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Suppose a basket contains 2 white balls and 1 black ball. We are going to pick balls as follows: (1) Pick out a ball equally likely at random from all balls in the basket and put it in your box. (2) Pick out a ball equally likely at random from the basket. If its color is the same as the last ball chosen then put in the box and repeat from step (2). If its color is not the same, put it back in the basket and repeat from step (1).

My Idea: There are 2 outcomes on the first pick. If it's a black ball (P=1/3), the second and last pick will be white. If it's a white ball (P=2/3), I'll see what the 2nd pick gives me. If the 2nd pick is W (P=1/2), the last one left is Black, which isn't the result we want. So, if the 2nd pick is B (P=1/2), I'll have to put it back in the basket and pick again. If I pick B (P=1/2), the last one is white.

Thus, Prob. = 1/3 + (2/3 * 1/2 * 1/2) = 1/2. Correct?

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  • $\begingroup$ What is the event whose probability we are trying to compute? $\endgroup$ – Zur Luria Oct 1 '15 at 13:51
  • $\begingroup$ We're trying to find the probability that the last ball in the basket is white. $\endgroup$ – LinAlg Oct 1 '15 at 13:54
  • $\begingroup$ I agree with your calculation. $\endgroup$ – lulu Oct 1 '15 at 14:06
  • $\begingroup$ Also look up 'coupon collection problem', this is quite relevant to what you are doing $\endgroup$ – Alex Oct 1 '15 at 14:08
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The correct probability is $5/9$.

Here's how to derive it:

The probability to draw the black ball first is $1/3$. In that case, you obviously draw a white ball first.

If you draw a white ball first, then you are left with a black and a white ball in the basket. However only if you draw the white ball (which you do with probability $1/2$), it goes to your box (in which case the last ball is obviously black). But if you draw a black ball, it goes back to the basket, and you're going to draw another ball.

However this time, it's the black ball that will go into the basket (because the previous ball — which you put back into the basket — was black), leaving the white ball as last one, and if you draw the white ball it will go back to the basket. After that you're back at the situation after drawing the first white ball.

Now it's not hard to see that after drawing a white ball, the total probability of putting the second white ball into the box first is twice as large as the total probability to put the black ball into the box first (you have a probability of $1/2$ of getting the white ball out right away, a probability of $1/2\cdot 1/2 = 1/4$ of getting the black ball out in the next step, and a remaining probability of $1/4$ of starting over, with the same probabilities again). That is, after picking a white ball first, you've got a conditional probability of $2/3$ of picking the black ball last, and $1/3$ of picking the other white ball last.

Therefore the total probability of picking the white ball last is $1/3 + 2/3\cdot 1/3 = 5/9$.

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