Balls are taken one by one out of an urn containing $w$ white and $b$ black balls until the first white ball is drawn. Prove that the expectation of the number of black balls preceding the first white ball is $\frac {b}{w+1}$

Attempt: Let $X_i$ be the random variable that denotes the number of black balls that are drawn at the $i_{th}$ step before a white ball is drawn.

Then, the total number of such balls $ X= X_1 + \cdots+X_n \implies E(X)=\sum E(X_i).$

$E(X_i)= 1 \cdot \dfrac {^bC_i}{^{b+w}Cr}\cdot \dfrac {^wC_1}{^{b+w-r}C_1}$

Thus, $\sum E(X_i) = \sum_{i=1}^{b} ~ 1 \cdot \dfrac {^bC_i}{^{b+w}Ci}\cdot \dfrac {^wC_1}{^{b+w-i}C_1}$

Could someone please tell me if I attempted this correctly? Because I get a very complicated answer in the end after evaluating the above.

Thanks a lot!

  • $\begingroup$ Please do not cross-post across sites. $\endgroup$ Oct 22, 2019 at 17:23
  • $\begingroup$ @StubbornAtom I will take care. Have been stuck for long on this problem. Could you give me a direction please? $\endgroup$
    – MathMan
    Oct 22, 2019 at 17:24
  • 1
    $\begingroup$ You're working too hard. By considering how things change when a black ball is withdrawn, it suffices to check that the formula is correct when $b=0$ and to verify that otherwise $$e_w(b) = \frac{b}{b+w}\left(1+e_w(b-1)\right)$$ where $e_w(b) = b/(w+1).$ $\endgroup$
    – whuber
    Oct 22, 2019 at 17:37
  • $\begingroup$ @whuber I am a bit confused because of one conceptual problem. Does, the probability of finding a black ball at the $ith$ step remain the same as at every step? Shouldn't it change because at the $i-1~th$ step, a black ball could have been recovered? Thanks $\endgroup$
    – MathMan
    Oct 22, 2019 at 18:04
  • 2
    $\begingroup$ That's right, which is why "$e_w(b-1)$" appears in the recursion. $\endgroup$
    – whuber
    Oct 22, 2019 at 19:38

1 Answer 1


Mathematical induction. It is easy to do @whuber's proof by mathematical induction to show that with $w$ white balls and $b$ black ones in the urn, the number $X$ of black balls drawn before the first white one has $E(X) = \frac{b}{w+1}.$ [Start the induction step with $1 + e_w(b-1) = \frac{w+1}{w+1} + \frac{b-1}{w+1}.]$

Simple case with seven balls in the urn. Also, in the specific case where $w=2$ and $b = 5,$ simple combinatorial arguments show that $P(X = k) = \frac{6-k}{21},$ for $k = 0, 1, \dots, 5.$ [For example, $P(X = 1) = \frac{2 \cdot 5}{7\cdot 6} = \frac{5}{21}.$]

Then you can use a calculator to find $$E(X) = \sum_{k=0}^5 k\frac{6-k}{21} = \frac{b}{w+1} = \frac{5}{3} = 1.6667.$$

k = 0:5; sum(k*(6-k)/21)
[1] 1.666667

Simulation of specific case. A simulation in R of a million such experiments (drawing without replacement and counting the draws before getting a white ball) approximates the distribution of $X.$ [The R function match finds the draw with the first white ball (1). The sample function draws all the balls in sequence without replacement.]

b = 5;  w = 2
balls = c(rep(0,b),rep(1,w))  # `0` for black, `1` for white
x = replicate(10^6, match(1,sample(balls))-1)
[1] 1.66686
[1] 1.666667
       0        1        2        3        4        5 
0.286038 0.238137 0.189911 0.142529 0.095611 0.047774 

These simulated probabilities are accurate to a couple of decimal places as shown in the histogram below. Histogram bars show simulated probabilities and dots show exact ones.

enter image description here

Note; If balls were drawn with replacement, then $X$ would have a geometric distribution.


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