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If you plan to RE X items, how do you calculate the odds you'll get at least 1 schem?

STAR WARS: The Old Republic > English > Crew Skills
If you plan to RE X items, how do you calculate the odds you'll get at least 1 schem?

Telanis's Avatar


Telanis
11.01.2012 , 04:13 PM | #21
Quote: Originally Posted by psandak View Post
His math is suspect, but he has the right idea. While the percentage chance of success never changes the probability of continued failure does go down. On the first attempt the odds of success versus failure are 20%/80%, however the odds of success versus seven failures in a row is all but a coin flip - 20% chance of success versus 20.97% chance of seven failures. And the odds of success are actually better than that of eight failures 20%/16.78%
There no such thing as "the probability of continued failure". There's only (a) the probability of a single attempt, and (b) the probability of learning m schems in n attempts. Assuming we consider m >= 1, then the probability mentioned in (b) is higher for high n than low n. But it doesn't matter at all whether you have previously failed; no probabilities change when you continue to fail (or succeed).

GnatB's Avatar


GnatB
11.01.2012 , 04:55 PM | #22
Yep. The odds of getting a success on THIS attempt may be about the same as the odds that the next 7 attempts will be failures.

But once you've failed 6 times, it isn't 50/50 odds on the next attempt. The next attempt is still only 20% chance of success... and ~20% chance that the next 7 attempts will be failures.

Banegio's Avatar


Banegio
11.01.2012 , 06:04 PM | #23
There maybe no "the probability of continued failure" but there is "the probability of continuious failure". We are talking about future events.

"The odds that one will get at least 1 schem" will always be less than 100% regardless of the number of RE.

When the number of RE approaches infinity, "the odds that one will get at least 1 schem" becomes closer to 100%

http://i48.tinypic.com/v2r9yw.jpg

The following table may mean nothing or it may help some to visualize.

Code:
Roll1	Roll2	Roll3	Roll4	Roll5		Pr
0.8	0.8	0.8	0.8	0.8		0.32768
0.8	0.8	0.8	0.8	0.2		0.08192
0.8	0.8	0.8	0.2	0.8		0.08192
0.8	0.8	0.8	0.2	0.2		0.02048
0.8	0.8	0.2	0.8	0.8		0.08192
0.8	0.8	0.2	0.8	0.2		0.02048
0.8	0.8	0.2	0.2	0.8		0.02048
0.8	0.8	0.2	0.2	0.2		0.00512
0.8	0.2	0.8	0.8	0.8		0.08192
0.8	0.2	0.8	0.8	0.2		0.02048
0.8	0.2	0.8	0.2	0.8		0.02048
0.8	0.2	0.8	0.2	0.2		0.00512
0.8	0.2	0.2	0.8	0.8		0.02048
0.8	0.2	0.2	0.8	0.2		0.00512
0.8	0.2	0.2	0.2	0.8		0.00512
0.8	0.2	0.2	0.2	0.2		0.00128
0.2	0.8	0.8	0.8	0.8		0.08192
0.2	0.8	0.8	0.8	0.2		0.02048
0.2	0.8	0.8	0.2	0.8		0.02048
0.2	0.8	0.8	0.2	0.2		0.00512
0.2	0.8	0.2	0.8	0.8		0.02048
0.2	0.8	0.2	0.8	0.2		0.00512
0.2	0.8	0.2	0.2	0.8		0.00512
0.2	0.8	0.2	0.2	0.2		0.00128
0.2	0.2	0.8	0.8	0.8		0.02048
0.2	0.2	0.8	0.8	0.2		0.00512
0.2	0.2	0.8	0.2	0.8		0.00512
0.2	0.2	0.8	0.2	0.2		0.00128
0.2	0.2	0.2	0.8	0.8		0.00512
0.2	0.2	0.2	0.8	0.2		0.00128
0.2	0.2	0.2	0.2	0.8		0.00128
0.2	0.2	0.2	0.2	0.2		0.00032
						
						1
Note statisics is all theoretical. It is possible to fail million times in a row even the probability of a single success is 99.9999999%. Statisics become accurate when the number of events approaches infinity

well's Avatar


well
11.03.2012 , 10:20 AM | #24
Quote: Originally Posted by Alec_Fortescue View Post
I don't calculate, I just R/E The more you calculate, the more you hesitate. Keep that in mind.
You ain't kidding.I have created more swear words REing then when I was in the military.I could get pretty creative then.

TheRealCandyMan's Avatar


TheRealCandyMan
11.03.2012 , 11:37 AM | #25
Quote: Originally Posted by Skodan View Post
what are your numbers?!?

its 20% each time.

