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Post by zaffe on Apr 18, 2011 5:27:17 GMT -5
so basically, it's a shiny pikachu wave (5 shiny pikachus in one wave)?
crazy
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Post by mawile on Apr 18, 2011 7:50:10 GMT -5
so basically, it's a shiny pikachu wave (5 shiny pikachus in one wave)? crazy no, only one of them will be shiny (well, there is an extremely slim chance when twoor more in a wave are shiny.. )
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Post by Calvin on Apr 18, 2011 8:52:20 GMT -5
Theres some dubious assumptions and not entirely right use of probability at work here... So, given an average of 1 wave of 5 pikachu's per 10/3 runs, and 0.01% chance of any given pikachu to be shiny, or alternatively, a 99.99% chance that a given pikachu is *not* shiny. This gives the probability of seeing 'n' pikachus in a row that are *not* shiny to be (9999/10000)^n. The probability that you will see a shiny pikachu within n pikachus is thus 1-(9999/10000)^n. In order to have a 50% chance of seeing a pikachu then, you'll need to have encountered n=log(0.5)/[log(9999)-4]=6932 pikachus, rounded up. That means, at 1 wave of 5 pikachus per 10/3 runs, you'll need to do 4622 runs (again, rounded up). At 2 minutes per run, that comes to just over 154 hours, or around 6.4 days of straight playing to even have a 50% chance of having seen a shiny pikachu. (Note that this is different from having a 50% chance that the 6932'nd pikachu is shiny, it's instead the chance that any one of the 6932 pikachu's you've seen is shiny, and it could well have been the first one you saw.) Notice that even at 10k pikachu sightings you don't have a 100% chance of having seen a shiny pikachu (in fact, you only have about a 63% chance at 10k sightings). It'd also take infinite pikachu sightings before you were garunteed to have seen one, which is in line with what you'd expect for something random. In order to have a 90% chance of having seen one, you'd need to have seen closer to 23k pikachus, which would take 21.3 days or so of straight playing. These numbers aren't any more encouraging that the ones given in the OP, but at least they take into account the random nature of things, since most of us won't be catching tons of shiny pikachus, so the long term average of 1/10k doesn't really apply to any one of us individually. Reply: You have just fried my brains T^T , mind if you simplified what you meant? And what's wrong with my way of calculation. Thanks for pointing out my mistake anyway so basically, it's a shiny pikachu wave (5 shiny pikachus in one wave)? crazy no, only one of them will be shiny (well, there is an extremely slim chance when twoor more in a wave are shiny.. ) Reply: The chance should be considered to be insignificant
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Post by Mysticman89 on Apr 18, 2011 11:22:47 GMT -5
You have just fried my brains T^T , mind if you simplified what you meant? And what's wrong with my way of calculation. Thanks for pointing out my mistake anyway Roughly speaking, you calculated how long it'd take to see 10000 pikachus, on the assumption that every 10000'th pikachu would be shiny, as opposed to the chance of seeing a shiny pikachu within 10k pikachus. It's a subtle difference, and something people commonly do when it comes to probability. As a simpler example, what's the probability that if you roll a 6 sided dice 6 times, that you'll get at least one 6? (It's not 100%, as some niave approaches might suggest.)
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Post by blaatschaap on Apr 18, 2011 11:29:38 GMT -5
As a simpler example, what's the probability that if you roll a 6 sided dice 6 times, that you'll get at least one 6? (It's not 100%, as some niave approaches might suggest.) The math for that is flawed to begin with. Human factor isn't taken into consideration for the 6 sided dice math. Give the dice to a seasoned DnD player and he'd manage to roll at least once a 6 every 6 rolls and sometimes even 2 or 3 times. Guys like that and myself know our dice after 20 years well enough to know the right technique to throw it on the table. Which does make it a 100% succes rate of rolling a 6 at least once every 6 rolls. Nothing naive about that. It is just a fact.
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Post by Mysticman89 on Apr 18, 2011 11:46:31 GMT -5
Fine, consider a random number generator that works on a quantum mechanical non-deterministic basis (i.e. is truly random rather than pseudorandom) that produces the numbers one through six at random. If the generator runs 6 times, what is the probability that at least one 6 is produced? Of course the game code is only pseduorandom, but I reckon it's close enough to being truly random for our purposes.
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Post by hirschi15 on Apr 18, 2011 12:29:44 GMT -5
HOLY COW I DON'T UNDERSTAND ANYTHING YOU ARE SAYING MYSTICMAN89!!! i like it
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Post by sirgrizz on Apr 18, 2011 12:42:58 GMT -5
Just caught one ;D
Quite a few people seem to have them and im not that convinced about 9.25 days
Is the maths not just that:
You have a 0.01 chance of having a pikachu wave You have a 0.01 chance of one of those pikachu's being a shiny (1/10,000 = "0.0001") But because pikachu's come in waves of 5 (You only get one shiny in a wave") That means your 1/10,000 chance of getting a shiny is 1/2000 waves
Viridian Forest 2 (2.15m) 30 waves 2000/30 = 67 (times you should need to play the level) 67*135s = 2hr31m
Anybody on the same level as me?
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Post by hirschi15 on Apr 18, 2011 12:45:16 GMT -5
Grats man!
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Post by sirgrizz on Apr 18, 2011 12:48:58 GMT -5
...and just put your attacking pokemon at the bottom of the map leveling up with a (Beedrill-string shot) at the top slowing them down
They have a red bar as soon as they appear at full heath
Blue bar= Can't catch Green bar= Needs pwned a bit Red bar= Catchable
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Post by Mysticman89 on Apr 18, 2011 12:52:42 GMT -5
Just caught one ;D Quite a few people seem to have them and im not that convinced about 9.25 days Is the maths not just that: You have a 0.01 chance of having a pikachu wave You have a 0.01 chance of one of those pikachu's being a shiny (1/10,000 = "0.0001") But because pikachu's come in waves of 5 (You only get one shiny in a wave") That means your 1/10,000 chance of getting a shiny is 1/2000 waves Viridian Forest 2 (2.15m) 30 waves 2000/30 = 67 (times you should need to play the level) 67*135s = 2hr31m Anybody on the same level as me? Not all 30 waves of a VF2 run are pikachus though, and it's not exactly every 10k-th pikachu is shiny. Grats on the shiny though!
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Post by mawile on Apr 18, 2011 12:53:25 GMT -5
Reply: The chance should be considered to be insignificant I would say, the chance of finding a shiny should be considered insignificant..
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Post by hirschi15 on Apr 18, 2011 12:54:15 GMT -5
haha maybe a little insignificant! They are way flipping rare!
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Post by sirgrizz on Apr 18, 2011 13:11:08 GMT -5
Dunno man...
I think that its 1/10,000 pokemon not 1/10,000 pikachus
but if your right you have to play vf2 6700 times it is like 1 in every 250 hrs...which is sheer madness (they shouldnt be extremely rare)
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Post by hirschi15 on Apr 18, 2011 13:15:32 GMT -5
unfortunately, sirgrizz, it is 1/10,000 pikachu...
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