OMG check out this RNG

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grindst0ne wrote:

70 Attempts... 1/500 per attempt... You lose an orb after one attempt. So to get 1 HH it would be 70/500. For the second, there are 69 orbs left, and so it would be 69/500. Thus, getting 3 successes would have a chance of (70/500)*(69/500)*(68/500) = 0.00263%.

However, if you look at the video, he gets the HHs and has some ancient orbs left. 17 Orbs are left. So, the actual chance is: (53/500)*(52/500)*(51/500) = 0.001124%.

This is incorrect. Consider if you had 500 Ancient Orbs. Your calculation would give this a 500/500 = 100% chance to get a Headhunter.

Compare to a similar but simpler case: flipping a coin. If you flip a coin twice, are you then guaranteed to get heads?

watching this thread is so depressing, seeing you guys think that the "RNG" is natural and not manipulated by whoever controls this game (GGG or tencent,i really dont know which one )

Please guys, by all means, keep flipping ancients orbs :D.
I'm no video expert so I will not bother studying the material but I tend to be defiant towards things that deviate too much from the expected, and this actually deviates quite a whole lot ^^'. As often in this forum, reading other players reactions and analysis is priceless.

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galuf wrote:

Please guys, by all means, keep flipping ancients orbs :D.
I'm no video expert so I will not bother studying the material but I tend to be defiant towards things that deviate too much from the expected, and this actually deviates quite a whole lot ^^'. As often in this forum, reading other players reactions and analysis is priceless.

ye idk whether to laugh or cry or both at its whole genius.

Never buying another supporter pack ever again

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taggedjc wrote:

"

grindst0ne wrote:

70 Attempts... 1/500 per attempt... You lose an orb after one attempt. So to get 1 HH it would be 70/500. For the second, there are 69 orbs left, and so it would be 69/500. Thus, getting 3 successes would have a chance of (70/500)*(69/500)*(68/500) = 0.00263%.

However, if you look at the video, he gets the HHs and has some ancient orbs left. 17 Orbs are left. So, the actual chance is: (53/500)*(52/500)*(51/500) = 0.001124%.

This is incorrect. Consider if you had 500 Ancient Orbs. Your calculation would give this a 500/500 = 100% chance to get a Headhunter.

Compare to a similar but simpler case: flipping a coin. If you flip a coin twice, are you then guaranteed to get heads?

Indeed. Ironically, people's imagining of what randomness looks like is often the exact opposite of randomness. They see streaks and suspect it's not random, when in fact true randomness is full of streaks. It's the relatively smooth switching of outcomes (T-H-T-H-T-H...) that's very non-random.

"

taggedjc wrote:

"

grindst0ne wrote:

70 Attempts... 1/500 per attempt... You lose an orb after one attempt. So to get 1 HH it would be 70/500. For the second, there are 69 orbs left, and so it would be 69/500. Thus, getting 3 successes would have a chance of (70/500)*(69/500)*(68/500) = 0.00263%.

However, if you look at the video, he gets the HHs and has some ancient orbs left. 17 Orbs are left. So, the actual chance is: (53/500)*(52/500)*(51/500) = 0.001124%.

This is incorrect. Consider if you had 500 Ancient Orbs. Your calculation would give this a 500/500 = 100% chance to get a Headhunter.

Compare to a similar but simpler case: flipping a coin. If you flip a coin twice, are you then guaranteed to get heads?

Nope.
With 500 orbs, there are 500 possible outcomes (yes/no). However, after one attempt, the orb is spent. So, for the next attempt, you'd have 499 possible outcomes. This boils down to: (500/500)*(499/500)*(498/500) ... (1/500) = (500+1)/2/500 = 0.501% chance.

With 500 orbs, there are more than 500 possible outcomes and I'm not really sure what you're meaning by that.

You first claim that your chance of getting a Headhunter would be 70/500 but then you just said that with 500 orbs you would have only a 0.501% chance of getting one.



Can you please explain where you're getting your numbers from? It may help to use a smaller, easier to intuit example. For instance, we can use dice instead. Have a 1 on a six-sided die represent a success (getting a Headhunter) and have some number of dice be your pool of Ancient Orbs.

If you have four dice, what's your chances of getting at least one 1?

If you have six dice, what's your chances of getting at least one 1?

If you have one dice, what's your chances of getting at least one 1?

If you have thirty dice, what's your chances of getting at least one 1? How about your chances of getting at least two 1s?

What's your chances of getting at least two 1s if you only have four dice like in the first example?

If you can show how your calculations would answer these questions, I may be able to help you find where the error in your reasoning lies (and provide a good example experiment you can do to see if your calculation truly is incorrect!) but it's really hard to do this for the big numbers (like Headhunters from Ancient Orbs) since people naturally have difficulty with intuition regarding large numbers and probability, and no "correcting of formulas" is going to help with that understanding more than just seeing and comparing real understandable results with what was predicted with the formulas.

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wullack wrote:

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Shagsbeard wrote:

nah... simple combinatorics

C(3,60) (1/500)^3 (499/500)^57 is the probability you want.

Unfortunately this is not correct.

No sir... that is exactly the probably of getting exactly what he specified. You might not understand what that was, but it doesn't make me incorrect.

I know where this is going... you're going to claim that I didn't take into account that a player could stop once he got three HHs, or that if he got 4 or 5 HHs, he would technically have 3. I've heard it all before. Trust me. The probability I was asked for was the probability of getting 3 headhunters with 60 tries with a probability of success of 1/500. That's what you got.

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Aim_Deep wrote:

Is this a calculus problem? I never even took algebra but there is no way it's 500:1 still. He basically won lotto 3x in a row.


Let me ask another way. Your given 60 ancients. What are odds you'll roll 3 HH if chance is 1 in 500 per ancient.

For problems like that you can Google a binomial probability calculator and plug your numbers, first result off Google: https://stattrek.com/online-calculator/binomial.aspx
Probability of success is 1/500 so 0.002, 60 trials, 3 successes and calculate, in this case the probability is 0.00024423666 or 1 in 4094 or times 100 that's 0.0244% or if you look at P(X>=x) the chance of at least 3 Headhunters 0.0251% (But I'm pretty sure 1 in 500 is lower than the actual probability).