I have a basket with 4 balls with different colors. What is the expected value of distinct colors I can see after picking 4 times from bin. I return the ball back after each try.

I tried computing probabilities of number of colors

$$\mathbb{P}(1) = \binom 41 \times \frac 1{4^4}$$

$$\mathbb{P}(4) = 1 \times \frac 34 \times \frac 24 \times \frac 14$$

But I struggle with computing $\mathbb{P}(2)$ and $\mathbb{P}(3)$. I was thinking:

$$\mathbb{P}(2) = \binom 42 \times \frac 1{2^4} - \mathbb{P}(1) = \frac 6{16} - \frac 1{64} = \frac {23}{64}$$ but I was told that I am wrong.

  • $\begingroup$ A useful tutorial on formatting can be found here $\endgroup$ – lulu Jun 20 '16 at 23:47
  • $\begingroup$ Thank you, I will try do better next time. :) $\endgroup$ – Tomáš Šíma Jun 20 '16 at 23:48

The probability of obtaining two colours twice each is $$\frac{4\times3\times3}{4^4}=\frac{9}{64}$$ (pick the colour which occurs first; pick another place for the same colour; pick the remaining colour). The probability of obtaining one colour once and another three times is $$\frac{4\times4\times3}{4^4}=\frac{12}{64}$$ (pick the colour to occur once; place it; pick the other colour). So $$P(2)=\frac{21}{64}\ .$$


The probability of seeing the $i$-th color when you have $n$ colors at least once in $k$ draws is $1-(1-1/n)^k$ (the latter term is the probability of not seeing color $i$ in $k$ draws). Then, the average number of colors you see is $E[\mathbb{1}_{\text{color 1 is seen in k draws }} + \mathbb{1}_{\text{color 2 is seen in k draws }} + \ldots + \mathbb{1}_{\text{color n is seen in k draws }}] = n E[\mathbb{1}_{\text{color 1 is seen in k draws }}] = n (1-(1-1/n)^k)$.

  • $\begingroup$ Using linearity of expectation, as done in this answer, very often provides simpler and more elegant calculations than determining the distribution explicitly. $\endgroup$ – Shagnik Jun 21 '16 at 9:24

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