1. ## Simple probability question

Hi everybody, I have a really simple question to ask.

Let us suppose to have a box with a button. This box contains an unknown number of white balls.
When I push the button a white ball is ejected. Then I take that ball and put it in the same box. Now, what is the probability that next time that I push the button the very same ball is ejected?

My opinion is since I do not have any way to estimate this probability (I do not know the balls number in the box) the only way is to consider favorable outcomes over possible outcomes: {the same ball}/{the same ball + another ball} which is 1/2.
Many people do not agree with me, but nobody is able to provide me a different way to estimate this probability. Essentially for most of them is higher the probability to have more than two balls in the box, so the probability to get the same ball is less than 1/2. I disagree with them because we have no information about the balls in the box.

Can somebody help me? either proving me that I am wrong or providing me a way to prove that I am right.

2. ## Re: Simple probability question

Well how can you disagree with them? You said you have no information on the number of balls in the box and yet your proposed method is implicitly assuming that there are *exactly* two balls in the box. If you aren't willing to assume anything at all then this is just unsolvable. Now if you're asking about how you could do this in practice and you had some way to mark the balls before you replaced them back in the box then you could estimate the number of balls in the box and in turn estimate your probability of interest. If you are willing to maybe put a distribution on the number of balls you believe could be in the box you could use that to in turn put a distribution on what you believe the probability to be (this is very Bayesian in flavor but since we aren't collecting any data we would basically just be relying on prior distributions the whole time).

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