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An urn contains R red balls, G green balls, and B blue balls. A player randomly selects $9$ balls, doing this one by one and withOUT replacement. The player wins a dollar every time he selects three red balls in a row. (Assume $R\ge 3$ and $R + G + B \ge 9$.) To be more precise, a player receives a dollar after the $i^{th}$ draw, if he selected red balls on draws $i − 2, i − 1$, and $i$. Note that under these rules, a player drawing $4$ red balls in a row receives a total of $2$ dollars; a player drawing $5$ red balls in a row receives a total of $3$ dollars, etc. Let $X$ be the player’s total winnings.

Neee to find $E(X)$ and $E(X^2)$.

My work: Suppose we had the same setup WITH replacement, I would have defined $A_i:=$ (indicator) the event that $i,i+1\ \& \ i+2$ the balls are red. Then $X=\sum\limits_1^7 A_i$. And we have $E(A_i)=\left(\frac{R}{R+G+B}\right)^3$ . Also we have $E(A_i^2)=E(A_i)$ (since indicator) and $E(A_iA_j)$ can be found out very easily. Hence we solve the problem.

However I cant find a way to proceed in the WITHOUT replacement setup.

Qwerty
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  • I think you can find $E[X]$. To find $E[A_1]$ without replacement, you need to find $$ P(\text{ball 1 is red})\times P(\text{ball 2 is red}\mid \text{ball 1 is red})\times P(\text{ball 3 is red}\mid \text{balls 1 and 2 are red}) $$ You know $P(\text{ball 1 is red})$. Can you find the conditional probability that ball $2$ is red, given that ball $1$ is red? Conditional on this, there is one fewer red ball available, so...$\tag*{}$The tricky party is analyzing $E(X^2)$, since $E[A_iA_j]$ is different depending on whether $|i-j|$ is $1$, $2$, or $\ge 3$. – Mike Earnest Aug 10 '21 at 16:13
  • @MikeEarnest Yes $A_1$ is easy. But when we get to $A_5$ and higher, a conditioning event comes whether Ball 1 was red or not... How to tackle such a case? – Qwerty Aug 10 '21 at 16:15
  • By symmetry, $E[A_5]$ is the same as $E[A_1]$, and all $E[A_i]$. – Mike Earnest Aug 10 '21 at 16:36
  • @MikeEarnest could you please explain how $A_5$ and $A_1$ are symmetric in the without replacement case? – Qwerty Aug 10 '21 at 16:54
  • Let $R_i$ denote the event that ball number $i$ is red. $$\begin{align} E[A_i] &=P(R_i)\times P(R_{i+1}\mid R_i)\times P(R_{i+2}\mid R_i\cap R_{i+1}) \&=\frac{R}{R+B+G}\cdot \frac{R-1}{R+B+G-1}\cdot \frac{R-2}{R+B+G-2} \end{align}$$ As you can see, the result does not depend on $i$. – Mike Earnest Aug 10 '21 at 17:16
  • @MikeEarnest In the WITHOUT replacement case , Consider a case where Ball 1 is red ball 2 is green . Then $P(R_3)\neq \frac{R}{R+G+B}$ It will be equal to $\frac{R-1}{R-1+G+B}$ – Qwerty Aug 10 '21 at 17:23
  • @Qwerty That only shows $P(R_3 | R_1, G_2) \ne \frac{R}{R+G+B}$. I think that the formula given by Mike should still work in the without replacement case. – user6247850 Aug 10 '21 at 17:32
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    No, $P(R_3)$ is still equal to $\frac{R}{R+B+G}$. Depending on what the the first two balls are, the conditional probability for the third ball being red is $\frac{R-\text{something}}{R+B+G-2}$, where something is $0,1$ or $2$. If you average over these possibilities, it works out to $\frac{R}{R+B+G}$. $\tag*{}$ The intuition is that if you choose the balls, and then permute them, then you have the same random distribution of balls, so first and third places behave the same. More discussion of this phenomenon here: https://math.stackexchange.com/questions/1287393/ – Mike Earnest Aug 10 '21 at 17:35

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