How to calculate total chance from different simultaneous chances

i had a brainfart and couldn't word the title quite well i think, i got this question wondering about dragon wisdom in mobs (basically, 5% chance to drain per number) but since it applies to several things in real life, i decided to ask here.

is there a way to calculate the total % chance of a specific event happening, when what would cause it is up to chance?

for example, rolling a 6-sided dice would mean 1/6 of rolling one, how would i go about calculating the chance of 2 dice rolling 1 if thrown simultaneously? or them getting a 3 and a 4, and such.

or, the example that got me wondering, crusher does 4 hits, each hit has a 5% chance to drain HP, and with each monster that's 4 more chances to regain HP, how would i go around calculating this? >.<

tried wiki, it threw a bunch of letters at me and didn't answer much, please explain me as if i was at pre-kinder thank you, i really want to know how to work this (no, not going to the casino)

It's always 5% per hit.
http://en.wikipedia.org/wiki/Gambler's_fallacy

You don't calculate them together, you calculate the two as separate events.
At least that's my understanding of it.

Not quite sure how it works, but if it's 5% per cast then it's...well, 5%.

However, if it's 5% per hit (regardless of whether the skill drains only once or otherwise)...

P(4 activations) = (0.05)^4 = 0.00000625
P(3 activations) = (0.05)^3 = 0.000125
P(2 activations) = (0.05)^2 = 0.0025

Also:

X ~ No. of activations, out of 4.
X ~ B(4, 0.05)

P(X greater than or equal to 1)
= 1 - P(X = 0)
= 0.18549375

p.s. simply put, the last one is the probability that Dragon Wisdom would activate at least once per 4-hit Buster.

Not quite sure how it works, but if it's 5% per cast then it's...well, 5%.

However, if it's 5% per hit (regardless of whether the skill drains only once or otherwise)...

P(4 activations) = (0.05)^4 = 0.00000625
P(3 activations) = (0.05)^3 = 0.000125
P(2 activations) = (0.05)^2 = 0.0025

Also:

X ~ No. of activations, out of 4.
X ~ B(4, 0.05)

P(X greater than or equal to 1) = 0.81450625

p.s. simply put, the last one is the probability that Dragon Wisdom would activate at least once per 4-hit Buster.

this is exactly what i'm asking, how to know the chance of at least one of them activating, i was also thinking of the gambler's fallacy, but for some reason i kept thinking there's a way, the above example would be about all of them activating, i am wondering the second part, how to calculate the chance that an individual one would activate.

i am still unsure if it's possible though.

EDIT: forgot to add, yes, it's 5% per hit, meaning crusher on one target has 4 hits, 4 chances of it to activate. 3 targets would be 12 hits, and so

It's always 5% per hit.
http://en.wikipedia.org/wiki/Gambler's_fallacy

You don't calculate them together, you calculate the two as separate events.
At least that's my understanding of it.

Yes, it's always 5% per hit, however there is a 100% chance that the next result -might- differ from the previous one.

However, that still keeps the chance at 5% .

But, since the random number generator in computers is never truly random, the chance of of the next result being different depends on how the RNG is made.

What you're talking about is known as the Binomial Probability Distribution.

The general formula for this is:

nCr p^(n-r) q^r

Where n is the number of times something is done
r is the number of successes
p is the chance of success
and q is the chance of failure

Let's take your example of rolling a 6-sided die. And you want to see if you roll 4 times, what is the probability of rolling a 3, two times.
Your n is 4, the number of times you will be rolling
Your r is 2, the number of times you want to roll a 3
Your p is (1/6), since there is a one-in-six chance of rolling a certain number on a 6-sided die, in this case, a 3
And your q is 5/6, the chance of not rolling a 3.

So now we just plug this in to the formula

4C2 (1/6)^(4-2) (5/6)^(2)
6 (1/36) (25/36)
=25/216, or a .0011% chance of success.

Hope that made sense.

What you're talking about is known as the Binomial Probability Distribution.

The general formula for this is:

nCr p^(n-r) q^r

Where n is the number of times something is done
r is the number of successes
p is the chance of success
and q is the chance of failure

Let's take your example of rolling a 6-sided die. And you want to see if you roll 4 times, what is the probability of rolling a 3, two times.
Your n is 4, the number of times you will be rolling
Your r is 2, the number of times you want to roll a 3
Your p is (1/6), since there is a one-in-six chance of rolling a certain number on a 6-sided die, in this case, a 3
And your q is 5/6, the chance of not rolling a 3.

So now we just plug this in to the formula

4C2 (1/6)^(4-2) (5/6)^(2)
6 (1/36) (25/36)
=25/216, or a .0011% chance of success.

Hope that made sense.

i'll go read about it, i just can't get my head around how it works.

