Guild Wars 2

But what if the coin bounces than rolls into a crack and ends up sticking strait up?

The chance of getting shrapnel to proc on 300 consecutive grenades is 0.67201^-249.
or 0.00000000000000000000000000000000000000000000000000000000000000000000000
0000000000000000000000000000000000000000000000000000000000000000000000000
0000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000067201%

.

DGraves.3720:

You do not double attribute behavior.

A coins chance is 50/50 no matter how many coins you flip. Double attribution specifically ruins outcomes by adding unnecessary data.

I might simply be misunderstanding what you mean by “double attribute”, don’t hesitate to clarify
To me, it was the fact he added the probability of having the proc on only grenade 1, 2 and 3 to get the probability of getting only one proc on a given throw. Which is the right way to do it, it just is a matter of case enumeration. You want cumulative probability, not single grenade probability. By the way, “cumulative” doesn’t mean “dependent”.
My point, Ging result concerning odds of proc is correct.

DGraves.3720:

It’s as simple as asking “what are the odds of getting heads on a fair coin if I flip it 900 times on any given throw?”

A: 50/50

This is NOT “what are the odds of getting at least one heads if I flip 900 fair coins?”

A: (.5^900)-1= A very, very small number.

Well, the odd of a given throw is not contested here. I don’t see what you want to say :/
And I have to note that the odds of getting at least one head in 900 throw are anything but a very small number. In fact, it is almost 100% (the correct equation being: 1 – (0.5^900)).

Now, you might feel old, but I think you might also be very tired, when saying:

It should be just 45% since they are independent events for “at least one”.

Basically it runs like any other form of basic dice game; if you have two twenty sided dice the odds are additive for rolling a twenty on either and multiplicative for rolling a twenty on both, I.E.

5% + 5% = 10%

Which is, let say, slightly less than accurate ^^‘. You get 9.5%.
Replace your 20-sided dice with coins. You obvioulsy won’t have 100% chance of getting head when throwing both. Nor 100/200 (going from another of your answers). The odds for the coin are 75%.
The event “getting at least a 20 on two independent throws of a dice” is an event distinct from “getting a 20 on a unique die throw”, and have different odds. The fact that the two throws are independant doesn’t change that.

Now, I’m not sure you still have in mind what Ging (and I) was answering to. In your own words, that are the very subject of our answers:

For instance, Shrapnel, if you throw one volley of grenades what is the probably that two of those grenades will proc Shrapnel?

This probability is 5.71% for strictly 2 proc, and 6.08% to get at least 2. Independent verification, with a very simple function. Nothing “overly complicated”.

At this point I wonder if, instead of math, it is another point you made you are trying to prove:

E_But_ what makes this possible is simply reputation. We will believe anyone not based on the merit of their claim but based on the warmth of their ideas. An easy to read spreadsheet, even if wrong, is the best way to win fans. Show a video of you doing something people consider difficult or some crazy large numbers and you’ve got people hooked. After that you can go the way of L. Pauling (the source Vitamin C myth) and just say whatever the heck you want and people will hang on your every word.

You should be happy people aren’t falling for a nice equation and are actually doing independant verification to make their own opinion :P
By the way, I share the feeling. Just not the math.

Elidath.5679:

Independent verification

Let’s just say you’re right. It’s easier than continuing this strange and petty back and forth that I can’t bring myself to read half of. Which is true; I know I’m skipping over it on purpose and that’s not fair to you.

Good luck with your 39%.

Ging.6485:

The chance of getting shrapnel to proc on 300 consecutive grenades is 0.67201^-249.
or 0.00000000000000000000000000000000000000000000000000000000000000000000000
0000000000000000000000000000000000000000000000000000000000000000000000000
0000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000067201%

This isn’t even in scientific notation… WTF

DGraves.3720:

Elidath.5679:

Independent verification

Let’s just say you’re right. It’s easier than continuing this strange and petty back and forth that I can’t bring myself to read half of. Which is true; I know I’m skipping over it on purpose and that’s not fair to you.

Good luck with your 39%.

Look, you don’t get to say

I do math all day and can see errors

Then resume making the very same, basic error people are pointing to you, then call them strange and petty for it. Then pile up more basic, dumbfounding error on it.

Now, if you wan’t a simple and crystal clear proof, compare the odds over 99 throws.
How many proc should you expect? 15% of them, right? So 99 * 0.15 = 14,85 total.
Now make the throw by batches of three (an in-game grenade throw). You’ll do 33 throws. You get statistically:
- 33*61.41% = 20.27 throws with 0 proc
- 33*32.51% = 10.73 throws with 1 proc
- 33*5.74% = 1.89 throw with 2 procs
- 33*0.34% = 0.11 throws with 3 procs
Total: 20.27 * 0 + 10.73 * 1 + 1.89 * 2 + 0.11 * 3 = 14.84 procs. The expected result (with a rounding error, because I’m lazy).
I don’t need any luck with “my” 39%. It is probability 101, really.

Elidath.5679:

DGraves.3720:

Elidath.5679:

Independent verification

Let’s just say you’re right. It’s easier than continuing this strange and petty back and forth that I can’t bring myself to read half of. Which is true; I know I’m skipping over it on purpose and that’s not fair to you.

Good luck with your 39%.

Look, you don’t get to say

I do math all day and can see errors

Then resume making the very same, basic error people are pointing to you, then call them strange and petty for it. Then pile up more basic, dumbfounding error on it.

