My friend has three chests, each chest contains a number of balls of color either red or green or blue. The first chest has 13 red balls, 12 blue balls, and 8 green balls. The second chest has 20 red and 3 green balls. The third chest has 13 blue balls and 10 green balls.
Each time I first pick a chest (equally likely) and then pick a ball and I don’t return the ball I’ve picked. If I do this three times, on average how many red balls will I end up with?
It look to me this problem is almost impossible to calculate by hand, at least not without a considerable amount of time, because the first pick gives me 7 different results considering all (chest, color) combinations (e.g. chest1 and green ball). After three times the number of different results expands to 343. Or is there any shortcut I’m not aware of? Please enlighten me.
It is difficult to calculate the exact answer to this problem by hand, because there are a large number of possible outcomes. However, we can use some simple probability principles to approximate the answer.
First, let's calculate the expected value for the number of red balls we will get from one chest. If we pick one chest at random, we have a 1/3 chance of picking the first chest, a 1/3 chance of picking the second chest, and a 1/3 chance of picking the third chest.
If we pick the first chest, we know that there are 13 red balls, 12 blue balls, and 8 green balls. We also know that we will not return the ball after we pick it, so our expected value for the number of red balls we will get from the first chest is 13/33.
Similarly, if we pick the second chest, we know that there are 20 red balls and 3 green balls. Our expected value for the number of red balls we will get from the second chest is 20/23.
Finally, if we pick the third chest, we know that there are 13 blue balls and 10 green balls. Our expected value for the number of red balls we will get from the third chest is 0.
Now, we can use the linearity of expectation to calculate the expected value for the number of red balls we will get in total. We know that we have a 1/3 chance of picking each chest, so the expected value for the number of red balls we will get in total is:
$$(13/33) + (20/23) + (0) = 959/759$$
Thus, on average, we will get about $1.26$ red balls, or slightly less than 1/2 of the 3 balls we pick.
In short:
Let $N_A,N_B,N_C$ the counts for balls drawn from each chest, and $R_A,R_B,R_C$ be the counts for red balls drawn from each chest. Then the expected total of red balls drawn is:
$$\small\begin{align}\mathsf E(R) &= \mathsf E(R_A)+\mathsf E(R_B)+\mathsf E(R_C) &&\text{By Linearity of Expectation}\\&=\mathsf E(R_A)+\mathsf E(R_B)+0&&\text{By no red balls there}\\&=\mathsf E(\mathsf E(R_A\mid N_A))+\mathsf E(\mathsf E(R_B\mid N_B)) &&\text{By Law of Iterated Expectation}\\&=\mathsf E(\tfrac{13}{33}N_A)+\mathsf E(\tfrac{20}{23}N_B)&&\text{By }R_A\mid N_A\sim\mathcal{Hypergeo}(33,13,N_A)\text{, et cetera.}\\&=\tfrac {13}{33}\cdot\tfrac 33+\tfrac{20}{23}\cdot\tfrac 33&&\text{By }N_A\sim\mathcal{Bin}(3,1/3)\text{, et cetera.}\\&=\tfrac{959}{759}\end{align}$$