A bag contains 100 black balls

Question

A bag contains 100 black balls. A ball is randomly picked from the bag and if it's a black ball , two white balls are added in its place and if its a white ball it is kept back. After how many turns should the probability of picking a white be closest to 50%.(if taken out,a black ball will be not kept back)

Answer 1

Hi Jazzel T.,

The language of your question makes it unclear. What does "kept back" mean? Is the ball put back into the bag, or is it kept out? Your 2nd sentence seems to imply that the black ball picked stays outside the bag, while 2 white balls are put in to replace it, but that a white ball picked is returned to the bag.

If that is the case, then making the equality happens implies white = black. Call the number of black balls removed = x. Then equality is when (100-x) (# of black balls remaining, since you started with 100 and removed x) = 0 + 2x (# of white balls consequentially substituted in). Solve: 100 = 3x ; x = 33 (leaves 67:66 ratio; if removed 34, that would be too much, since the ratio would become 66:68) . So you must make incremental progress towards picking out 33 total black balls, leaving 66+2/3 ideally. Incremental progress is made in an exponential fashion, so to speak. For ease of visualization, designate a new variable, let's say r, to track the instantaneous number of black balls (r is just 100-x, but it's easier to grasp the idea of tracking the number of actual objects as a variable, than the number of objects that have disappeared -- don't you think?). Never be afraid to change variables to get the best set to work with!

Then the instantaneous number of white balls is 2(100-r) [b/c every time a black ball disappeared, 2 white balls took its place]. So the probability of picking a black ball, from all the balls in the bag, per pick (dr/dt) is -r/(r+2(100-r)) or -r/(200-r). You must integrate that for t sufficient to make r reach 66+2/3.

So how do you do that? If dr/dt = -(r/(200-r), then separate your incremental variables:.

dr(-(200-r)/r) = dt . you can integrate each side of this equation, the left side for r, the right side for t.

Find the form of the indefinite integral for r by standard calculus = -d(200-r)/r dr = dr/dr - 200 int (dr/r), or something like that = C - 200 ln r . Figure your boundary conditions for C, then plug in for desired value of r and calculate t (number of draws). I'll leave that piece of work to you. You should calculate it that way, at least once, to make sure you can.

I personally wouldn't bother to solve analytically, if I had a worksheet program available -- just fill/series a column for t (# draws), and a column for r, and put a repetitive formula in for r(t) = r(t-1) + dr where dr is as above. That would give you r = 66.8741 (closest approach to 66.6666....) at t=47.

-- Cheers, -- Mr. d.