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There isn't really much to add but since the site insists at least 30 chars body, I am asking again : If something happens to 1 in 100 persons, is the chance of that person being you 50/50?

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Clearly not. If you’re one of a group of $100$ people, and one of you is to be chosen at random to be shot at dawn, your probability of surviving past dawn is $0.99$, not $0.5$. –  Brian M. Scott Dec 26 '12 at 13:38
    
There are 200 people in one of my classes. (horribly many). The teacher (uniformly) picks random students and asks him to answer a question. After about 60 students, I was picked. So for most people, it'snot 50/50. If your teacher has $1/2$ chance of picking you and $1/2$ chance picking from the rest, then it's 50/50. So, it depends on your distribution and how you ask questions. –  FrenzY DT. Dec 26 '12 at 13:39
    
ok, so if I have no knowledge of the distribution, is it wrong to assume a 50/50 chance? What will it be then? –  phil Dec 26 '12 at 13:45
    
@phil Nbody an tell you what you are supposed to believe. –  Michael Greinecker Dec 26 '12 at 13:54

1 Answer 1

Being that person with a 50/50 chance means, that almost every second person in a group will be concerned because you have a chance of 50% being the one or 50% not being the one. The probability you are looking for is

$$\frac{1}{100}=0.01=1\%.$$

If we think of any disease you will be affected with the probability of 1% and with probability 99% you will be healthy.

An example for the 50/50 chance would be a model with balls and bins. Assume we have one bin and $n$ blue and $n$ red balls in there. You are allowed to take one ball without explicitly looking which color it has - the probability of it being blue/red will be both $50\%$ because $n$ out of $2n$ balls have the same property (here the color). If you want to repeat this experiment keep in mind to put the ball back, otherwise you would get other probabilities!

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Suppose I don't know there are n number each of red and blue balls. Is it wrong to assume a 50/50 chance? What to assume then? –  phil Dec 26 '12 at 13:53
    
@phil: So let's say you have $n$ blue and $k$ red balls without knowing what $n$ and $k$ are... yeah, we can't do anything because we only know that we have $n+k$ balls, but we need at least one of the numbers to be able to provide information about the problem and how to handle it. –  Christian Ivicevic Dec 26 '12 at 13:55
    
@phil: Assume we have a room where a few people are in there. You have no idea how many of them are female or male and therefore there is no way to say that the chance of one being male is 50/50. You have absolutely no information to rely on, and that is the issue with this problem. –  Christian Ivicevic Dec 26 '12 at 13:57

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