2
$\begingroup$

I have found this topic (Average number of trials until drawing $k$ red balls out of a box with $m$ blue and $n$ red balls) which gives the answer but I'd like to know what is wrong with my reasoning (see below).

Here is the problem: let's have $R$ red balls and $B$ black balls in a box, and let $N=R+B $ be the total number of balls.

The balls are drawn one by one without replacement. What is the expected number of draws needed to have $M \leq R$ red balls?


I have had the following reasoning, but it is obviously false since the sum of probabilities is not 1.

Let $L=R-M$ be the number of red balls left over in the box when the success condition is reached.

Let $n$ be the number of draws and $p(n)$ be the probability of reaching the success condition after these draws.

If $n<M$ (not enough draws) or $n>N-L$ (even in the worst case where only red balls would be left in the box, we would still have drawn enough of them), $p(n)=0$

Otherwise, we succeed after $n$ draws if there is left in the box exactly $L$ red balls and $N-n-L$ black balls, which I think gives:

$$p(n) = pr^L * pb^{N-L-n} * \binom {N-n} {L}$$

with $pr = R/N$ (respectively $pb = B/N$) the probability of drawing a red (respectively black) ball.


So the question is: what is wrong with my reasoning? And what would be the correct answer to get the expected number of draws needed?

$\endgroup$
0
$\begingroup$

For $n\in\{M,\dots,M+B=N-L\}$ we succeed if the $n$-th ball drawn is red and exactly $M-1$ of the preceding draws result in a red ball.

This leads to:

$$p(n)=\Pr(n\text{-th draw red}\mid M-1\text{ of preceding draws red})\Pr(M-1\text{ of preceding draws red})$$

where:

$$\Pr(M-1\text{ of preceding draws red})=\frac{\binom{R}{M-1}\binom{B}{n-M}}{\binom{N}{n-1}}$$ and:$$\Pr(n\text{-th draw red}\mid M-1\text{ of preceding draws red})=\frac{R-M+1}{N-n+1}=\frac{L+1}{N-n+1}$$

The expectation can be found now as $\sum np(n)$ but there is a much nicer way. For that see this answer on the linked question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.