More help required with FE and math.

[Hanes]

Suppose I want to know what Lowen's chance of leveling up Speed in 4 levels is, his growth is that of 30% and he has a base of 7 so I'm aiming to have 8 by level 5.

How do I calculate my chances for that?

Note: I'm not asking for how many points of speed he should have by then, just what my chance of him getting a point of speed during 4 levels is.

Edited February 3, 2019 by Critical Sniper

[ping]

In this particular case, you can solve the problem relatively easily by calculating the chance of him not leveling Spd at all: 0.7^4 = 0.24 = 24%. [0.7 being 1 minus his Spd growth of 0.3] So the chance that Lowen procs Spd at least once would be 100% - 24% = 76%.

Note that this doesn't say anything about his chance to proc Spd two, three, or all four times. Well, the last one is easy, too (.3^4 = 0.0081 = 0.81%), but to get the other answers, you'll need the Binomial Distribution, which isn't all that easy to understand if you never worked with it before. Excel can help you with that, if you're so inclined - BINOM.DIST is your friend there with number of procs, number of level ups, growth rate, and FALSE (in that order) being the numbers to put into it.

[Hanes]

2 minutes ago, ping said:

In this particular case, you can solve the problem relatively easily by calculating the chance of him not leveling Spd at all: 0.7^4 = 0.24 = 24%. [0.7 being 1 minus his Spd growth of 0.3] So the chance that Lowen procs Spd at least once would be 100% - 24% = 76%.

But this is incremental isn't it? Like doesn't it eventually go over 100% if I add more levels?

[eclipse]

16 minutes ago, Critical Sniper said:

But this is incremental isn't it? Like doesn't it eventually go over 100% if I add more levels?

Not really.  I'll try to break it down in words.

Lowen's speed growth = 30% = 0.3

That means he has a 70% chance NOT to level up speed.  Think of it as a coin flip, except you'll hit tails 70% of the time.  That means that in one level, Lowen has a 30% chance to level speed.  Easy, right?  Now, let's throw two levels in there.  Every single possible outcome of these two levels are:

- Lowen levels speed twice
- Lowen levels speed the first time and doesn't the second time
- Lowen doesn't level speed the first time and levels speed the second time
- Lowen doesn't level speed twice

We know what the odds are, so let's put numbers to those possibilities:

- Lowen levels speed twice (0.3 * 0.3 = 0.09)
- Lowen levels speed the first time and doesn't the second time (0.3 * 0.7 = 0.21)
- Lowen doesn't level speed the first time and levels speed the second time (0.7 * 0.3 = 0.21)
- Lowen doesn't level speed twice (0.7 * 0.7 = 0.49)

Or, to put it in words, Lowen has a 9% chance of leveling speed twice, a 49% chance of not leveling speed at all, and a 42% chance of leveling speed once.

If you add another level, and write it out, you'll see a pattern regarding how the numbers work.  If you add four levels (what you're looking for), and want him to level speed at least once, you'll take the "Lowen doesn't level up speed four times in a row" possibility, and subtract that from 100% (which is every possibility).  That will leave you with the odds of Lowen getting at least one speed level. . .which, in this case, is 76%.  Thus, the odds are in your favor.  Good luck~!

[Slumber]

24 minutes ago, Critical Sniper said:

But this is incremental isn't it? Like doesn't it eventually go over 100% if I add more levels?

It'll approach 100%, but never cross it.

[ping]

Probabilities never go above 100%. 100% means "always" and "more than always" doesn't make any sense.

The probability for Lowen to proc Spd at least once converges towards 100%, which techically means that it will never reach 100%, but "technically" is the operative word here:

Column A is the number of level-ups, B the probability for Lowen to get at least one Spd proc.

As you can see, a Lv.15 Lowen (i.e. 13 level-ups) has a chance of about 99% to get at least one Spd proc. Again, he's quite likely to proc it more often than once - that 99% is the probability that he procs Spd between 1 and 13 times in those 13 level-ups. If you go through it one by one, a Lv.15 Lowen has a chance of...

