Unexpected Christmas monster -> "The Three-Headed-Mouboo"

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Dicekid
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Unexpected Christmas monster -> "The Three-Headed-Mouboo"

Post by Dicekid »

Very rare, but I got some screenshots. Don't know if the drop chances for aqua ornament is higher then for normal mouboos, because I just met this one. You can see it staying around and then a few attacks... It also takes three times longer to kill this one, but it gets easier the more heads you killed... :P :lol: :P :lol:

Image

Greetings,
Dicekid

P.S.: Don't be too serious.... :wink:
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Amuk
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Re: Unexpected Christmas monster -> "The Three-Headed-Mouboo"

Post by Amuk »

Nice, I too have encountered this mythical beast. Unlike you however, instead of killing it I decided to start a religion with it as the main figure head.

You are all welcome to join and worship the three-headed mouboo!
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meway
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Re: Unexpected Christmas monster -> "The Three-Headed-Mouboo"

Post by meway »

I think if this is what I think :P lol you have 3% instead of 1% chance.
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Re: Unexpected Christmas monster -> "The Three-Headed-Mouboo"

Post by TheBasilisk »

meway wrote:I think if this is what I think :P lol you have 3% instead of 1% chance.
Not quite. You have a 1.444444 (recurring) % chance.

That's 1% * ((1/3)^3)

I think those calculations are right. xD
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radiant
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Re: Unexpected Christmas monster -> "The Three-Headed-Mouboo"

Post by radiant »

That would be 13/900, which I don't see how you get from that calculation. Multiplying that way would give lower odds than that of a single monster, which is clearly wrong.

The correct odds in this case are 2.9701%, a bit less than the naive 3% because there's a chance that two of the three would both drop ornaments, and that outcome has to be treated as no better than a single drop.
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Re: Unexpected Christmas monster -> "The Three-Headed-Mouboo"

Post by bigglesworth »

Buying three lottery tickets doesn't increase your odds of winning the lottery unless they sell a fixed number of tickets.

If it did exist, wouldn't the same odds of 'winning' an ornament from the hyrda-mouboo be the same as a single-headed mouboo, since there are an infinite amount of ornaments (until the quest ends)?
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enchilado
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Re: Unexpected Christmas monster -> "The Three-Headed-Mouboo"

Post by enchilado »

Actually it does increase the odds of winning. The Lottery uses numbers, right? So you're covering more number combinations.
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Re: Unexpected Christmas monster -> "The Three-Headed-Mouboo"

Post by DarkLord »

I believe this is fake. I can see sets of legs. That's just 3 mouboos.
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Re: Unexpected Christmas monster -> "The Three-Headed-Mouboo"

Post by Crush »

DarkLord wrote:I believe this is fake. I can see sets of legs. That's just 3 mouboos.
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Re: Unexpected Christmas monster -> "The Three-Headed-Mouboo"

Post by meway »

lmao p owned
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Re: Unexpected Christmas monster -> "The Three-Headed-Mouboo"

Post by DarkLord »

Crush wrote:
DarkLord wrote:I believe this is fake. I can see sets of legs. That's just 3 mouboos.
Image
...
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Dicekid
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Re: Unexpected Christmas monster -> "The Three-Headed-Mouboo"

Post by Dicekid »

Hi,

to get some light in the discussion and darkness of the theory of probabilities...
Assuming that the chances for one mouboo to drop an aqua is 1% (which means 0.01) and that we can deal every head like a single mouboo, which is a good assumtion but has to be proven to be correct... :mrgreen:
Now the chances to get at least 1 (in words one) aqua includes the following three cases:

1. every head drops one aqua -> 0.01 x 0.01 x 0.01 = 0.000001 :shock:
2. two of the heads drop one and the last one drops none-> 0.01 x 0.01 x 0.99 = 0.000099
this can happen three times (because we don't care which two heads drop the aqua, do we?)
-> 0.000099 x 3 = 0.000297 :shock:
3. just one of the heads drops one and the other two drop nothing -> 0.01 x 0.99 x 0.99
obviously this also can happen in three cases (as stated above :roll: ) -> 0.009801 x 3 = 0.029403 :(

Now because we would be happy with one, two or three aquas we add up all three and the result is:
0.029403 + 0.000297 + 0.000001 = 0.029701 which is 2.9701% which is a little bit less than 3% .

I hope you enjoyed the excourse in probabilities... :D

Greetings,
Dicekid

P.S.: I don't like to quote myself, but please don't be too serious... :wink: :mrgreen:

EDIT: I recognised a tiny error in my calculation(which is no longer existing, so this here shows just another approach to calculate...). One has to calculate the chance that no aqua drops which is
0.99 x 0.99 x 0.99 = 0.970299 and then subtract this from 1 -> 0.029701 -> 2.9701%. I will think about how to get the same result the other way round or where the logical mistake was. But this one is at least correct. Because even if you fight a million mouboos there is a chance that you won't get any aqua from them and this would not be the case with the former calculation (as you can easily see...) :shock:
So sorry for the confusion... :oops: :mrgreen:

2nd EDIT: Sorry... Ok, I corrected the things above, the red things are new and this is the correct way to calculate the correct possibility. :-) I hope you don't mind.... :oops: :mrgreen:
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Re: Unexpected Christmas monster -> "The Three-Headed-Mouboo"

Post by enchilado »

Image
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Re: Unexpected Christmas monster -> "The Three-Headed-Mouboo"

Post by JackDeth »

You know what?

In the time it took you guys to go through all this crap.....

...you could have found 3 aqua ornaments, leveled up 5 new characters, eaten lunch, gone to the bathroom, had a snack, drank a beer, ran for a government position as an elected official, saved the whales, solved world hunger, farted in someone's general direction, did the hokey-pokie, and then turned yourself around.

Just play the game.

Merry Christmas.

8)
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Re: Unexpected Christmas monster -> "The Three-Headed-Mouboo"

Post by Maxxie »

:lol: LOL :lol: dang Crush that was good :D Im still laughing ...... lmao, you too poison_ivy this is too much :lol:
...Oh and nice one Deth :)
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