About the probability of inperias rex christals

beeredpapabeeredpapa Member Posts: 27 ★
in General Discussion
I took the 100 avobe christals and I got no tier 6 hero, so I calculated these probability(P).
P = (1-0.015)^100 ≒ 22%
I think if you pay about $100, the probability of getting no tier 6 hero is 22%. Moreover, the probability of getting no tier 5 or 6 pick up hero is 45%.
I think some gift is given to a person pay $100 for supporting these game.
What do you think about that?

Comments

  • beeredpapabeeredpapa Member Posts: 27 ★
    sorry, I miss it. $100→$1000.
  • beeredpapabeeredpapa Member Posts: 27 ★
    No way, I'm more interested in the probability of the christals in this game than Rosetta Stone.
  • WoogieboogieWoogieboogie Member Posts: 340 ★★
    beeredpapa said:

    No way, I'm more interested in the probability of the christals in this game than Rosetta Stone.

    Your spelling is atrocious. It’s not a Christ al.
  • beeredpapabeeredpapa Member Posts: 27 ★
    oh, I see, crystal〜
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  • beeredpapabeeredpapa Member Posts: 27 ★
    Do you know poisson distribution? It is basic statistic for small samples. We can estimate the probability using it.
  • WoogieboogieWoogieboogie Member Posts: 340 ★★
    beeredpapa said:

    oh, I see, crystal〜

    Thank you 😘
  • beeredpapabeeredpapa Member Posts: 27 ★
    In poisson distrbution, the probability of no tier 6 in 100 crystals is below:
    P = exp(-0.015×100) ≒ 22%
  • FreeToPlay_21FreeToPlay_21 Member Posts: 1,594 ★★★★
    I'm sorry, I'm usually not the guy to judge people from their grammar/spellings but man, that title is painful to read.
  • Bear3Bear3 Member Posts: 996 ★★★
    Could be that it’s not his first language, in which case people are being kind of insensitive and unreasonable.
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  • shadow_lurker22shadow_lurker22 Member Posts: 3,245 ★★★★★
    beeredpapa said:

    In poisson distrbution, the probability of no tier 6 in 100 crystals is below:
    P = exp(-0.015×100) ≒ 22%

    Kid loves math. I am assuming you are young because you spell things like I did when I was a kid (not being judgemental just wanted to point it out so that people on here wont judge you for you spelling mistakes).
  • beeredpapabeeredpapa Member Posts: 27 ★
    OK, you are lack of knowledge of statistics lol.
  • DemonzfyreDemonzfyre Member Posts: 22,024 ★★★★★
    beeredpapa said:

    In poisson distrbution, the probability of no tier 6 in 100 crystals is below:
    P = exp(-0.015×100) ≒ 22%

    Or in your case it was 100%....
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  • DNA3000DNA3000 Member, Guardian Posts: 19,658 Guardian
    beeredpapa said:

    Do you know poisson distribution? It is basic statistic for small samples. We can estimate the probability using it.

    Actually, you don't use Poisson approximations for small samples. Poisson would only be useful in situations where the true situation is calculable as a binomial expansion and the number of events in the sequence is sufficiently large relative to the margin of error of the calculated event density.

    In this case, the interval window is large enough for the Poisson approximation to be reasonably valid. Poisson says the odds of seeing zero events would be e^(-a)[a^0)/0!] = e^(-a)[1] = e^(-1.5) ~= 0.22313 or 22.313%. The actual odds are [1-0.015]^100 ~= 0.22061 or 22.061%.

    Fairly close, but I don't know why anyone would do the lambda calculation instead of the direct one.
  • beeredpapabeeredpapa Member Posts: 27 ★
    DNA3000 said:

    beeredpapa said:

    Do you know poisson distribution? It is basic statistic for small samples. We can estimate the probability using it.

    Actually, you don't use Poisson approximations for small samples. Poisson would only be useful in situations where the true situation is calculable as a binomial expansion and the number of events in the sequence is sufficiently large relative to the margin of error of the calculated event density.

    In this case, the interval window is large enough for the Poisson approximation to be reasonably valid. Poisson says the odds of seeing zero events would be e^(-a)[a^0)/0!] = e^(-a)[1] = e^(-1.5) ~= 0.22313 or 22.313%. The actual odds are [1-0.015]^100 ~= 0.22061 or 22.061%.

    Fairly close, but I don't know why anyone would do the lambda calculation instead of the direct one.
    I need you, thank you for your adding for my calculation. You are right.
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