Post B7Qc8Z9gdRuUJM1arY by mitsunee@mk.absturztau.be
 (DIR) More posts by mitsunee@mk.absturztau.be
 (DIR) Post #B7Qc8Z9gdRuUJM1arY by mitsunee@mk.absturztau.be
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       random math question, let's say I have a dynamic chance for something (increases every 50 trials) and would like to calculate my total chance for 75 trials. Can I still use BINOMDIST for that? How do I combine the two chances properly?edit: bonus question since this is for pokemon stuff. Each trial actually does n amount of checks of a dynamic target value against a roll [0,1000[ and if those fail there's still the regular 1/4096 shiny chance too. Could I factor that in as well?#maths #duckduckfedi ​:boost_appreciated:​
       
 (DIR) Post #B7Qexc6PdYxC82xj28 by jacky@fedi.absturztau.be
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       @mitsunee chance of no success in the first 50 trials = (1-p)^50 where p is the chance of success per trial chance of no success in the next 25 trials = (1-q)^25 where q is the chance of success per trial chance of no success in the first 75 trials = (1-p)^50 * (1-q)^25 chance of at least 1 sucess in the first 75 trials = 1 - ((1-p)^50 * (1-q)^25)
       
 (DIR) Post #B7QfPGa7FF8tUW2eIK by jacky@fedi.absturztau.be
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       @mitsunee the bonus question i'm not sure if i understand; if it's just about increasing the chance of success (p and q) by a separate 1/4096 chance of success, it's 1 - (((1-p)*r)^50 * ((1-q)*r)^25) where r = 4095/4096
       
 (DIR) Post #B7Qg7ZfPeAbBQQeAQi by jacky@fedi.absturztau.be
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       @mitsunee or in other words 1 - (binomdist(0, 50, p) * binomdist(0, 25, q))