Probability: discarding things without knowing their identities You have a bag of 4 white chips and 6 red chips. Two chips are drawn and discarded, without you knowing the identities of the chips. What is the probability that the third chip drawn is red?
I am aware of a short cut to this and that the solution is the same as drawing the first chip as red. This means that the result is 0.6.
I have tried to work it out using tedious conditional probability. Is there a theorem that generalises this statement?
 A: Using the Law of Total Probabilities will be tedious.  Well, not too tedious for this, but it can get worse the longer the list.
Letting $R_1,R_2,R_3$ be the events for drawing red on draw $1,2,3$, respectively, and likewise $W_1,W_2,W_3$ be their complements (aka drawing white on the relevant draws).
$$\begin{align}\mathsf P(R_3)&=\mathsf P(R_1,R_2,R_3)+\mathsf P(W_1,R_2,R_3)+\mathsf P(R_1,W_2,R_3)+\mathsf P(W_1,W_2,R_3)\\&=\dfrac 6{10}\dfrac 5{9}\dfrac 4{8}+\dfrac 4{10}\dfrac 6{9}\dfrac 5{8}+\dfrac 6{10}\dfrac 4{9}\dfrac 5{8}+\dfrac 4{10}\dfrac 3{9}\dfrac 6{8}\\&=\dfrac{6}{10}\dfrac{4\cdot(5\cdot 3+ 3)}{9\cdot 8}\\&=\dfrac{6}{10}\end{align}$$
That is why it is nice to be able to reason from first principles: "Each of the ten balls has an unbiased chance to be drawn third, and six of the ten are red.  Therefore the probability that the third drawn ball will be red is $6/10$."
That's not a theorem, it is just modelling this particular system.

Still, it is nice to see that the two approaches are consilient.
A: You can't call it a theorem, perhaps but consider this:
With no a priori information, the probability that a particular type of object occupies a particular position in a sequence is equal to the probability that it occupies any position in the sequence.
