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chance of getting 6 on a dice is 1/6 = ~17% to calculate that

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chance of getting 6 on a dice is 1/6 = ~17%
to calculate that we get at least one 6 in 10:

chance of not getting 6 = 100% - 17% = 83%
for 10 tries 0.83^10 = 0.15%
so chance of getting at least one six in 10 rolls is 85%

But how would I go with the question, of throwing at least two sixes?
>>
(1/6) * (1/6) * 1^8
>>
>>8844214
p(X>1) X=B(10,1/6)
>>
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>>8844225
this does not seem correct

>>8844227
this does not make much sense to me
>>
[eqn]{\binom {n}{k}}p^{k}(1-p)^{n-k} [/eqn]
n number of rolls, k number of 6es, p probability of a 6.
>>
>>8844227
>>8844258
>Binomial distribution
Fucking undergrads.
>>
>>8844260
?
>>
>>8844231
http://stattrek.com/probability-distributions/binomial.aspx
git gud
>>
OP here, reading shit some

should not it be
1/6 * 1/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6

or in short 1/6^2 * 5/6^8

logic being that those are the chances of events occurring and we dont care about order
but actual results are so tiny that its absolutely bullshit
>>
>>8844277
no, order does matter, wo you have to multiply it by all the ways it ca happen, which is 45.

so its

45 * 1/6 * 1/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 ~ 29%
>>
>>8844285
and you get the 45 from 10 * 9 * 1/2

(10 * 9) because its the first 2 of 10,9,8,7,... and 1/2 because you want 2 6es.
>>
>>8844214
>chance
Lrn2probability fgt pls
>>
>>8844285
but when I calculated probability of single 6 in 10 throws I did not need to do that shit, even when 6 could come on first or third or last throw...
>>
>>8844648
>single 6 in 10 throws
that's 10 * 1/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 ~ 32%


that for exactly 1 six, for one or more its 1 - chance for 10 non sixes as you said, which is 1 - (1 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6) ~ 84%

the 10 in the first one is because there is 10 ways to pick 1 six out of the 10 rolls, the 1 in the second one is because theres only 1 way to pick 0 sixes out of 10 rolls.
>>
>>8844214
1 - chance of throwing exactly one six - chance of throwing 0 sixes
>>
>>8844679
well, I dont feel like I understand, but whatever
numbers seem plausible so I take your word for it
>>
probability is pseudoscience.
>>
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>>8844214
>>
>>8845151
upload?
http://www.zippyshare.com/
>>
>>8845171
>http://www.zippyshare.com/
http://www78.zippyshare.com/v/KagrmIga/file.html
>>
>>8844896
fuck off, probability is literally measure theory
>>
>>8845187
probability utilizes measure theory, but is still pseudoscience, the same as astrology utilizes astronomy.
>>
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>>8844214

You can just calculate the probability of getting 2 sixes, the probability of getting 3 sixes etc all the way through 6 and then just sum up all the probabilities

In other words punch pic related into wolfram

I am 99 + 0.999... percent sure this is true
>>
>>8845202
only good post in the thread
>>
>>8845202
and then you can roll five hundred die and not get a single six.

probability is psuedoscience.
>>
Basically you are asking whats the chance of getting one six in 5 rolls.
And thats 5/6^5 = 0.40
>>
>>8845218
>probability is psuedoscience
tell that to Las Vegas
>>
>>8845218
is this a /sci/ meme or are people this retarded?
>>
>>8845220
no, it's 1-P(no sixes)-P(one six)

about 51.5%
>>
>>8845221
las vegas has just been fortunate with consistency up until today. someday, they will experience the equivalent of five hundred non-six rolls in a row and there will be a localized economic crisis. inevitability.

>>8845225
demonstrate probability as a model of reality.
>>
>>8844214
>a dice
>>
>>8845239
>I don't understand Statistics so it's fake

Okay, bud.
>>
>>8844214
I'm not sure that's correct. Something about the dice not being effected by any other roll. nevermind
>>
>>8845202
>>8845213
No, it's retarded

Just take 1 and subtract the probabilities of getting 0 sixes and 1 six
>>
>>8845151
>>8845183
I am kinda confused on the k=0 row
k is number of events that succeed? Right?
but k = zero, it should not be just 0 and 1 in following columns?
while the 64.56% actually fits when the k is = 1 and I do the classic 1-((1-p)^n) that I wrote in the opening post
>>
>>8845320
Well yeah that would be faster

Even better though, now OP can do it both ways and if he gets the same answer then it's legit
>>
>>8845346
k is the number of sixes you rolled
0.16 = (5/6)^10
>>
>>8845346
>what k=0 means
with 10 throws, the probability of getting no sixes
is a bit over 16%
>>
>>8845346
>while the 64.56%
idk what that is

in 10 throws, the probability of getting 2 or more sixes is about 51.5%
>>
>>8845246
do you have to rely on ad hominem or do you have an argument?
>>
>>8844214

n(2, 10, 0.17)

1(6)x1(6). Binomial probability, theres tables to look up values in standard variance, though; unconditional probability an be equated to the multiplication of probability

(Unconditional probability is associative but not communicative)
Thread posts: 41
Thread images: 4


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