This question is about the intuition behind the classic "picking the balls out of a bag" problem. It feels like it flattens out time, which is what I want to understand. First i will layout an example, and then explain my confusion later. Many thanks!
Example
Taking an existing example on MSE:
In a bag there are 6 red balls and 10 green balls and 3 balls are drawn at random then what is the probability that the first 2 balls are green given that the third ball is red?
- We can get the answer via Bayes theorem:
P($G_1 \cap G_2 | R_3$) = $\frac{P(R_3|G_1\cap G_2 )P(G_1\cap G_2)}{P(R_3)}$ = $\frac{3}{7}$
- We can [Copying the quote from the previous answer]:"'Rearrange the combinatorial tree', so that the first ball is known to be red, and we are asking for the probability that after this known first ball we further extract two green balls. Which is:"
$\frac{\binom{10}2}{\binom{15}2}=\frac {10\cdot 9/2}{15\cdot 14/2}=\frac 37\ . $
- We can note [Copying the answer from previous]: "Once you know that the third draw is red, we have 5 reds and 10 greens left to distribute amongst 15 slots. Each ball is equally likely to wind up in any of the available slots:"
$\frac {10}{15}\times \frac 9{14}=\frac 37$
Questions
Taking the 3rd approach. I believe it's saying: We can layout our 16 slots, and then adjust the probability once a slot is filled, no matter what slot it is e.g.
If we were to know that one red ball is picked from one slot (e.g. 2nd or 3rd etc). the probability of picking a green ball first would be the same each time:
P($G_1|R_2$) = P($G_1|R_3$) = P($G_1|R_4$) = $\frac {10}{15}$
The intuition I'm struggling with here is why, if we fill in the 3'rd or 4'th slot etc. why don't we have to factor in the fact that the previous slots must have been filled [even though we don't know what with]? Why can we treat filling in the 3rd slot the same as the 2nd, when the third implies their are only 13 slots left, and the second implie stheir are 14 left?
It feels like we are jumping to a point in time, i.e. the 3rd slot...filling it in...syaing okay now their are 15 slots left, because we have only filled in 1 slot...but without considering that for the person picking....they must have filled in the 1st and 2nd slot, to get to the 3rd...we just don't know what with...so actually there are only 13 slots left. This is why I said it kind of feels like it's just laying out time flat, and picking and chosing points, rather than viewing it sequentually as it actually happens.
Approach 2 makes this idea even more explicit. Claiming that we can literally rearrange it so that our 3rd, 4th etc. pick is 1st. And then solve for the other ball/balls as if it was sequential.
So, what's the intution behind viewing this as individual picks and individuals slots, to be rearranged and viewed without context of previous unknown picks?
I think my confusion can also be described by the fact that $P(G_1) = P(G_2)$
$P(G_1) = P(G_2) = P(G_2|G_1)p(G_1) + P(G_2|R_1)P(R_1)$ = $(\frac{9}{15} * \frac {10}{16}) + (\frac {10}{15} * \frac {6}{16}) = \frac {10}{16} $
I think I'm unclear how any individual pick can have the same probability, even though later picks must assume prior picks...?
Thank you for reading!
It works because there is a kind of symmetry, because the various conditional probabilities balance out regardless of which order you examine them in.
Consider an extreme case - there are 100 balls in the bag, of which 1 is green and 99 are red. I draw all 100 balls out and lay them in a line, but I cover all of them up.
If I ask you what the probability is of the 49th ball being green, hopefully you will agree that it's 1/100 - every ball in the line is equally likely to be the green one if you have no information to the contrary.
If I reveal the 50th ball, and it's red, then when I ask you again about the colour of the 49th ball then since you now know that there's one red ball it can't be, you can assign a 1/99 probability of it being green.
If I then reveal that the 100th ball is green, then what's the probability the 1st ball is green? Does it matter that I drew them out in the opposite order?
In more mathematical terms, you can think about listing out the events like $G_1 \cap G_2 \cap R_3 \cap G_4$ and so on. The $\cap$ operator is commutative, meaning that you can rearrange them as you like, so that $G_1 \cap G_2 = G_2 \cap G_1$ etc. It doesn't mean that $G_2$ is the same event as $G_1$, and they certainly aren't independent of each other, but it does mean that it makes as much sense to condition each one on the other - i.e. $G_2 | G_1$ is as meaningful as $G_1 | G_2$.