How to calculate total chance from different simultaneous chances

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.

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.

no problem, so that was my only mistake before, at least i know how to do the thing now, using buster example, the letters should be:

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 = 4, 8, 12 (depending on targets)
r = 0 (i want to find out the chance of it activating at least once, so i use zero for chance of fail, and then 100-answer for chance of success)
p = 0.05 (5/100)
q = 0.95 (95/100)

and just formula it out, correct? can be used for impale and so on.


i'd keep asking about the card example, but i think it's better not >.< since each new statement confuses me.

Yep. Seems pretty optimistic for 12 target, huh?

Yep. Seems pretty optimistic for 12 target, huh?

what do you mean?

n = 4, 8, 12 (depending on targets)
r = 0 (i want to find out the chance of it activating at least once, so i use zero for chance of fail, and then 100-answer for chance of success)
p = 0.05 (5/100)
q = 0.95 (95/100)

and just formula it out, correct? can be used for impale and so on.


i'd keep asking about the card example, but i think it's better not >.< since each new statement confuses me.

The numbers you listed are correct.

For probability, when events are completely independent, meaning that the outcome of one event does not depend on the one that happened before, you can just multiply the probabilities together.

It's much easier when you want only one thing to happen rather than combinations. For example, if you want to find the probability of all successes on 4 targets. prob = 0.05^(4). Or the probability of all failures, prob = 0.95^(4).

The binomial formula is just more general in that in can account for combinations. For example, how many ways can you have 1 success in 4 targets: (where S is success, F is fail)
SFFF, FSFF, FFSF, FFFS
That is what is captured in the 'X choose Y' term.
If you're only looking for all successes or all fails, there's only one possible combination.

The card picking example can get quite complex. There's a difference between:
1. Picking and then replacing (all events are independent, 1/4 chance to get hearts each time.)
2. Picking, but not replacing (events are now conditional upon what happened before: if you get hearts the first time, then the next card only has a 12/51 chance, not 13/52 = 1/4 like before; but if you didn't get hearts, the next card has a 13/51 = 1/4 chance.)
3. Picking all 5 cards at once. (In this case, the probability of picking hearts for the cards is not 1/4 for each card. The hypergeometric distribution needs to be used)

And @above question: I assume that he means that with 12 targets, the chance of getting a 'success' are pretty good. ( prob of at least one success = 1 - prob(all fails) = 1 - 0.95^12 = 0.46 )

perfect, that cleared all the doubts i had about the card problem (well, except one, the method #2 implies i replace the card i draw from the deck with a dummy right? to keep 52 cards on it)

and i could have used the ^(hits) method, nice to know there's an easier way to part of it, speaking of...

1- (0.95^30) = 78.5%
1- (0.95^24) = 70.8%
1- (0.95^18) = 60.2%
1- (0.95^12) = 45.9%

so impale on 2 targets already has a better chance to heal (albeit less) thanks a lot for the help, gotta love math. (not sarcasm) gotta save this page for the large formula though.

perfect, that cleared all the doubts i had about the card problem (well, except one, the method #2 implies i replace the card i draw from the deck with a dummy right? to keep 52 cards on it)

Nice catch on the 52 thing. You're right, it should be 51, since a card was removed. My mistake. Will update post to revise it.

Math is pretty cool when you can apply it. But for the most part, it's quite dry =.=

Nice catch on the 52 thing. You're right, it should be 51, since a card was removed. My mistake. Will update post to revise it.

Math is pretty cool when you can apply it. But for the most part, it's quite dry =.=

i enjoy math (except geometry, i hate it and i don't know why.)

if anyone wants to add anything or try to explain something else, feel free, i like the topic.

