What is the distribution of choosing kids from a class?

Question

Given a class of $m$ kids, $\frac{m}{2}$ boys and $\frac{m}{2}$ girls. Their teacher Erica randomly choosing kids from her class one by one. Define success in the experiment when Erica chose at least one boy and at least one girl. Suppose $Y$ is the number of kids that Erica chose until she was successful in the experiment. What is the distribution of $Y$?

My attempt: Divide the class into two groups, $A$ of boys and $B$ of girls. Erica chooses one child. Suppose, without loss of generality, that Erica chose a boy from group A. Because Erica chose $Y$ kids in total, she has another $Y-2$ kids to choose from group $A$, and one kid from group $B$. Therefore: $$P(Y=i)=(0.5)^{i-2} * 0.5 = 0.5^{i-1} , i >1 $$

I have a feeling that I'm wrong, but I can't think about another solution to this problem.

Edit: @Brian M. Scott has been brought my attention to the fact that the solution I have proposed relates to the case where kids beign chosen with replacement. Therefore, because creative solutions were provided for both cases, I separate the original question into two questions, the first with replacement, and the second without replacement.

Answer 1

For convenience let $m=2n$, so that we have $n$ girls and $n$ boys.

Clearly $P(Y=0)=P(Y=1)=0$. Assuming that $n\ge 2$ there are $n(n-1)$ ways to begin by choosing $2$ girls, $n(n-1)$ ways to begin by choosing $2$ boys, $n^2$ ways to begin by choosing a girl and then a boy, and $n^2$ ways to begin by choosing a boy and then a girl, so

$$\begin{align*} P(Y=2)&=\frac{2n^2}{2n^2+2n(n-1)}\\ &=\frac{n}{2n-1}\\ &=\frac12+\frac1{4n-2}\;, \end{align*}$$

a bit more than $\frac12$. (If $n=1$, of course, $P(Y=2)=1$.)

Now suppose that $n\ge 3$. The successful sequences are $GGB$ and $BBG$, each of which can be made in $n^2(n-1)$ ways. Each of the other $4$ sequences with two children of one sex and one of the other can also be made in $n^2(n-1)$ ways, and each of the sequences $GGG$ and $BBB$ can be made in $n(n-1)(n-2)$, so

$$\begin{align*} P(Y=3)&=\frac{2n^2(n-1)}{6n^2(n-1)+2n(n-1)(n-2)}\\ &=\frac{n}{4n-2}\\ &=\frac14+\frac1{8n-4}\;. \end{align*}$$

Let’s look at one more: suppose that $n\ge 4$. A similar analysis shows that

$$\begin{align*} P(Y=4)&=\frac{2n^2(n-1)(n-2)}{8n^2(n-1)(n-2)+6n^2(n-1)^2+2n(n-1)(n-2)(n-3)}\\ &=\frac{n(n-2)}{4n(n-2)+3n(n-1)+(n-2)(n-3)}\\ &=\frac{n^2-2n}{8n^2-16n+6}\\ &=\frac18-\frac3{8(4n^2-8n+3)}\;, \end{align*}$$

which pretty well puts paid to any hope of finding a simple pattern for the simplified expression.¹

It is, of course, possible to write an expression for the original fraction: assuming that $n\ge k$,

$$P(Y=k)=\frac{2nn^{\underline{k-1}}}{\sum_{i=0}^k\binom{k}in^{\underline i}n^{\underline{k-i}}}\;,$$

where $x^{\underline k}$ is the falling factorial,

$$x^{\underline k}=x(x-1)(x-2)\ldots(x-k+1)=\prod_{i=0}^{k-1}(x-i)\;.$$

¹ As a quick check I evaluated these at $n=4$; they sum to $\frac{34}{35}$, which is correct, since in this case

$$P(Y=5)=\frac{2\cdot4!\cdot4}{8\cdot7\cdot6\cdot5\cdot4}=\frac1{35}\;.$$

Answer 2

Let $N$ be the random variable for the number of kids, such that its a success, we can write the distribution of $N$ as follows: $$P(\text{atleast one boy and one girl}) = 1- P(\text{picking all girls}) - P(\text{picking all boys})$$ $$P(N=n) = 1 - {n\choose 0,n}(0.5)^{0}(0.5)^{n}- {n\choose 0,n}(0.5)^{n}(0.5)^{0}$$ $$P(N=n) = 1 - 2(0.5)^n$$

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Edited the answer to my new attempt $$P(N=2) = 1 - 2\frac{1}{4}$$ $$ = \frac{1}{2}$$ Which is as expected. Now the probability of picking all girls or all boys will indeed tend to zero. Note: ${n \choose i,j} = {n \choose r} $, where $i$ = $n-r$ and $j = r$