Can you calculate a statistics question for me?

If, when event E1 happens, there is a probability, p, that a second event, E2, will be triggered, and E1 is repeatable, how often, on average, must event E1 be repeated for E2 to happen?





I know that there is no guarantee that event E2 will ever happen, but I want to know what's the average number of times event E1 has to happen before event E2 is triggered.





To put this in World of Warcraft terms, if a given epic drops off of a given boss with droprate p, how many times on average does a player have to kill the boss before he sees the item drop?





The more precise your answer, the better. The answer should be given as an integer, because you cannot perform an event, say, 6.93411723 times.

Can you calculate a statistics question for me?
would calculate your chance of me calculating a statistics question for you at 1234.367:1
Reply:What the hell?? Report Abuse

Reply:I will have to agree with a few other answers here: the answer would be 6 over 18.





In term of what question this answers, please read the first two answers. I like them the best.





But then to get my two points here, I'll have to make some sense here, right?





How about p(E1,E2) = P(E1)^pi * p(E2)/p(E1)*pi^2 = 6/18
Reply:Now this is the veay thing you have to have when a Twister is formed in a Thunder storm cloud.


So it can happen very offen even by chance.


It is all calculated in the weather chanel web site.
Reply:I would calculate your chance of me calculating a statistics question for you at 1234.367:1
Reply:50%
Reply:Lets look at it from this way


the chance of something ocuring is 50% but there is a problem, Like u said in WOW the boss does not know what the previous boss dropped nor does it matter so mathematicaly speaking u need 2 runs to get it pactically u can kill him 1000000000 times and still not see it because each time u kill the server rolls out of a 100 and if u are below lets say, u dont see it, so mathematically look at the formulas above if u can understand them cause i can't, practically there is no "if i do it X ammount of times I am guaranteed". GL
Reply:The answer that you are looking for is 1234.367:1
Reply:Let's assume that the chance E2 happens given E1 occurs is random trial after trial. The probability that E2 won't occur grows smaller as the number of trials increases, but it remains non-zero, so perhaps a better way of thinking about your answer is as follows:





The chance that E2 won't happen the first time E1 does is (1-p). The chance that E2 won't happen in two occurrences of E1 is (1-p)^2. The chance that E2 won't happen in N occurrences of E1 is (1-p)^N. Note that if p is strictly greater than 0, then this probability of E2 never happening is decreasing as N increases. However, for 0 %26lt; p %26lt; 1 and finite N, it's never zero. It can be made as small one wishes by making N large enough.





The probably that E2 must happen by N trials is 1-(1-p)^N.





Suppose you want the probability of E2 never happening to be less than q. Then





(1-p)^N = q





so N = log(q)/log(1-p)





For example, suppose q = 1/1000 and p = 1/10.





Then N = -3/-.0458 = 65





If q = 1/1,000,000 and p = 1/10, then





N = -6/-.0458 = 130





If q = 1/1000 and p = 1/2, then





N = -3 / -.301 = 10





If q = 1/1,000,000 and p = 1/2, then





N = -6 / -.301 = 20
Reply:This all depends on what p equals. Think of it this way: When E1 happens, there is a (100*p)% chance that E2 will also happen. Thus, on average, E2 happens 100*p times for every 100 times E1 happens. Thus, E2 should happen once every 100/(100*p) = 1/p times that E1 happens.





Remember, p is a probability. So, it is a decimal between 0 and 1. Thus, if p = .15, then E2 should happen once for every 1/.15 = 6.6666666666 times that E1 happens. Since probabilisits always like to round up, E2 should happen once for every 7 times that E1 happens. If p = .29, then E2 should happen once for every 1/.29 = 3.44827586207 = 4 times that E1 happens.





Good luck.
Reply:This is a classic conditional probability problem. Remember, that P(A|B)=P(A B)/P(B).
Reply:The answer is to be found in a standard distribution: In this case it would be the geometric distribution. This deals with a waiting time for an event and is explained fully in the link below.





The expected number of trials for a geometric distribution with parameter p, as you requested, would be 1/p.





Hope this helps, I remember how difficult first-year probability used to be!





EDIT: As you have not given p explicitly in the question I can't give integer answers; however, for example, if p=0.1, then expected number of trials = 10.