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.

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.

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.

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.