What is the probability of your next choice that make your choices uniformly distributed

Question

Given an #n of balls. you are only allowed to pick between two balls at a time with a probability $P_{i}=0.5$ with i being the number of the ball.

For example: you're presented with a black ball and a red ball.. so to be uniformly distributed the probability would be $P_{black}=P_{red}=0.5$ so you pick the red ball and forget about the black ball.

but then you're introduced to a blue ball. what is the probability of picking the blue $P_{picking}$ ball so that the set would be uni-formally distributed? $P_{black}=P_{red}=P_{blue}=1/3$

after each pick you're given a new ball what is the $p_{picking}$ to make $p_{ball}=1/n$

How do you go about solving this?

Answer 1

As we are picking in a sequential order either to pick the next ball or not with the main goal of making the distribution of result uniform which means that for all balls it's equally likely for all other balls to be selected.

so let's analyze the first three balls.

$P_{picking}=100\%$ for the first ball $P_{1st} = 100\%$

$p_{picking}=50\%$ for the second ball $P_{1st} = 50\% , P_{2nd}=50\%$

$P_{picking}= ??$ for the third ball, for this we have to consider

1. sequence of our actions $50\%$ to pick one of the previous balls and $X\%$ to pick the third ball
2. The number of ways the ball can be picked {(1st,third),(2nd,third)} $n_{ways} = 2$
3. our goal is to make $p_{3rd} = \frac{1}{3}$

$P_{3rd} = P_{picking,3rd} \times p_{picking,2nd} \times n_{ways}$

$P_{ball} = \frac{1}{n}$ Target

$ways = i-1$ where i equals the number of total balls

General form would be $P_{picking,new} = \frac{1}{i} / P_{picking,old} / (i-1)$

Credit: thanks for dsillman2000 for pointing out that the probability of the ball newly added ball equals to the sum of it's ways

Answer 2

welcome to Math stack exchange!

Suppose we're dealing with $n$ balls, with ball $i$ having a probability of being chosen denoted $0\leq p_i\leq 1$. We also know that all of our probabilities need to sum up to $1$, so we have that

$$ p_1 + p_2 + p_3 +\cdots+p_n=1 $$

Because we want our probabilities to be evenly distributed, we want all of our probabilities to be equal, so

$$ p_1 = p_2 = p_3 =\cdots=p_n $$

The solution to this problem (which you can verify rigorously by solving the system equations, if you want) is to set each probability $p_i = 1/n$. This works because

$$ \frac{1}{n} + \frac{1}{n} + \frac{1}{n} + \cdots \frac{1}{n} = n\cdot\frac{1}{n} = 1 $$ and $$ p_1=\frac{1}{n}=p_2=\cdots=p_n $$ So the case with $n=3$ balls means we have a probability of each ball being chosen at $p_i=1/3$.