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Nyo wrote on 2011-02-26 20:54
Quote from Andy-Buddy;342956:
Darn, I had a spreadsheet of this a while back, too. :l
Here it is.
Repair.xls
Notes: The blessing stat is unconfirmed. leave it on no for an accurate representation.
Here is my formula that calculates the chance of failure, appended to your spreadsheet. I didn't touch any portions of what you did, so feel free to clean it up.
http://puu.sh/143I
(wow that is a neat program)
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Grifter wrote on 2011-02-27 03:04
I said it in that other thread, and I'll say it again: Purrz needs to take a course in basic probability.
B> derep button
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Retard wrote on 2011-04-13 22:29
Idk maybe i am a retard. I keep re-running my logic in my head and i keep coming to the conclusion that
80% of the time you will have a perfect repair. T-T
Edit: Scratch the above. I think I know now but seriously no need to ask for a Derep Button ;~;
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casshem wrote on 2011-04-14 00:57
If you're doing 10 points at 98%, you will have a 81.7% chance of no fails. That's .98^10. Let's say you come back to repair 10 points again; now it becomes .98^20; that's a 66.7% chance of no fails on your second trip to the blacksmith. BUT, remember this: once the blacksmith fails, you stop adding to the exponent. So let's do a quick sum up:
You have a 0/10 weapon and you head over to Edern. The chance of Edern successfully repairing your weapon is 81.7%. Let's say he does not fail any points and you're happy your sword is now at 10/10, so you go spam some SMs instead of cutting your wrists.
After a while of spamming SMs, your sword is now 0/10 :T_T: The probability of Edern pulling off another successful FULL repair is .98^20 or 66.7%. Let's say you're at 5/10 going for 6/10 and he FAILS. You're now at 5/9 :T_T: The probability that Edern will not mess up any further now becomes .98^4 or 92.2%.
Wow-...that wasn't a quick sum up. Oh well
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qaccy wrote on 2011-04-14 01:32
Quote from casshem;408522:
If you're doing 10 points at 98%, you will have a 81.7% chance of no fails. That's .98^10. Let's say you come back to repair 10 points again; now it becomes .98^20; that's a 66.7% chance of no fails on your second trip to the blacksmith. BUT, remember this: once the blacksmith fails, you stop adding to the exponent. So let's do a quick sum up:
You have a 0/10 weapon and you head over to Edern. The chance of Edern successfully repairing your weapon is 81.7%. Let's say he does not fail any points and you're happy your sword is now at 10/10, so you go spam some SMs instead of cutting your wrists.
After a while of spamming SMs, your sword is now 0/10 :T_T: The probability of Edern pulling off another successful FULL repair is .98^20 or 66.7%. Let's say you're at 5/10 going for 6/10 and he FAILS. You're now at 5/9 :T_T: The probability that Edern will not mess up any further now becomes .98^4 or 92.2%.
Wow-...that wasn't a quick sum up. Oh well
Uh...why would the exponent be 20? You're repairing another 10 points, so it's again a .98^10 chance of having no fails. The success rate doesn't get lower simply because you've already full repaired once before...it's the same odds.
However, if you meant it as in figuring out the odds of, say, a full 0/20 repair with no fails, then the exponent would be 20.
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casshem wrote on 2011-04-14 01:45
It becomes .98^20 because you are going for 20 succeeds in a row. You only stop piling on the successes to the exponent once the successful repairs in a row streak is broken by a fail (as I demonstrated later on in my post). This is why so many people think the Mabi %'s are wrong.
Just to hammer the nail into the coffin:
Repair#(% success @ 98% )[S= success; F= fail] - #1(.98)[S], #2(.98^2)[S], #3(.98^3)[S], #4(.98^4)[S], #5(.98^5)[S], #6(.98^6)[F], #7(.98)[S], #8(.98^2)[S], #9(.98^3)[F], #10(.98)[S]~and so on.
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TheKartheus wrote on 2011-04-16 13:49
This all only works in probability, not in actuality.
It's like having babies. If you have 4 babies that are all boys, does that mean you have a 6.25% of having another boy child? No, you still have a 50% chance of getting a boy or a girl. The same concept works for this.
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casshem wrote on 2011-04-16 14:30
Probability is the guess of what will happen and actuality is what actually occurred. Probability is not affected by the last outcome, unless you want one outcome in a consecutive fashion.
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Omegatronic wrote on 2011-04-18 11:55
past events do not affect the probability of future events unless they eliminate a possible future outcome.
I can flip a coin 100 times and have it come up heads 100 times (past events), and the probability that the next flip will be heads is 50%.
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casshem wrote on 2011-04-18 12:36
I give up. Y'all's dumb.
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Omegatronic wrote on 2011-04-18 18:49
Quote from casshem;414446:
I give up. Y'all's dumb.
http://en.wikipedia.org/wiki/Gambler%27s_fallacy
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casshem wrote on 2011-04-18 20:04
You're not catching the message. You're saying the odds of a single toss is 50% no matter what, which is of course CORRECT. What I'm saying is: what are the odds that you'll get 10 heads OR tails in a row, consecutively, which would be .5^10 or roughly .001%.
When repairing a 10/10 weapon, you want 10 success in a row ideally. At 98%, you have a .98^10 chance, or roughly 82.%, of that happening.
What part do you not comprehend?
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WheeeE wrote on 2011-04-18 20:07
Oh nu basic probability turned into a huge post :(
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arklian wrote on 2011-04-18 20:30
Quote from Omegatronic;414791:
http://en.wikipedia.org/wiki/Gambler%27s_fallacy
We can see from the above that, if one flips a fair coin 21 times, then the probability of 21 heads is 1 in 2,097,152. However, the probability of flipping a head after having already flipped 20 heads in a row is simply 1â„2. This is an application of Bayes' theorem,
From your link.
Quote from casshem;408522:
If you're doing 10 points at 98%, you will have a 81.7% chance of no fails. That's .98^10. Let's say you come back to repair 10 points again; now it becomes .98^20; that's a 66.7% chance of no fails on your second trip to the blacksmith.
The place where you went wrong is that you said "let's say you come back to repair 10 points
again." This implies that the first 10 already occurred in the past and you shouldn't factor these in.
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ikotomi wrote on 2011-04-19 03:36
I can't believe people are still arguing over this stuff.
Quote from casshem;414862:
You're not catching the message. You're saying the odds of a single toss is 50% no matter what, which is of course CORRECT. What I'm saying is: what are the odds that you'll get 10 heads OR tails in a row, consecutively, which would be .5^10 or roughly .001%.
When repairing a 10/10 weapon, you want 10 success in a row ideally. At 98%, you have a .98^10 chance, or roughly 82.%, of that happening.
What part do you not comprehend?
The part that is wrong is that if you repair 10 more times after succeeding in 10 repairs, you do not say that the probability of succeeding all 10 times is .98^20. If it were an arbitrary 20 repairs, that would be true, but we already know that the first 10 repairs succeeded, so they do not factor into our probability calculation--there is no longer any randomness in the first 10 repairs.
To formulate the situation pedantically, we want to know specifically the probability that Edern succeeds 20 times conditional on the first 10 time being successes. The actually computation is 1^10*.98^10. This is a completely different question than what is the probability that Edern will succeed the next 20 times.