# If something happens to 1 in 100 persons, is the chance of that person being you 50/50? [closed]

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?

-

## closed as off-topic by Jonas Meyer, Alexander Konovalov, Julián Aguirre, JChau, pizzaApr 3 at 23:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jonas Meyer, Julián Aguirre, JChau, pizza
If this question can be reworded to fit the rules in the help center, please edit the question.

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

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