Odds getting specific quanta from a quantum tower.

[jmdt]

yugenotaht got 6 entropy quanta from a single quantum tower on his first turn.   After that we started having a math discussion about the probability of this occuring.  Sadly its been 6-7 years since I had statistics so I can't remember the correct way to calculate the probability.

Part of me thinks its  1/12^N, but that doesn't seem quite right either.  That would make the odds of getting 6 quantum of one element 1 in 2,985,984.

So what is the probability of getting 1, 2, 3, 4, 5, or 6 quanta of one element from a quantum tower?

[Mastermind79]

I remember somebody posted (maybe zanzarino himself) saying it was like this:

1. Choose random 1-12
2. Repeat 1 until the number of quanta has been reached.


Edit: here it is:

0. Check if there is enough quanta doing the summation of all pool. If yes:

1. Generate a N random number between 1 and 12
2. Is there quanta in the N pool?      Yes: quanta-1          No:GoTo1
3. Enough quanta removed?             Yes: The end            No:GoTo1

[Appawesome]

That's all I know.

.	Specific	Non-Specific
1	50%	600%
2	??	??
3	??	??
4	??	??
5	0000037%	0000405%
6	0000003%	0000040%

Rounded.
That's all I know.
Anyone fill in 2,3,4?

[catalyzeme]

The odds of getting 6 of a single element should be (1/12)^5, or 1/248,832, if it is random. (1/12)^6 would be the odds of getting 6 Entropy specifically, or any other individual element.

[Appawesome]

The odds of getting 6 of a single element should be (1/12)^5, or 1/248,832, if it is random. (1/12)^6 would be the odds of getting 6 Entropy specifically, or any other individual element.

No there are 12 possible ways of getting 12 of a single element.

6 Entropy, 6 death, 6 gravity, 6 earth, 6 life, 6 fire...

[Dragoon1140]

The odds of getting 6 of a single element should be (1/12)^5, or 1/248,832, if it is random. (1/12)^6 would be the odds of getting 6 Entropy specifically, or any other individual element.

No there are 12 possible ways of getting 12 of a single element.

6 Entropy, 6 death, 6 gravity, 6 earth, 6 life, 6 fire...

The possibility of each of those happening is also (1/12)^6, as stated by catalyzeme.  If you are talking about the possibility of getting 12 quanta of one element, then the equation would be (1/12)^12.

[Zeru]

Is it really hard to open wikipedia/google/book?

It's really simple. We are starting with our 1st quanta. There are 12 possible events (each represented by it's elements).
Let's say Entropy was chosen. The odds for this was 1/12.
Now what are the odds for a second entropy quanta. Again you have 12 possible events, form with only 1gives you  . Again we have 1/12
However, we already did something that had 1/12 chance of appearing. This means that the total chance is 1/12 * 1/12 = 1/144.

Again what are the odds for n  . It's 1/12, so each times you want more of the same thing you just multiply the odds by 1/12

So for 6  the odds are (1/12)^6 = 1/2985984

But lets assume that we play a game where the winning event is 6 of any element.
The 1st round is 100% win, because each elements can potentially win. Now for the next 5 rounds you must use the reasoning posted higher.
That gives you 1 * (1/12)^5 = 1/12)^5

Please don't criticize catalyzeme when he is right.


Appawesome: The odds of any event in any game is between 0% and 100%. You either had it totally wrong or invented a brand new math theory.


Hopefully this post helps.

[teffy]

That could help:
http://stattrek.com/Tables/Binomial.aspx

It´s the same website as in the thread with the hypergeometrical distribution, but here, it´s the binomial distribution

Here:
1 Tower, 6 trials, probability of success: 1/12 or 0.0833333...
Number of success: whatever you want.

[Unholy Spirit]

Zeru is right, sort of. Though at the most technical level, there's no such thing as chances or odds. So to answer your question, I'd need to know the exact specifics of the situation, which nobody knows. In your situation, the "odds" were obviously 100%. In most other situations, it would have been 0%.