• 5.4% to have 8 Spd (i.e. one proc in 13 tries)
• 13.9% to have 9 Spd
• 21.8% to have 10 Spd
• 23.4% to have 11 Spd
• 18.0% to have 12 Spd

...and so on, with decreasing percentages for Spd values above 12. And all those numbers add up to the 99% probability of "at least one Spd proc". (Also, as you can see, the value closest to his average stat (10.9 Spd) is indeed the likeliest)

[Hanes]

2 minutes ago, eclipse said:

Not really.  I'll try to break it down in words.

Lowen's speed growth = 30% = 0.3

That means he has a 70% chance NOT to level up speed.  Think of it as a coin flip, except you'll hit tails 70% of the time.  That means that in one level, Lowen has a 30% chance to level speed.  Easy, right?  Now, let's throw two levels in there.  Every single possible outcome of these two levels are:

- Lowen levels speed twice
- Lowen levels speed the first time and doesn't the second time
- Lowen doesn't level speed the first time and levels speed the second time
- Lowen doesn't level speed twice

We know what the odds are, so let's put numbers to those possibilities:

- Lowen levels speed twice (0.3 * 0.3 = 0.09)
- Lowen levels speed the first time and doesn't the second time (0.3 * 0.7 = 0.21)
- Lowen doesn't level speed the first time and levels speed the second time (0.7 * 0.3 = 0.21)
- Lowen doesn't level speed twice (0.7 * 0.7 = 0.49)

Or, to put it in words, Lowen has a 9% chance of leveling speed twice, a 49% chance of not leveling speed at all, and a 42% chance of leveling speed once.

If you add another level, and write it out, you'll see a pattern regarding how the numbers work.  If you add four levels (what you're looking for), and want him to level speed at least once, you'll take the "Lowen doesn't level up speed four times in a row" possibility, and subtract that from 100% (which is every possibility).  That will leave you with the odds of Lowen getting at least one speed level. . .which, in this case, is 76%.  Thus, the odds are in your favor.  Good luck~!

Who knew eclipse was good at math? Thank you! Thank you!!

Ok but I don't understand the bolded, why do I take the doesn't level up 4 times in a row possibility and substract that from 100% instead of the he doesnt get 3 speed level ups possibility?

[eclipse]

Just now, Critical Sniper said:

Who knew eclipse was good at math? Thank you! Thank you!!

Ok but I don't understand the bolded, why do I take the doesn't level up 4 times in a row possibility and substract that from 100% instead of the he doesnt get 3 speed level ups possibility?

I think you were looking for the odds of Lowen to get speed at least once in four levels?

[Hanes]

7 minutes ago, eclipse said:

I think you were looking for the odds of Lowen to get speed at least once in four levels?

Yes but that's the thing. So from what I understoof drom you; it's way easier to figure out percentages by subtracting 100% from the opposite of what you want because the sum of both gives 100% right? So if I want to get 1 speed level up in 4 shouldn't I then subtract the probability of getting 3 in a row because that's the opposite and 3+1=4?

EDIT: wait a minute. The exact opposite of getting 1 speed level up is getting 4 right?

Edited February 3, 2019 by Critical Sniper

[eclipse]

1 minute ago, Critical Sniper said:

Yes but that's the thing. So from what I understoof drom you; it's way easier to figure out percentages by subtracting 100% from the opposite of what you want because the sum of both gives 100% right? So if I want to get 1 speed level up in 4 shouldn't I then subtract the probability of getting 3 in a row because that's the opposite and 3+1=4?

 

Bolded: EXACTLY!  Think of 100% as every single possibility.  Rather than try to figure out the scenarios where Lowen gains speed once, twice, three times, and four times, it's faster to eliminate the one future you don't want.

[Hanes]

1 minute ago, eclipse said:

Bolded: EXACTLY!  Think of 100% as every single possibility.  Rather than try to figure out the scenarios where Lowen gains speed once, twice, three times, and four times, it's faster to eliminate the one future you don't want.

