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Consider an urn which initially has w white balls and b black balls. Draw one of the balls in the urn at random, then put this ball back into the urn and add another ball to the urn of the same color of the one just drawn. Continue to draw and add balls in this manner indefinitely.

Show that in fact, the probability the n-th ball drawn is white is also $\frac{w}{w+b}$

Prob(Ball #1 is white) = $\frac{w}{w+b}$

Prob(Ball #2 is white) = $P( W_{2}| W_{1})P(W_{1}) + P(W_{2}| B_{1})P(B_{1})$ = $\frac{w+1}{w+b+1}$ $\frac{w}{w+b}$ + $\frac{w}{w+b+1}$ $\frac{b}{w+b}$ = $\frac{w}{w+b}$

How can I prove it for Ball #n without induction?

Nick
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Leaderboard281923
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3 Answers3

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As requested, this is providing further explanation on my favorite answer to this question, a symmetry-based argument by Ned.

Instead of starting with $w$ white balls and $b$ black balls, imagine instead you start with $t=w+b$ balls all of different colors, red, green, yellow, etc., $1$ ball per color. Clearly by symmetry (among the $n$ colors),

$$P(\text{first ball is red}) = P(\text{$n$-th ball is red}) = \frac1t$$

because any sequence of draws that caused the $n$-th ball to be red could equally likely have caused it to be any other specific color, say yellow (if you had swapped red with yellow to begin with). There is simply no way at all to argue that for the $n$-th ball, a specific color (say, red) is more probable than another specific color (say, yellow).

Do you buy the above argument? I can't really explain the symmetry any better.

If you buy the above argument, then consider you have a special form of "color blindness" which causes you to see the first $w$ out of $t$ colors as white, and the remaining $b$ out of $t$ colors as black. Then the $n$-th ball is white-to-you iff its color is one of the $w$ white-to-you colors, so the prob is

$$P(\text{$n$-th ball is white-to-you}) = w \times \frac1t = {w \over w+b}$$

antkam
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Note that on average adding a ball of the color you just drew does not change the average mix of balls. It is like you add $\frac w{w+b}$ white balls and $\frac b{w+b}$ black balls so the mix stays the same. You can define a variable that is the expectation value of the fraction of white balls and show that it does not change from one step to the next.

Ross Millikan
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Let us think of a basic Probability concept. Suppose, the original contents of the urn are w white balls and b black balls, and that after a ball is drawn from the urn, it is replaced along with d more balls of the same color. Then the probability that the second ball drawn is white is $\frac{w}{w+b}$, i.e. that this probability does not depend on d. Again, the probability that the second ball drawn is Black is $\frac{b}{w+b}$, i.e. this probability also does not depend on d. Please go through the diagram, where, the required probability does not depend on additional balls denoted by d has been established. Therefore, the probability that the 100th or nth ball you pick is black is $\frac{b}{w+b}$ and White is $\frac{w}{w+b}$

Soudipta Dutta
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