Is this a sign or something?

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User Info: crunchy612

crunchy612
4 years ago#11
You can't play Enigma?

>:(
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GC IGN: BlazenFury. Sig also dedicated to the great Gamefaqs GC community. :D

User Info: vegetaken

vegetaken
4 years ago#12
I just get heroes I don't like it seems. Very common to get obsidian destroyer, riki and viper or god forbid drow in which everyone is pissed then.

Today was good though I got earthshaker and then a CK which I traded for a void.
axe<great axe<greater axe<greatest axe. no need to thank me. ~_Emissary_
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User Info: DARKlegend64

DARKlegend64
4 years ago#13
crunchy612 posted...
You can't play Enigma?

>:(


I liked Tide better.
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User Info: blade6321

blade6321
4 years ago#14
Annuit_Coeptis posted...
Heh, trivia time. What's the chance of you getting the same hero (any hero) at least three times out of ten?


Ugh...I'm supposed to be working on functions and you've got me doing probability :l

I did an exam a few weeks ago on statistics and was hoping it would be a long time before I got back to it again >.<
"Kids these days huh? Can't live with them, cant chop them up into tiny pieces." -Mcbeth
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User Info: ReadOnlyAccnt

ReadOnlyAccnt
4 years ago#15
Divide 1 by the number of heroes in the game then multiply that probability by its self 3 times. This is assuming you we're always the first to random and all hero choices were still in the pool. Is that right?
Bleh

User Info: DragooneerZero

DragooneerZero
4 years ago#16
From: ReadOnlyAccnt | #015
Divide 1 by the number of heroes in the game then multiply that probability by its self 3 times. This is assuming you we're always the first to random and all hero choices were still in the pool. Is that right?


you have to also account for the probabilities of picking the same hero 4,5,...,10 times, and for each of those probabilities, multiply them by the number of permutations of 10 elements (i.e. the different orders in which those events happen)
Is there much left to enumerate? What won't idle, well-fed, unfeeling people invent?
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User Info: ReadOnlyAccnt

ReadOnlyAccnt
4 years ago#17
I accounted for getting the Lina 3 times hence cubing the original probability. I thought you just raised the original probability to the Nth power. N being how many times in a row Lina was randomed
Bleh

User Info: blade6321

blade6321
4 years ago#18
ReadOnlyAccnt posted...
Divide 1 by the number of heroes in the game then multiply that probability by its self 3 times. This is assuming you we're always the first to random and all hero choices were still in the pool. Is that right?


What this is is the chance of randoming Lina three times in a row. You might random her in the last three games, or one of the many variations of it, which I believe you would get by doing:

10!/(7!*3!)

(ten factorial over seven multiplied by three factorial)

Which comes out at 120 possible variations

In the case of a single, specific hero such as Lina determined before the first random it would be:

10C3*(1/95^3)*)(94/95)^7)

Which comes out at 0.00013 (aprox).

I also assumed the question was getting any hero three times in the ten games, which is much more complicated.

For example, it doesn't matter what the first hero randomed is. So to figure out the chance of getting the first hero you random would be the same but r=2 n=9

But that doesn't take into account that you might random a different hero the second time, and the chance of them being chosen three times, and the third time etc.

If it was twice it'd be a lot easier; that way you can assume that you'll either: A get the same hero twice and thus you're good, or B you'll get a different hero, which we can use nCr to figure out the prob' of. With three times you can get the same hero twice and this means you have to find all the probabilities regarding that possibility.

Of course, this all assumes the question is what I think it is: that after ten randoms you'll have three of a single hero. Even if it's not what he originally intended I'm going to have to work it out now >.>
"Kids these days huh? Can't live with them, cant chop them up into tiny pieces." -Mcbeth
-Crabdom Ambassador.

User Info: RealYorae

RealYorae
4 years ago#19
ReadOnlyAccnt posted...
I accounted for getting the Lina 3 times hence cubing the original probability. I thought you just raised the original probability to the Nth power. N being how many times in a row Lina was randomed


Yours is the way we all thought and is correct in that regard. The other way is more work for an unnecessary answer. 1/100ish(# of heroes) 3 times = 1/1000000ish or 1 in a millionish. Though that would be 3 times in a row opposed to 3 out of ten times, which should be higher.
It is not things, but opinions about things that have absolutely no existence, which have so deranged mankind! - Nietzsche

User Info: MusashiExtra

MusashiExtra
4 years ago#20
I always get SK, who I'm terrible at. I think I have like 10 losses and 1 win with him.
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