Calling MATHamagicians! :)

[GM Rex Nihilo]

Math minds please help with an odds calculation,

I would like to know the odds of getting a particular psychic discipline power with a ML 3. Once the formula is known it can be used for all ML's

I know rolling a particular number w 3 dice is aprox 40% but I can't mathematically factor in is fact that you can reroll if you get the same number.

Well mathies GO!! I plead ignorance.

GM Rex Nihilo

[Chaoschrist]

Wouldn't it be something along the lines of

1/6 * 1/5 * 1/4

Since it's a 1 in 6 chance to get the first, then a 1 in 5 chance because of the re-roll and then a 1 in 4 chance for 2 re-rolls. To me that would make up an 84% chance.

But it's been a while since I've been to school and actually did math. So if anything I'd be interested to see if I'm right or if I'm not, how it does work.

[hyudun]

EDIT: LMFAO I was waaaaaaay over-complicating and over-thinking things with my initial proof, though it got me to the right answer at the end. If there was a real statistician keeping up with my updates, he/she would be laughing him/herself to death. I'm definitely laughing at myself.

The answer is 50%. Why? Because you are guaranteed to have 3 distinct, separate powers after 3 rolls, so the probability of you having any one particular power out of the 6 powers in your 3 powers is 50%. I will put the full, overly-complicated proof in a new post below.

[hyudun]

[This post was used to warn that my initial proof was incorrect; correct proof has been posted further down.]

[hyudun]

Wouldn't it be something along the lines of

1/6 * 1/5 * 1/4

Since it's a 1 in 6 chance to get the first, then a 1 in 5 chance because of the re-roll and then a 1 in 4 chance for 2 re-rolls. To me that would make up an 84% chance.

But it's been a while since I've been to school and actually did math. So if anything I'd be interested to see if I'm right or if I'm not, how it does work.

No offense, but if you added those fractions together it would only come out to 37/60 or 61.66% (repeating 6). [URL="https://www.google.com/webhp?sourceid=chrome-instant&rlz=2C1ARAB_enUS0537US0537&ion=1&espv=2&ie =UTF-8#q=1%2F6%2B1%2F5%2B1%2F4"]You can check it in Google[/URL].
If you multiplied them as per your notation you would only get 1/120.

[hyudun]

Ok, I am 100% this proof is correct:

Conceptually, we need to add:
Probability of succeeding on the first die + probability of failing the first die, but succeeding on the second die + probability of failing the first two die, but succeeding on the third die.

1. First die - easy: 1/6

2. Failing first die and succeeding on second die. Bear with me here, it gets complicated.
Success on the first try of the second die (after failing the first) is straightforward - (5/6)(1/6).
However, we need to account for re-rolls.
The chances of you failing the first roll, triggering a re-roll on your second roll, and succeeding that is:
(5/6)(1/6)(1/6)
Triggering yet another re-roll, but succeeding on it would be:
(5/6)(1/6)(1/6)(1/6)
We can see the general formula for triggering N number of re-rolls, but succeeding on the Nth one is:
(5/6)((1/6)^N)(1/6)
So to account for all re-rolls, we need to sum (5/6)((1/6)^N)(1/6) for N = 1 to infinity and then add the probability of a first-roll-success on the second die probability of (5/6)(1/6).
Long story short, go read-up on limits and calculus if you don't remember what that means or how to do it, but adding that up gives us 1/6. So 1/6 chance that you fail the first die, but succeed on the second die.

Which means 1/6+1/6=2/6 chance of succeeding on the first two die, OR
1-2/6 = 4/6 chance of needing the third die. So, onto the third die.

Failing the first TWO die, but succeeding on the third die on the first try is:
(4/6)(1/6)
Failing the first TWO die, triggering a re-roll on the third die and succeeding on it is:
(4/6)(2/6)(1/6)
Similar to re-rolls of the second die, the general formula for triggering N re-rolls on the third die and succeeding on the Nth re-roll is:
(4/6)((2/6)^N)(1/6)
So to get the probability of failing the first TWO die, but succeeding on the third die on any attempt is:
(4/6)(1/6)+SUM((4/6)(2/6)^N(1/6)) for N = 1 to infinity, which also gives us 1/6

So for three die, 1/6+1/6+1/6 = 3/6 = 50%

Basically each ML adds 1/6 chance of getting any particular power. This makes sense because at (theoretical) ML6, you reach 100% of getting the power as you just keep re-rolling until you've got all powers 1-6 since you're not allowed duplicates.

[Chaoschrist]

No offense, but if you added those fractions together it would only come out to 37/60 or 61.66% (repeating 6). [URL="https://www.google.com/webhp?sourceid=chrome-instant&rlz=2C1ARAB_enUS0537US0537&ion=1&espv=2&ie =UTF-8#q=1%2F6%2B1%2F5%2B1%2F4"]You can check it in Google[/URL].
If you multiplied them as per your notation you would only get 1/120.

*slowly backs away from anything relating to math*

Good thing I don't rely on my math skills on a daily basis. Thanks for pointing it out though.

[allisdust]

It's easier than all that, if you work out your odds of failing instead:

5/6 that you miss it on the first go.
4/5 that you miss it on the second.
3/4 that you miss it on the third.

(5/6) * (4/5) * (3/4)
= 60/120
= 1/2
= 50%

... Not that Hyudun is wrong.

[hyudun]

It's easier than all that, if you work out your odds of failing instead:

5/6 that you miss it on the first go.
4/5 that you miss it on the second.
3/4 that you miss it on the third.

(5/6) * (4/5) * (3/4)
= 60/120
= 1/2
= 50%

... Not that Hyudun is wrong.

Hah, that's so much more elegant. Once you reduce your pool of available outcomes down to 5, you have an equal chance of rolling any of them, even if you factor in re-rolls, so even my method could've been simplified to (1/6)+(5/6)(1/5)+(5/6)(4/5)(1/4) = 0.5.

[iplaythisgame]

Lol, now calculate Tigurius!