Explaining a probability for a repeated drawing without replacement Consider the following experience : a bag contains some balls, say 3 blue balls and 4 red balls. A ball is drawn from the bag, then a second one without replacement of the first.
It's a classic result that the probability of drawing a blue ball the second time is the same as if the first ball had been replaced in the bag :
$${\rm P}(B_2) = {\rm P}(B_1){\rm P}(B_2|B_1) + {\rm P}(\overline{B_1}){\rm P}(B_2|\overline{B_1}) = \frac37\frac26 + \frac47\frac36 = \frac37$$
My question is : how do I explain to my students why the probability doesn't change ? I have a computation to answer the question, but no "common probability sense" answer.
Any thoughts about this ?
Thank you for your help.
\bye

Here is a way I think many students would have an easy time understanding.
Think if we were to apply the numbers $1$ through $7$ to the balls at random. Perhaps we draw them on the balls with a marker. We could put the numbers on in whichever order we wanted, so long as the numbers are applied randomly. If we were to apply #$2$ first, we obviously have a $3/7$ chance that #$2$ goes to a blue ball. 
This applies to any #$k$, the probability that the $k$th ball will be blue is still $3/7$.
