Consider an urn containing $c$ distinguishable balls, $\alpha$ of which are red, $\beta$ of which are blue, and $\gamma$ of which are green, and $\alpha+\beta+\gamma=c$. We assume $\alpha,\beta,\gamma>0$.

We introduce the non-negative, integer-valued random variable $X$, defined as "the number of independent trials (i.e. with replacement) needed to get at least one red ball and at least one blue ball", and the non-negative, integer-valued random variable $Y$, defined as "the number of independent trials needed to get at least one red ball and at least one blue ball and at least one green ball".

What are the expected values of $X$ and $Y$?

My attempt:

The solution for the case of the variable $X$, as suggested here, is based on the fact that the probability to get at least one red ball and at least one blue ball, in $i$ independent trials, is

$$ P(L_{1,1}^{A,B};i)=1-\left(\frac{\alpha+\gamma}{c}\right)^i-\left(\frac{\beta+\gamma}{c}\right)^i+\left(\frac{\gamma}{c}\right)^i. $$

Since $X$ is a non-negative and integer-valued variable, its expected value $E[X]$can be written as

$$E[X]=\sum_{i=0}^{\infty} P(X>i).$$

However, $P(X>i)=1-P(L_{1,1}^{A,B};i)$. Therefore, since we are dealing with the geometric series, $$ E[X]=\sum_{i=0}^{\infty}\left[ \left(\frac{\alpha+\gamma}{c}\right)^i+\left(\frac{\beta+\gamma}{c}\right)^i-\left(\frac{\gamma}{c}\right)^i\right]=\frac{c}{\beta}+\frac{c}{\alpha}-\frac{c}{\alpha+\beta}. $$

My problem here is: Are we correctly considering the fact that the event $L_{1,1}^{A,B}$ cannot occur at the first trial?

I would say "yes", and I am generally confident on this result, but I cannot really extend this reasoning to the variable $Y$.

My solution for the variable $Y$, following this suggestion, would be the following: The probability to get at least one red ball and at least one blue ball and at least one green ball, always in $i$ trials, is $$P(L_{1,1,1}^{A,B,G};i)=1-\left(\frac{\alpha+\gamma}{c}\right)^i-\left(\frac{\beta+\gamma}{c}\right)^i-\left(\frac{\alpha+\beta}{c}\right)^i+\left(\frac{\alpha}{c}\right)^i+\left(\frac{\beta}{c}\right)^i+\left(\frac{\gamma}{c}\right)^i$$.

Since, again, $Y$ is a non-negative, integer-valued random variable, $$E[Y]=\sum_{i=0}^{\infty} P(Y>i).$$

And, again, since $P(Y>i)=1-P(L_{1,1,1}^{A,B,G};i)$, and applying again the property of the sum of the geometric series, $$ E[Y]=\frac{c}{\beta}+\frac{c}{\alpha}+\frac{c}{\gamma}-\frac{c}{\alpha+\beta}-\frac{c}{\alpha+\gamma}-\frac{c}{\beta+\gamma}. $$

The problem I mentioned before appears here more evident: what about the case of $i=0,1,2$? Is the index $i$ correctly treated in the expression of $E[Y]=\sum_{i=0}^{\infty} P(Y>i)$?

As observed in this other post, I would say "no". But then I do not understand how to deal also with the index $i$ in the expression $E[X]=\sum_{i=0}^{\infty} P(X>i)$.

Thanks for your help!

EDIT: A further question:

Since we can extract at least once a ball of each of the three colors only if we have already extracted at least once two balls of two different colors, is it correct to assess $E[Y]≥E[X]+1$?

up vote 1 down vote accepted

For each $j=1,2,3$, let $p_j$ denote the probability of getting a red, blue, green ball (respectively) in one trial. Let $(p,q,r):=(p_1,p_2,p_3)$, so that $p+q+r=1$. For $x=0,1,\dots$ and $j=1,2,3$, let $N_{x,j}$ denote the number of trials with outcome $j$ among the first $x$ trials. Then for $x=0,1,\dots$ \begin{align} P(X>x)&=P(N_{x,1}=0\text{ or }N_{x,2}=0) \\ &=P(N_{x,1}=0)+P(N_{x,2}=0)-P(N_{x,1}=0\text{ and }N_{x,2}=0) \\ &=(q+r)^x+(p+r)^x-r^x \end{align} and similarly \begin{align} P(Y>x)&=P(N_{x,1}=0\text{ or }N_{x,2}=0\text{ or }N_{x,3}=0) \\ &=(q+r)^x+(p+r)^x+(p+q)^x-r^x-q^x-p^x+0^x, \end{align} with $0^0:=1$.

Thus, \begin{equation} EX=\sum_{x=0}^\infty P(X>x)=\frac1p+\frac1q-\frac1{p+q} \end{equation} and \begin{equation} EY=\sum_{x=0}^\infty P(Y>x)=\frac1p+\frac1q+\frac1r-\frac1{q+r}-\frac1{p+r}-\frac1{p+q}+1. \end{equation} It seem that your answer for $EY$ misses the summand $1$.

  • Thank you Iosif! I am studying your answer only now. Sorry for late reply. I'll come back to you soon! – Andrea Prunotto Nov 27 at 4:51
  • Yes. I think this solves my doubts. Thanks again! A further question: Since we can extract at least once the three colors only if we have already extracted at least once two colors, is it correct to assess $EY\geq EX+1$? – Andrea Prunotto Nov 27 at 5:09
  • 1
    @AndreaPrunotto : Concerning your latter comment: This reasoning would be incorrect in general, because one can get at least two colors of the three faster (or even much faster, depending on $p,q,r$), than both the red and blue ones. The conclusion $EY\ge EX+1$ will not hold in general either, when $p$ and $q$ are small enough as compared with $r$; e.g., compare the expressions for $EX$ and $EY$ for $p=q=r/2$. – Iosif Pinelis Nov 27 at 13:26
  • Thanks again, Iosif. I am thinking about your observation. – Andrea Prunotto Nov 27 at 16:22
  • Sure. Your example is illuminating. I still have some difficulties with some aspects of this problem, but I will post another question maybe. – Andrea Prunotto Nov 28 at 9:47

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