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In the bag there are $n \geq 1$ black balls.At every second we randomly choose $1$ ball and replace it with white ball(even if ball that we took was white). Let $T$ be first time that all balls are white. Find expected value of $T$.

My work.

There is a hint to look into $T_k, k=1,2,\cdots n$ where it is first time that in the bag there is exactly $k$ white balls. We need to find $E(T_n)$.

So another hint is to show that $T_k - T_{k-1}$ has geometric distribution and use that to solve the problem

So $$T_k - T_{k-1} = \frac{\binom{n}{n-k+1}}{n}(1-\frac{\binom{n}{n-k+1}}{n})^{m-1}$$

because at $k-1$-th step we already have $k-1$ white balls so we can look as a geometric distribution. If we take a black ball then we replace with white ball, probability of this is first part and second part is if we take white ball.

But from here how find $E(T_k - T_{k-1})?$ and did I calculate right geometric parameter?

joriki
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unit 1991
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  • This looks like a version of the coupon collector's problem – Henry Mar 19 '23 at 10:11
  • If your geometric parameter is $\frac{\binom{n}{n-k+1}}{n}$ then no, you have not calculated it correctly: it is much simpler than that. Think about the probability that $T_k-T_{k-1}=1$ – Henry Mar 19 '23 at 10:14
  • Between $\ T_{k-1}\ $ and $\ T_k\ $ the probability of drawing a white ball is $\ \frac{k-1}{n}\ $ and the probability of drawing a black ball is $\ \frac{n+1-k}{n}\ $. The event $\ T_k-T_{k-1}=r\ $ occurs if and only if the first $\ r-1\ $ balls drawn after time $\ T_{k-1}\ $ are white and the $\ r^\text{th}\ $ one is black. What, therefore, is the probability that $\ T_k-T_{k-1}=r\ $? – lonza leggiera Mar 19 '23 at 11:54
  • @lonzaleggiera $P(T_k - T_{k-1} = r) = (\frac{k-1}{n})^{r-1} * \frac{n+1-k}{n}$. And then $E(T_k - T_{k-1}) = \sum_{r=0} ^n r (\frac{k-1}{n})^{r-1}(\frac{n+1-k}{n})$ – unit 1991 Mar 19 '23 at 19:16
  • And then $E(T_n) = E(T_n - T_{n-1}) + E(T_{n-1} - T_{n-2}) + \cdots + E(T_1)$. – unit 1991 Mar 19 '23 at 19:19
  • @lonzaleggiera is this right? Computing sum I don't think is trivial – unit 1991 Mar 19 '23 at 19:20
  • @unit 1991 Your expressions for $\ E(T_k-T_{k-1})\ $ isn't quite correct. Since $\ T_k-T_{k-1}\ $ has a positive probability of $\ \left(\frac{k-1}{n}\right)^{r-1}\left(\frac{n+1-k}{n}\right)\ $ of taking any positive integer value $\ r\ $, the upper limit of the sum should be $\ \infty\ $, not $\ n\ $. – lonza leggiera Mar 20 '23 at 05:29
  • While I wouldn't call the evaluation of that infinite sum "trivial", it's nevertheless well-known, and not particularly difficult. It's the expected value of a geometric distribution with parameter $\ p=\frac{n+1-k}{n}\ $, which is known to be $\ \frac{1}{p}\ $, as stated in the above-linked Wikipedia article. – lonza leggiera Mar 20 '23 at 05:34

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