Urn problem and expected values

Question

First of all, I would like to say that my English is bad. Sorry for that, I'll do my best. On top of that, my math level is even worse. Thank you for you patience !

Here is my problem : I have an urn containing X balls and i have to make Y successive draws with replacement. Is it possible to know the expected value of the number of balls drawn Z times ?

Example : I make 10 successive draws in an urn of 10 balls (I chose values ​​arbitrarily).

• 3 balls are draw 1 times
• 2 balls are draw 2 times
• 1 ball are draw 3 times

I could probably do it iteratively, but is there not a formula that gives me the set of values ​​that are expected directly ? A big thanks in advance !

Answer 1

Instead of $X,Y,Z$ I will write $n,m,k$ for number of balls, draws and times respectively.

Give the balls numbers $1,2,\dots,n$.

For $i=1,2,\dots, n$ let $D_{k,i}$ take value $1$ if ball $i$ is drawn exactly $k$ times and let it take value $0$ otherwise.

Then the number of balls drawn exactly $k$ times equals: $$D_k:=D_{k,1}+\cdots+D_{k,n}$$ Applying linearity of expectation and symmetry we find:$$\mathbb ED_k=n\mathbb ED_{k,1}=nP(D_{k,1}=1)=n\binom{m}{k}\left(\frac1{n}\right)^k\left(1-\frac1{n}\right)^{m-k}$$

Answer 2

You have $X^Y$ different outcomes, let say $P_Z(k)$ is probability that there are k balls that have been drawn Z times. Expected value you are looking for is $E=\sum_{k=0}^nkP_Z(k)$ where n is largest integer such that $nZ\le Y$.

Now you have the hard part to find $P_Z(k)$

So you have $k$ balls which were drawn $Z$ times and any number of rest of $X-k$ balls which were drown any number of times but Z.

Lets calculate it for $n$ first, the maximum possible $k$.

1. Choose the $n$ balls ${X \choose n}$
2. Choose $nZ$ positions to put those balls ${Y \choose nZ}$
3. Count all possible orders of them $\frac{(nZ)!}{(Z!)^n}$
4. Put rest of balls on rest of positions $(X-n)^{Y-nZ}$

$P_Z(n)={X \choose n}{Y \choose nZ}\frac{(nZ)!}{(Z!)^n}(X-n)^{Y-nZ}/X^Y$

For others $k$ it is similar but we remove cases for higher $k$

$P_Z(k)={X \choose k}{Y \choose kZ}\frac{(kZ)!}{(Z!)^k}(X-k)^{Y-kZ}/X^Y-\sum_{i=k+1}^nP_Z(i)$

See if you can do something with this...