.20
.20
.20
.20
.20
.20
.20
.20
.20

this never changes
He is talking in the eyes of statistics and probabilities.

Corporal_Armstro's Avatar


Corporal_Armstro
11.06.2012 , 07:03 PM | #26
Quote: Originally Posted by Skodan View Post
what are your numbers?!?

its 20% each time.

.20
.20
.20
.20
.20
.20
.20
.20
.20

this never changes
Did you ever do statistics and probability at school? The odds will always stay the same, but probability suggests that the chance of successive failures reduces every time. For Eg.

I have 2 dice and my aim is to roll a 4 ,5 ,6. My odds are 50% per dice, but by probability which is
2 x .5 = 1

Say if my goal is unchanged but instead of rolling two dice at once, I get a reroll. In this case it would then be
1 x .5 = .5
1.5 x .5 = .75

So in the realms of probability I have a greater chance of rolling a 4 ,5 ,6 with 2 dice than with rolling the one twice. So in which the odds for an RE is always .2 or .1, each successive roll would then be 1.2 x the proceeding. The odds are ALWAYS the same, my probability just increases.

If we put the above example into the question of RE probability it is:

1 x .2 = .2
1.2 x .2 = .24
1.24 x .2 = .248

And so on.

I'm a glass half full kind of guy. I like to think positive.

WooliestWorm's Avatar


WooliestWorm
11.06.2012 , 07:59 PM | #27
Quote: Originally Posted by Corporal_Armstro View Post
Did you ever do statistics and probability at school? The odds will always stay the same, but probability suggests that the chance of successive failures reduces every time. For Eg.

I have 2 dice and my aim is to roll a 4 ,5 ,6. My odds are 50% per dice, but by probability which is
2 x .5 = 1

Say if my goal is unchanged but instead of rolling two dice at once, I get a reroll. In this case it would then be
1 x .5 = .5
1.5 x .5 = .75

So in the realms of probability I have a greater chance of rolling a 4 ,5 ,6 with 2 dice than with rolling the one twice. So in which the odds for an RE is always .2 or .1, each successive roll would then be 1.2 x the proceeding. The odds are ALWAYS the same, my probability just increases.

If we put the above example into the question of RE probability it is:

1 x .2 = .2
1.2 x .2 = .24
1.24 x .2 = .248

And so on.

I'm a glass half full kind of guy. I like to think positive.
In terms of theoretical possibility don't you add the probability for an or situation.
So for your dice scenario
You want the odds of rolling 4,5,6 on dice one OR rolling 4, 5, 6 on dice 2
0.5 each time so 0.5 + 0.5 = 1
The odds of rolling 4, 5, 6 on the first roll of a dice OR 4, 5, 6 on the second roll of the dice is exactly the same.

The best way to work out odds of success to RE'ing is probably to calculate the odds of failing a given number of times. If you subtract the result from 1 that is your chance of succeeding within any one of those attempts.
So if a companion maxes their crafting slots (5 crafted items) each item has a 0.8 chance of failing.
It's an and function so the chance of failing all 5 of those is:

0.8 * 0.8 * 0.8 * 0.8 * 0.8 = 0.8^5 = 0.32768
1 - 0.32768 = 0.67232

So with one companions crafting slots maxed you have a 67% chance of reverse engineering one schematic from that.

You could use the or approach of adding 0.2 + 0.2 + 0.2 + 0.2 + 0.2 = 1 but as I'm sure you realize that's far from a surefire way of estimating the odds.
All of this if only estimation there is no definitive way of saying if you RE a certain number of items you'll get a schematic but hopefully the numbers help.

Corporal_Armstro's Avatar


Corporal_Armstro
11.06.2012 , 10:29 PM | #28
Quote: Originally Posted by WooliestWorm View Post
In terms of theoretical possibility don't you add the probability for an or situation.
So for your dice scenario
You want the odds of rolling 4,5,6 on dice one OR rolling 4, 5, 6 on dice 2
0.5 each time so 0.5 + 0.5 = 1
Yes, example one correct.

Quote: Originally Posted by WooliestWorm View Post
The odds of rolling 4, 5, 6 on the first roll of a dice OR 4, 5, 6 on the second roll of the dice is exactly the same.
Though a reroll, generally defined as a second chance in the event of failure is resolved differently, The effect this has on the outcome is as follows. First, you have the odds to succeed on the first roll. If you pass the first roll, the second roll doesn't matter. If you fail, though, you get to try again. So, the odds can be described as, the chance that you succeed, plus the chance that, if you fail, you succeed on the second attempt. Mathematically, this looks like (chance to succeed) + (1-chance to succeed)*(chance to succeed). If you want to use algebra, you can also write this as 2*(chance to succeed) - (chance to succeed)squared.