I am trying to figure if drawing order has an influence on probability.
There are 100 balls in a box, 10 of them are red, 90 of them are black. 100 different person are taking turn to draw a ball.
Is there a prefered position to take to draw a red ball (before knowing any of the draws) or is the probability constant (0.1) ?
First person has 10 out of 100 = 0.1 chance of drawing a red ball. My intuition is that since its 0.1 for the first one, in 9 cases out of ten, the second drawer will have 10 out of 99, and in the other 9 out of 99, which would translate in 0.9*(10/99) + 0.1*(9/99) = 0.1
Does this remain constant over the course of the drawing ? How can i generalise this formula with P(drawing a red ball), N the sample size (here 100)
Intuitively, by counting over all of the combinations of balls possible, the probability of the $j^{th}$ ball being red is
$$\frac{\binom{99}{9}}{\binom{100}{10}}$$
because $\binom{n-1}{k-1}=\frac{k}{n}\binom{n}{k}$, this equals $0.1$.
More technically, the probability of the $j^{th}$ ball being red can be determined by looking at all the previous cases.
Assume $k$ red balls have already been drawn.
$$\frac{(100-j)!}{100!}\frac{10!}{(9-k)!}\frac{90!}{(90-j+k)!}\binom{j-1}{k}$$
Collecting the straight factorials together gives
$$\frac1{\binom{100}{10}}\frac{(100-j)!}{(9-k)!(90-j+k)!}\binom{j-1}{k}$$
and summing over $k$
$$\frac1{\binom{100}{10}}\sum_{k=0}^j\frac{(100-j)!}{(9-k)!(90-j+k)!}\binom{j-1}{k}$$
$$\frac1{\binom{100}{10}}\sum_{k=0}^j\binom{100-j}{9-k}\binom{j-1}{k}$$
and applying Vandermonde's identity gives the result.