You gotta love how I messed up understood as understoof and from as drom -.-

Ok but that doen't answer my other question, isn't the opposite of getting 1 speed level getting 3? instead of 4 because 3+1=4 and thats the amount of levels I am planning this in?

[eclipse]

Just now, Critical Sniper said:

You gotta love how I messed up understood as understoof and from as drom -.-

Ok but that doen't answer my other question, isn't the opposite of getting 1 speed level getting 3? instead of 4 because 3+1=4 and thats the amount of levels I am planning this in?

Hmmm. . .I guess, if you look at it that way.  Did you want the probability of getting exactly one speed level, or getting at least one speed level?  They're different numbers!

[Hanes]

1 minute ago, eclipse said:

Hmmm. . .I guess, if you look at it that way.  Did you want the probability of getting exactly one speed level, or getting at least one speed level?  They're different numbers!

Hmm what would the difference in calculating both would be?

 

Spoiler

I've never been good a tprobabilities and I'm ashamed of it because I'm very good at multiplication. Don't look, my face ain't red. 

 

[ping]

3 minutes ago, Critical Sniper said:

EDIT: wait a minute. The exact opposite of getting 1 speed level up is getting 4 right?

Nope, at least not in terms of stochastics.

If you're asking for the chance for exactly one Spd proc, the complementary event is all other outcomes combined - so the opposite of "exactly one Spd proc" is "zero OR two OR three OR four Spd procs)

Your initial question (at least how I understand it) was about Lowen getting at least one Spd proc, which means "one OR two OR three OR four", so the complementary event is only "zero Spd procs", which is a lot easier to figure out than the actual question. But since the probabilities for an event and its complementary event always add up to 100% (because per definition, they make up all possible outcomes), you can just calculate "100% minus the probability for zero Spd procs" to get the probability for at least one Spd proc.

Calculating the chance for exactly one Spd proc is actually a bit more difficult because there are multiple ways (four in this case) to accomplish this: only on the first level-up and never again; only on the 2nd level-up and not before and after that; only on the 3rd level-up and not before and after that; only on the last level-up and not before that.
Because you can rearrange numbers when multiplying (e.g. 2*3 = 3*2), all of these four paths have the same probability: 0.7 * 0.7 * 0.7 * 0.3 = 0.1029 = 10.29%. And because it's four paths, the total probability is four times that, so it's 41.16%.

Counting those possible paths gets a lot more difficult when the number of successes and trials (or, in this context, procs and level-ups) increases, though - good luck counting the possible ways to get 6 procs in 10 level-ups:

• 1st, 2nd, 3rd, 4th, 5th, 6th level-up
• 1st, 2nd, 3rd, 4th, 5th, 7th level-up
• 1st, 2nd, 3rd, 4th, 5th, 8th level-up
• 1st, 2nd, 3rd, 4th, 5th, 9th level-up
• 1st, 2nd, 3rd, 4th, 5th, 10th level-up
• 1st, 2nd, 3rd, 4th, 6th, 7th level-up

And so on, and so on. This is why I brought up Binomial Distributions earlier - its main formula includes a term that does this counting for you.

[Hanes]

3 minutes ago, ping said:

Nope, at least not in terms of stochastics.

If you're asking for the chance for exactly one Spd proc, the complementary event is all other outcomes combined - so the opposite of "exactly one Spd proc" is "zero OR two OR three OR four Spd procs)

Your initial question (at least how I understand it) was about Lowen getting at least one Spd proc, which means "one OR two OR three OR four", so the complementary event is only "zero Spd procs", which is a lot easier to figure out than the actual question. But since the probabilities for an event and its complementary event always add up to 100% (because per definition, they make up all possible outcomes), you can just calculate "100% minus the probability for zero Spd procs" to get the probability for at least one Spd proc.

Calculating the chance for exactly one Spd proc is actually a bit more difficult because there are multiple ways (four in this case) to accomplish this: only on the first level-up and never again; only on the 2nd level-up and not before and after that; only on the 3rd level-up and not before and after that; only on the last level-up and not before that.
Because you can rearrange numbers when multiplying (e.g. 2*3 = 3*2), all of these four paths have the same probability: 0.7 * 0.7 * 0.7 * 0.3 = 0.1029 = 10.29%. And because it's four paths, the total probability is four times that, so it's 41.16%.

