduate background in discrete math and a little bit of series (what you learn from calc II or so), I recommend taking a look at a book called generatingfunctionology. You should be able to find a pdf by googling that if it sounds like it'd be of use.
This is a negative binomial distibution with r=2, p=1/2. This gives us a mean of 2, which means 2 successes, leaving us with an average of 4 seconds until failure. I'm sure there's a calculator online to compute exact values for each number of success. en.wikipedia.org/wiki/Negative_binomial_distribution
Long story short, I believe the mathematical equation for this calculation would be 6x × £°I + 80 ÷1.330 ₩^5/y^8(910.73 × 56) + i/729.284 - ¥~||~453 = 68% = 0.68 + i ÷ 94 < 8281 ___ 762.382 ^ 5e10 + 72€ / 73z(792 + 920x^y×12) × 91n ÷625i % × ◇T^5 - 3x^2 + 2y - 5 [8×92.3728(72x - 1z^2 - 33y) 10%)=> 9+10. For reference, variable x would equal 72-58^9(123.270 × 63.1) the rest should be self explanatory. So in conclusion, this man's calculations are indeed correct.
Wow. Hoffster is so cool.
Agreed :D
😎
Wow, 4 seconds on average? That's like the T1 prism...
Wow! 3:20 u know how work in this cmd! genius!
I've hit 10s plus before, some random gave me a deca he was so surprised lol
Unofficial world record
The coin has blessed you. You are the chosen one
I did a nest on a level 12 knight with nothing but a t6 sword, t0 sheild and t8 armor, not even a ring. My fight or flight was all da way up.
Am I the only dumb here in comments ?
No don't worry I'm here too
Same
Video on the bard ppe when
Don't even start with me
this man threw a coin on my guild leader
Everyone must be judged by the coin.
50% chance of any number of spins because it either happens or it doesn’t
lol. Everything in this world is either a peanut or not a peanut
What is Hoffster’s resolution?
I think he's playing at 1440x900 right now
Yeah 1440x900 is the best since the GUI isn't too large and you can see the whole playable area
wasnt there someone who got a minute in shatts?
@@pqr2673 yup, maybe that person kept using the coin in the same spot
What in the heck is that thumbnail
It's a dabbing Gemsbok
@@TheMotherShip What in the heck is that thumbnail
pog.
So does that mean it’s possible to last literally forever? Or is there a cap?
It could last 1 entire hour if you're lucky enough
It's possible for it to last "forever" if it's truly random. However, there's a good chance that the values aren't truly random since they are computer generated. So it's probably not possible for this to last forever, but theoretically it could if the randomness were truly random and you were infinitely lucky
What new nest reconstruction? Wtf
What do you mean?
first?!? pog
THIRD POGGGGGGGGGGGG