Now, if you wan’t a simple and crystal clear proof, compare the odds over 99 throws.
How many proc should you expect? 15% of them, right? So 99 * 0.15 = 14,85 total.
Now make the throw by batches of three (an in-game grenade throw). You’ll do 33 throws. You get statistically:
- 33*61.41% = 20.27 throws with 0 proc
- 33*32.51% = 10.73 throws with 1 proc
- 33*5.74% = 1.89 throw with 2 procs
- 33*0.34% = 0.11 throws with 3 procs
Total: 20.27 * 0 + 10.73 * 1 + 1.89 * 2 + 0.11 * 3 = 14.84 procs. The expected result (with a rounding error, because I’m lazy).
I don’t need any luck with “my” 39%. It is probability 101, really.

Did you read your own calculator?

Because part of me wonders if you understand why your own source says the mean probability is 45% under the summary tab. That’s what the (.45/.62) means.

I had two options:

1. Call you out on it and wonder what is wrong with you.

2. Concede.

I choose the high road. Let it go. You win.

DGraves.3720:

Did you read your own calculator?

Yes. I did read the premise too. You know:

For instance, Shrapnel, if you throw one volley of grenades what is the probably that two of those grenades will proc Shrapnel? Simple probability:

.15^2 = .0225

All three? .15^3 = .003375.

The odds of getting two procs are in the normal tab. They clearly states “exactly 2 procs : 5.7%”. Not 2.25%.

DGraves.3720:

Because part of me wonders if you understand why your own source says the mean probability is 45% under the summary tab.

The summery tab indicate a mean value of 0,45 success in three rolls.
With the repartition of 61/33/5.7/0.3 (so 39% of at least one success) you have a mean value of (61*0 + 33*1 + 5.7*2 + 0.3 * 3) / (61 + 33 + 5.7 + 0.3) = 0.453 success (again, rounding error). It does add up, it is the very basic “computing a mean” mathematic operation, which as nothing to do with computing odds.
You can’t equate that to 45% chance of at least one success since those values are weighted, so some outcomes result in more than one success, shifting the mean value.
Take the coin toss with head set as “success”. The mean value for the double coin toss is one success. But the odds of getting at least one head is 75%. You can’t just swap those values, they don’t have the same meaning.

Elidath.5679:

DGraves.3720:

Did you read your own calculator?

Yes. I did read the premise too. You know:

For instance, Shrapnel, if you throw one volley of grenades what is the probably that two of those grenades will proc Shrapnel? Simple probability:

.15^2 = .0225

All three? .15^3 = .003375.

The odds of getting two procs are in the normal tab. They clearly states “exactly 2 procs : 5.7%”. Not 2.25%.

But it isn’t. The reason why is because you are duplicating outcomes by looking at all variables which produce the outcome itself multiple times. Do you know what .225^2 is? .050625 and when you round that it is .057 or 5.7%. This isn’t a coincidence. The matrix:

Y Y N
Y N Y
N Y Y

Since Y Y N and N Y Y are actually the same result you have Y Y N and Y N Y. So yes, it does clearly state that 5.7% is the outcome. But it doesn’t matter if you don’t know why it does. Since we are not looking for all the matrix possibilities we would only take the odds of success ( .15^2 ) and apply them but if you were to apply them in a distribution (which you are for whatever reason) you’re going to end up with more than one output.

You can prove this by looking at “at least one”, .15 × .85 × .85 or “Y N N” =10.8375%, which is one third of 32.5125% which is the basis for your “32%” because when we move the result of Y (getting a Shrapnel) we have three combinations:

“Y N N”
“N Y N”
“N N Y”

These are all distinct outputs therefore resulting in three elements multiplied by the base of 10.8375.

Okay, so I did my part. We are officially done. If you don’t understand at this point why it is 45%, 2.25% or .34% you’ll have to take it up with god and your math teachers. But yes, in conclusion, you’ve no idea what you’re talking about.

But…but…all math aside….Engi still good? PVE, raids, etc.?

DGraves.3720:

But it isn’t. The reason why is because you are duplicating outcomes by looking at all variables which produce the outcome itself multiple times.
(…)
These are all distinct outputs therefore resulting in three elements multiplied by the base of 10.8375.

You say yourself why it is so. They are distinct outputs that you have to take into account separately, even when their individual probability is the same. I don’t even see where your problem is with that, since you basically takes my initial equation as your conclusion. Yeah, it works.

DGraves.3720:

Do you know what .225^2 is? .050625 and when you round that it is .057 or 5.7%. This isn’t a coincidence.

Well, considering it is the first appearance of .225 in the discussion, and you provide no context, I fail to see the meaning of that. Now, I do see that rounding 0.050625 doesn’t make 0.057 but 0.051. Quite a big rounding error.

DGraves.3720:

If you don’t understand at this point why it is 45%, 2.25% or .34% you’ll have to take it up with god and your math teachers. But yes, in conclusion, you’ve no idea what you’re talking about.

Look, just apply the formula for Bernoulli trials. It is a tool designed to compute exactly what we are doing: chances of exactly X success in N independent, fair throws. What do you get ? 32/5.7/0.3.
And when you yourself accepted that 61% is the odd for no success, how can your odds of “at least one” not add up to 39%? It means you have cases when you have simultaneously no result and some results. It doesn’t make any sense!

Now, happens I have a friend who is a university math teacher, specialized in probabilities (part of the reason I can call you on your mistakes). Just to be extra sure I wasn’t going crazy, I presented the problem to him. And got the very same answer I got by myself. So yes, you are wrong. Don’t take my word for it: math teachers call you wrong ; Wikipedia calls you wrong ; Bernoulli calls you wrong ; math softwares call you wrong. Maybe it is time for some self reflection.

@Joxer: Engi never was bad per se in PvE. There just isn’t any incentive to take one over something else most of the time.