Counting those possible paths gets a lot more difficult when the number of successes and trials (or, in this context, procs and level-ups) increases, though - good luck counting the possible ways to get 6 procs in 10 level-ups:

• 1st, 2nd, 3rd, 4th, 5th, 6th level-up
• 1st, 2nd, 3rd, 4th, 5th, 7th level-up
• 1st, 2nd, 3rd, 4th, 5th, 8th level-up
• 1st, 2nd, 3rd, 4th, 5th, 9th level-up
• 1st, 2nd, 3rd, 4th, 5th, 10th level-up
• 1st, 2nd, 3rd, 4th, 6th, 7th level-up

And so on, and so on. This is why I brought up Binomial Distributions earlier - its main formula includes a term that does this counting for you.

Eduardo is confused. It hurt itself in confusion.

[eclipse]

14 minutes ago, Critical Sniper said:

Hmm what would the difference in calculating both would be?

 

  Reveal hidden contents

I've never been good a tprobabilities and I'm ashamed of it because I'm very good at multiplication. Don't look, my face ain't red. 

 

This is where binomial distribution comes in (was linked earlier).  And lucky us, the example shows a coin with a 30% chance to hit heads!  According to this example, you're looking for the probability of exactly one speed level up in four levels.  The equation's kinda complicated, so I'll do my best to illustrate it here:

(4,1) = 4! / 1!(4 - 1)! = 24 / (1 * 6)  = 24 / 6 = 4

4 * (0.3) * (1 - 0.3) ^ 3 = 0.4116 = 41.16%

(equation shamelessly stolen from Wikipedia)

So you have a 41.16% chance to get exactly one speed level in four levels.

[ping]

2 minutes ago, Critical Sniper said:

Eduardo is confused. It hurt itself in confusion.

That's way too unspecific for me to give a concrete answer.

The basic principle (before binomials hit) is that you
a) identify all possible outcomes of a random experiment (in this case: in four level ups, Lowen can proc Spd 0, 1, 2, 3, or 4 times. There is no other possibility) and
b) check which outcomes fulfill my criteria. In this case, the criteria was "at least one Spd proc" which is fulfilled if that number is 1, 2, 3, or 4.

And in this particular case, we notice that the complementary event to "at least one Spd proc" is a lot easier to calculate than the event itself - it's just "zero Spd procs" because that's the only outcome that does NOT fulfil my criteria. And since P(Event) + P(Complementary Event) is ALWAYS 100%, we can use this to get our answer faster.

[Hanes]

16 minutes ago, ping said:

That's way too unspecific for me to give a concrete answer.

The basic principle (before binomials hit) is that you
a) identify all possible outcomes of a random experiment (in this case: in four level ups, Lowen can proc Spd 0, 1, 2, 3, or 4 times. There is no other possibility) and
b) check which outcomes fulfill my criteria. In this case, the criteria was "at least one Spd proc" which is fulfilled if that number is 1, 2, 3, or 4.

And in this particular case, we notice that the complementary event to "at least one Spd proc" is a lot easier to calculate than the event itself - it's just "zero Spd procs" because that's the only outcome that does NOT fulfil my criteria. And since P(Event) + P(Complementary Event) is ALWAYS 100%, we can use this to get our answer faster.

Oh I get that last part! So if I say I atleast want one speed proc in 4 levels because I dont want to get 0 then I use 100% - "I don't want 0 speed procs" right?

[ping]

1 minute ago, Critical Sniper said:

So if I say I atleast want one speed proc in 4 levels because I dont want to get 0 then I use 100% - "I don't want 0 speed procs" right?

Right.

[Hanes]

3 minutes ago, ping said:

Right.

Well that is one thing I understood during the time my Math class was going through probability. The only other one being "Al the chances add up to 100%" honestly I already forgot what the complementary was I understood the mutually exclusive and inclusive stuff. Never understood complementary.

Edited February 3, 2019 by Critical Sniper