The best way to learn about it (and anything mathematical) would be to try examples, not read about it.

Try this example:

You draw 5 cards from a standard deck of cards putting them back after looking at them. What is the probability of two of them being hearts?

Figure out what your n and r are and then figure out your p and q (Remember, there's 52 cards in a deck and 4 different suits.)

If you get stuck I'll try to help you out

i'll go read about it, i just can't get my head around how it works.

Well, it's exactly what I did for the "greater than or equal to 1" part, except that I was using a cumulative distribution function.

An easy way to calculate this particular case is to do this:

P(No. of activations greater than or equal to 1)
= 1 - P(No. of activations = 0) <- this one's a probability distribution function

Use a tree diagram to help you out.

Anyway, here's the answer to baconator's question. Do NOT look at it until you've attempted it yourself.

 DO NOT OPEN UNTIL QUESTION HAS BEEN ATTEMPTED

X ~ No. of hearts drawn, out of 5.
X ~ B(5, 0.25)

P(X = 2)
= 0.263671875
= 0.264 (3 s.f)

just a few things:

1)thanks for the help, really apprecciate it, will try to get this, i got the main trouble out of the way, which was:

2)understanding the probability, i didn't understand why the chance of it activating at least once was so low, after staring at the wiki article for a while i understood it isn't the chance of activating at least but of it activating exactly one time.

lessee about the card example...

nCr p^(n-r) q^r

Where n is the number of times something is done
r is the number of successes
p is the chance of success
and q is the chance of failure

n should be 5, since i am drawing 5 cards
r should be 2, since i am trying to get 2 successes (hearts)
p= 4/52 = 1/13
q= 12/13

nCr p^(n-r) q^r

5C2 (1/13)^(5-2) (12/13)^2

5C2 1/2197 144/169

5C2 144/371293

and i'm lost as to what to do with the 5C2 :X

and i'm lost as to what to do with the 5C2 :X

http://en.wikipedia.org/wiki/Choose_func...al_formula

5C2 is just 5!/(3!)(2!) = 120, nCr is n!/(r!)((n-r)!).
If you didn't know, you can work out the probability of activating at least once is 1 - P(not activating at all).

@Shinku: I can probably work out the answer if you give me the conditions to be fulfilled
EDIT: Ninja'd

Wisdom has 5% chance of activating.
p = 0.05
q = 0.95

Buster is 4 hits, so n = 4
If we want to work out at least one proc, work out for r = 0
4C0 (0.05)^0 (0.95)^4
= 0.8145
~81.45% chance of not triggering at all, which means 18.55% of triggering at least once per 4 hits of buster.

Also it should be nCr p^r q^(n-r), baconator25 kinda derped there.

5C2 is just 5!/(3!)(2!) = 120, nCr is n!/(r!)((n-r)!).
If you didn't know, you can work out the probability of activating at least once is 1 - P(not activating at all).

@Shinku: I can probably work out the answer if you give me the conditions to be fulfilled
EDIT: Ninja'd

Wisdom has 5% chance of activating.
p = 0.05
q = 0.95

Buster is 4 hits, so n = 4
If we want to work out at least one proc, work out for r = 0
4C0 (0.05)^0 (0.95)^4
= 0.8145
~81.45% chance of not triggering at all, which means 18.55% of triggering at least once per 4 hits of buster.

You got ninja'ed pretty badly lol.

Look at post #3: it was posted 11 hours and 41 mins before your post. :x

EDIT: Oops. I made a mistake in that one. How on earth did I get the wrong answer using a calculator?

Ninja'd with the 5C2 part, 3 mins apart

and i'm lost as to what to do with the 5C2 :X

Well instead of jumping right into the formulas, try understanding how it works first.

You draw 5 cards, they can all be Hearts, Diamonds, Spades, or Clubs.
You draw the first card, it can be any of the 4.
You draw the second card, it can be any of the 4.
...
To have exactly 2 hearts, your hand will look like one of the following combinations, where X means some other suit, and H means Hearts:
- HHXXX, HXHXX, HXXHX, HXXXH
- XHHXX, XHXHX, XHXXH
- XXHHX, XXHXH
- XXXHH
A total of 10, which is the same number as 5C2 (there are 10 ways to choose a combination of 2 items from 5 unique items). The mathematical formula for finding 5C2 is, as shown above, 5!/(2!*(5-2)!), but that's just nuances that don't help with the concept.
Then you find the probability of landing one those combinations. Which means 2 successes and 3 failures, strict.
For cards, it's (0.25)^2 * (0.75)^3 = 0.0263671875 = 2.64%, there are 10 ways this final result can be accepted, so 2.64% * 10 = 26.4%
Note, in your previous post, you said p = 4/52, which is the rate of getting a specific rank, not suit. There are 13 Cards of Hearts to choose from in a deck.
Also note that we're not exactly dealing with recurring cards, so the numbers might be off from a real-life case. If you want the events to be dependent, the probability will be: 5C2 * 13/52 * 12/51 * 39/50 * 38/49 * 37/48, I'm not sure if it should matter when you draw those hearts cards.

