>>11070694>I'd still like to know whether there is an intuitive way to obtain the formula.You can derive the whole thing if you get a recurrence relation, convert it to a PDE, then solve the PDE.
Let P(b,r) be the probability that the last pick is red given b blue balls and r red balls.
You get a recurrence by doing the first ball removal and writing P(b,r) in terms of P(b-1,r) and P(b,r-1).
You end up with
P(b,r)*(b+r)^2 = P(b-1,r)*b^2 + P(b,r-1)*r*(2b+r)
and initial conditions P(0,r)=1 (r>=0), P(b,0)=0 (b>=1).
Write p(x,y) = sum{P(b,r)*exp(bx + ry) : b,r>=0}
Differentiating p with respect to x and y will allow you to get the right coefficients of r and b.
Solve the second order PDE.
It might be easier to use P(b,r)(x^b)(y^r) or P(b,r)(x^(b+r))(y^r) for the starting point.