Well, to add on to the topic, the general formula which I gave is to calculate something happening in exactly a certain amount of occurrences. (For example, the probability of rolling a 2 on a 6-sided die exactly 3 times out of 4.) You could also use this to calculate the probability of at least and at most events. For example, the probability of rolling a 2 on a 6-sided die at most 3 times out of 4. The at most includes rolling a two 0 times, 1 time, 2 times and 3 times. This could be done using the binomial probability distribution as well by using the formula for each number (in this cases there's 4; 0, 1, 2 and 3) and adding them together. Of course, this process can become rather tedious, which is why you'd probably want to use the Cumulative Distribution function mentioned earlier. I just find it interesting how they both work. (In case you were wondering, the probability of rolling a 2 at most 3 times in 4 rolls is 99.9%.)

Oh, also, the Cumulative Distribution only calculates "at most" situations. If you wanted to calculate an "at least" situation using it, you'd need to do "1 - at most value + P®"
So if you wanted to calculate the probability of rolling at 2 at least 3 times in 4 rolls, it'd be:
1 - .999 + .015 = .16%

I hope my explanations make sense. I feel like I tried to throw too much information into that one post... I'm a math major and can do stats all day.

Well, to add on to the topic, the general formula which I gave is to calculate something happening in exactly a certain amount of occurrences. (For example, the probability of rolling a 2 on a 6-sided die exactly 3 times out of 4.) You could also use this to calculate the probability of at least and at most events. For example, the probability of rolling a 2 on a 6-sided die at most 3 times out of 4. The at most includes rolling a two 0 times, 1 time, 2 times and 3 times. This could be done using the binomial probability distribution as well by using the formula for each number (in this cases there's 4; 0, 1, 2 and 3) and adding them together. Of course, this process can become rather tedious, which is why you'd probably want to use the Cumulative Distribution function mentioned earlier. I just find it interesting how they both work. (In case you were wondering, the probability of rolling a 2 at most 3 times in 4 rolls is 99.9%.)

Oh, also, the Cumulative Distribution only calculates "at most" situations. If you wanted to calculate an "at least" situation using it, you'd need to do "1 - at most value + P®"
So if you wanted to calculate the probability of rolling at 2 at least 3 times in 4 rolls, it'd be:
1 - .999 + .015 = .16%

I hope my explanations make sense. I feel like I tried to throw too much information into that one post... I'm a math major and can do stats all day.

Which means P(X greater than or equals to 3) equals to 1 - P(X = 2) - P(X = 1) - P(X = 0), then?

i.e. P(X greater than or equals to 3) = P(X = 3) + P(X = 4) + P(X = 5) + ...

^is that it? o.O

Anyways, I use a graphic calculator so the actual formula is none of my business.

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?!

Dude. Note the difference in time zone.

Yes, both of those formulas are correct.

I use a graphing calculator too, as it drastically speed things up. It's just good to know where the answers that the calculator gets come from.

so impale on 2 targets already has a better chance to heal (albeit less) thanks a lot for the help, gotta love math. (not sarcasm) gotta save this page for the large formula though.

As for how much you'll heal, on average, it gets even more simple:
Chance of success * amount healed
amount healed = damage done * percent heal
So you have 5%*5%*X, where X is just... add up all the hits from your attack. If my DrK is doing about 30k * 12 with Buster, then it averages 900 hp healed per attack - of course the actual distribution of the healing depends how many hits succeed a particular time. But in the long run on average it works out.

From there you can figure out how fast you're healing per second, and decide what the max mob damage you can go to without burning pots is.

As for how much you'll heal, on average, it gets even more simple:
Chance of success * amount healed
amount healed = damage done * percent heal
So you have 5%*5%*X, where X is just... add up all the hits from your attack. If my DrK is doing about 30k * 12 with Buster, then it averages 900 hp healed per attack - of course the actual distribution of the healing depends how many hits succeed a particular time. But in the long run on average it works out.

From there you can figure out how fast you're healing per second, and decide what the max mob damage you can go to without burning pots is.

but i have to wonder, isn't that the chance of healing only from one strike out of 12? wouldn't you have a 46% chance of healing 900 hp or more? unless you are restricted to one healing per attack no matter how many trigger, but that's more about the skill than the math

Nah, that's an average - the amount healed when it works is gonna be damage*%heal, ie. 30000*0.05 = 1500 hp. It's just over time (technically an infinite number of attacks) that you reach the average.