Damn math nerds...I wanted to come in and help...Can't I guy sleep and go to school and come home without you people being up?! Don't you people have lives or jobs?!

Hanabira pegged it I think.

5C2 is just 5!/(3!)(2!) = 120, nCr is n!/(r!)((n-r)!).
If you didn't know, you can work out the probability of activating at least once is 1 - P(not activating at all).

@Shinku: I can probably work out the answer if you give me the conditions to be fulfilled
EDIT: Ninja'd

Wisdom has 5% chance of activating.
p = 0.05
q = 0.95

Buster is 4 hits, so n = 4
If we want to work out at least one proc, work out for r = 0
4C0 (0.05)^0 (0.95)^4
= 0.8145
~81.45% chance of not triggering at all, which means 18.55% of triggering at least once per 4 hits of buster.

Also it should be nCr p^r q^(n-r), baconator25 kinda derped there.

thanks a lot, but i posted this in the rubik's cube to know how to do it myself, something relating to probability like this is prone to be useful to at least have an idea later on.

Well instead of jumping right into the formulas, try understanding how it works first.

You draw 5 cards, they can all be Hearts, Diamonds, Spades, or Clubs.
You draw the first card, it can be any of the 4.
You draw the second card, it can be any of the 4.
...
To have exactly 2 hearts, your hand will look like one of the following combinations, where X means some other suit, and H means Hearts:
- HHXXX, HXHXX, HXXHX, HXXXH
- XHHXX, XHXHX, XHXXH
- XXHHX, XXHXH
- XXXHH
A total of 10, which is the same number as 5C2 (there are 10 ways to choose a combination of 2 items from 5 unique items). The mathematical formula for finding 5C2 is, as shown above, 5!/(2!*(5-2)!), but that's just nuances that don't help with the concept.
Then you find the probability of landing one those combinations. Which means 2 successes and 3 failures, strict.

thanks, i figured that before when i said "after staring long enough at the formula", when i said that i didn't know what to do with the 5C2, it's because i literally had no idea what it meant, i haven't done any serious math in ages, so i kinda have forgotten pretty much everything (had to wiki what the ! meant, that's how serious it is)

n should be 5, since i am drawing 5 cards
r should be 2, since i am trying to get 2 successes (hearts)
p= 4/52 = 1/13
q= 12/13

nCr p^r q^(n-r)
5!/(2!)(5-2)! (1/13)^2 (12/13)^(5-2)

120/(2)(6) (1/169) (1728/2197)

(120/12) (1/169) (1728/2197)

(207360/4455516)

0.04 so this would be the answer correct?

n should be 5, since i am drawing 5 cards
r should be 2, since i am trying to get 2 successes (hearts)
p= 4/52 = 1/13
q= 12/13

That would be the probability of drawing a One Pair* hand. There are 4 cards of the same rank and 13 cards of the same suit in a deck.

*fixed

That would be the probability of drawing a Four of a Kind hand (pretty generous margin of error, though, the real probability is 0.024%). There are 4 cards of the same rank and 13 cards of the same suit in a deck.

well great (sarcasm) in two ways, calculator gave me 0.0465... not 0.024, although this brings the question, since i am drawing them all from the same deck, wouldn't the chance change with each card drawn? compared to drawing each card from a different deck? and r would have to be 4?

and i think i know what i did wrong, but am unsure, p should be 1/4 'th right? 1 out of 4 chance to draw a heart?

well great (sarcasm) in two ways, calculator gave me 0.0465... not 0.024, although this brings the question, since i am drawing them all from the same deck, wouldn't the chance change with each card drawn? compared to drawing each card from a different deck? and r would have to be 4?

and i think i know what i did wrong, but am unsure, p should be 1/4 'th right? 1 out of 4 chance to draw a heart?

I think I overcomplicated it with my subtleness (and incorrect assumptions). Yes, your previous calculation was the probability of landing a specific ONE PAIR hand. Since you can have 13 different pairs, the answer you were looking for would be 13*0.04 = 52%+, which is quite a big leap from the true figure (42.25%), but it's not the problem here (if you're wondering, the gap is because of the problem mentioned in the last line).

Secondly, yes, p = 1/4 and q = 3/4, so to speak.

As for the same-deck thing, the problem was constructed with the assumption that you put the card back after drawing, so each draw is independent of the rest.

I don't know why but after a while being exposed to this kind of stuff, I start asking myself "When I flip a coin, does it land on heads via pure randomness?"