Mathematical expectation and variance of the dice game

Question

On the table there are N regular six-faced cubes. John calls a number from 1 to 6, rolls the dice,and takes all the dice that have this number. Then Petr repeats this operation. This procedure continues until all the cubes are disassembled. Find the average and variance of the number of cubes that will get John

I don't have any ideas. How is it possible to divide this random variable into more convenient and simple ones? I would like at least some hint I do not know how to approach the problem

Answer 1

Here is one way to "divide" the random variable into easier-to-manage variables.

Imagine that we are able to label each die in a way that does not affect the outcome of the roll, but allows us to identify the first die, the second die, the third die, and so forth all the way up to the $N$th die. For each integer $k$ from $1$ to $N$, you can ask whether John has the $k$th die at the end of the game. Let $X_k = 1$ if John gets the $k$th die, but $X_k = 0$ if Petr gets the $k$th die.

Then the number of dice that John has at the end is a random variable $Y$ defined by $$ Y = X_1 + X_2 + X_3 + \cdots + X_N. $$

A good thing about $X_k$ is it is a Bernoulli variable, so if you can just figure out the probability that John gets the $k$th die, you know all you need to know about $X_k$ (its expectation and variance).

There are a number of other difficult-looking problems for which this kind of decomposition is useful.

As for the probability that John gets the $k$th die, just consider that the die will be rolled repeatedly until either it rolls a number that John calls out or it rolls a number that Petr calls out. The probability for that particular die to end up in John's possession (or Petr's) on a particular roll is simply a matter of whether it is still on the table and whether it shows the same number John (or Petr) called out. That is not affected by how many other dice were already won by John or Petr or how many other dice are still on the table.

Answer 2

Avg no of cubes $\varepsilon(avg)=\frac{N}{6}+\frac{1}{6}\left(\frac{5}{6}\right)^{2} N+\frac{1}{6}\left(\frac{5}{6}\right)^{4} N--$

$\varepsilon\left(avg\right)=\frac{6 N}{11}$

Variance by symmary = var (cubes John get) = var ( cubes Petr get ) $=$ var $(\text { any } \text { number })$

$\operatorname{var}\left(Joh n\right)=\operatorname{var}(\text { Normal } 6 \operatorname{side} \operatorname{cube})$

$=\frac{1}{6} \sum_{i=1}^{6}\left(i-3 \cdot 5^{2}\right)$

$\operatorname{var}(\operatorname{John})=2.5$