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I have 3 mutually exclusive events, probability of each one happening is $\frac13$. What's the probability none of the events happens 3 times out of 3 tries.

My approach is this : $(\frac23*(1+\frac13+\frac19))^3=0.893$

Simulation shows 0.889, That 0.004 makes me think I'm incorrect.

Note:I'm looking for general formula, these numbers in my real problem aren't equal and even change during process.

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  • $\begingroup$ What was your final answer to the question? $\endgroup$ – Greenparker Aug 18 '17 at 11:00
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    $\begingroup$ I find the wording of your question somewhat confusing. I think Lukasz Derylo probably interpreted it correctly given the result he came up with, but I'd just like to point out that in that interpretation, A, B and C are not independent events; rather, they are mutually exclusive. $\endgroup$ – Ruben van Bergen Aug 18 '17 at 11:59
  • $\begingroup$ Yep you are right. $\endgroup$ – luka25 Aug 18 '17 at 12:09
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Let's call your events A, B and C. You have $3^3=27$ possible results here (AAA, AAB, AAC, ABA and so on). 3 of them are not of your interest (e.g. these are situations where one of the events happens 3 times: AAA, BBB, CCC). So you have 24 possible "good results". $24/27=0.888888...$

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  • $\begingroup$ This time it's simple because every result have similar chances of happening, I'm more interested in formula(when numbers aren't so good) $\endgroup$ – luka25 Aug 18 '17 at 12:08
  • $\begingroup$ A slightly different way of getting there would be: the probability of none of these events happening 3 (out of 3) times, is complementary to the probability that any one event happens exactly 3 times. For each one of your 3 events, the probability of occuring 3 times is $\frac{1}{3}^3=\frac{1}{27}$. Thus, the probability that any one of your 3 events is repeated three 3 is $3\times\frac{1}{27}=\frac{3}{27}=\frac{1}{9}$. So the probability that none of the events is repeated 3 times is $1-\frac{1}{9}=\frac{8}{9}=0.888888...$. But this is fundamentally the same as Lukasz Derylo's answer. $\endgroup$ – Ruben van Bergen Aug 18 '17 at 12:49
  • $\begingroup$ Nah, the probability that any one of your 3 events is repeated three is (1-1/27)^3 $\endgroup$ – luka25 Aug 18 '17 at 12:57
  • $\begingroup$ No I'm afraid that's wrong. We're considering a disjunction of three possible outcomes: event 1 is repeated three times (AAA), OR event 2 is repeated three times (BBB), OR event 3 is repeated three times (CCC). Thus the probability of any of these events happening is simply the sum of the individual probabilities of these events (that is, $P(AAA \vee BBB\vee CCC)=P(AAA)+P(BBB)+P(CCC)$). $\endgroup$ – Ruben van Bergen Aug 18 '17 at 13:06
  • $\begingroup$ Ah, you're right $\endgroup$ – luka25 Aug 18 '17 at 13:18

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