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<blockquote data-quote="Tervin" data-source="post: 4280965" data-attributes="member: 66491">I am sorry, but your result is still wrong. I know this sounds cocky, but I know my method (which takes longer to do, but uses math that I can get most high school students to do) works, and produces the correct answer. To try and explain in other words than the OP, hopefully making it easier to follow:We want to know the probability of getting 4 successes before we get 2 failures. This can only happen within 5 attempts, as we will always have reached 2 failures or 4 successes by then.To keep the math easy to understand I will calculate the probability of the two possible successful cases, and then add them together. Case 1:We get 4 successes in a row, wohoo!0.55^4=0.0915Case 2:We get a failure in 1 of our 4 first rolls, but luckily succeed using our 5th roll.0.55^4*0.45*4I guess I should stop and explain this one... Out of 5 rolls 4 are successful. Those four rolls are described just as in Case 1. Then we take into account that one roll is a failure, which means we have to multiply the result by 0.45. Finally all we know as that 1 of the first 4 rolls failed. To take into account that there are 4 such cases, we multiply the result by 4.Total result is 0.55^4+0.55^4*0.45*4=0.256125Edit: I see that Eldorian, who is clearly my superior when it comes to math, reached exactly the same answer as I did.</blockquote>

[QUOTE="Tervin, post: 4280965, member: 66491"] I am sorry, but your result is still wrong. I know this sounds cocky, but I know my method (which takes longer to do, but uses math that I can get most high school students to do) works, and produces the correct answer. To try and explain in other words than the OP, hopefully making it easier to follow: We want to know the probability of getting 4 successes before we get 2 failures. This can only happen within 5 attempts, as we will always have reached 2 failures or 4 successes by then. To keep the math easy to understand I will calculate the probability of the two possible successful cases, and then add them together. Case 1: We get 4 successes in a row, wohoo! 0.55^4=0.0915 Case 2: We get a failure in 1 of our 4 first rolls, but luckily succeed using our 5th roll. 0.55^4*0.45*4 I guess I should stop and explain this one... Out of 5 rolls 4 are successful. Those four rolls are described just as in Case 1. Then we take into account that one roll is a failure, which means we have to multiply the result by 0.45. Finally all we know as that 1 of the first 4 rolls failed. To take into account that there are 4 such cases, we multiply the result by 4. Total result is 0.55^4+0.55^4*0.45*4=0.256125 Edit: I see that Eldorian, who is clearly my superior when it comes to math, reached exactly the same answer as I did. [/QUOTE]

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