An urn contains $n$ white and $n$ black balls. We draw them one by one without replacement. We pay $£ 1$ for any black ball drawn but receive $£ 1$ for any white one. Denote by $X_i$ our money after the $i^{\text {th }}$ draw $\left(X_0=0\right)$. Let $$ Y_i=\frac{X_i}{2 n-i} \quad(1 \leq i \leq 2 n-1), \quad \text { and } \quad Z_i=\frac{X_i^2-(2 n-i)}{(2 n-i)(2 n-i-1)} \quad(1 \leq i \leq 2 n-2) \text {. } $$ (a) Show that both $Y_i$ and $Z_i$ are martingales.
Could I get some advice for trying to find the probability of X. I know the expectation of X is 1p - 1q where p+q =1 but I am struggling to quantify the probability of X as since the balls aren't being replaced. I think it might be a function along the lines of $\frac{k}{2n-i}$
We have of course to specify a filtration on the set $\Omega$ of all possible one-by-one extractions of the given white and black balls. The number $n$ is fixed. The chain of extraction has $N=2n$ places. We extract the balls in order, and at the end of the time line, in $2n$ we have a chain $c=(c_1,c_2,\dots,c_N)$. The components are white (W) or black (B), but instead of $W$ and $B$ i prefer $1$ for white and $0$ for black. We may also want to write $c$ as a word with $N$ letters, $c=c_1c_2\dotsc_N$. Seen as an indicator function pattern, $c$ is determined by and determines the set of the $1$-places, let us denote it by $w(c)$ a set of $n$ places among $1,2,\dots,N$. So $\Omega$, the modelling space, has $\binom Nn=\frac{N!}{n!n!}=\frac{(2n)!}{n!n!}$ elements.
Which is the filtration? The time line is $T=\{0,1,2,\dots ,N\}$.
At step $k$ in $T$ we have a $\sigma$-algebra $\mathcal F_k$ on $\Omega$ covering all information available at time $k$. For $k=0$ we have the trivial $\sigma$-algebra $\mathcal F_0=\{\emptyset,\Omega\}$. At point one, we "see" $c_1$, so $F_1$ has two atoms, $A(0*)$ and $A(1*)$, corresponding to all words that start with $0$, the first atom, and respectively all words that start with $1$, the second atom. So $\mathcal F_1=\{\emptyset, 0*, 1*,\Omega=*=0*\sqcup1*\}$, it has four elements. At time $k$ we have $\mathcal F_k$, the $\sigma$-algebra with atoms $A(c'*)$. Here the atom $A(c'*)$ is the set of words starting with a word $c'=c_1c_2\dots c_k$ of length $k$, which does not contain more than $n$ zero entries, and no more than $n$ one entries. I will omit now the star in the notation, use only $A(c')$ intead.
Let us show now that $Y$ is a martingale. This means, that passing from step $k$ to step $k+1$ is a "fair" step. To see this, we may and do assume the information in $k$ is the one corresponding to an atom $A(c')$, $c'=c_1c_2\dots c_k$. Let $w$ be the number of ones, $b$ the number of zeros in $c'$. Then on the atom $A(c')$ we have a constant value $Y(k) = \frac 1{N-k}X(k)=\frac{w-b}{N-k}$.
Constrained to $A(c')$, which is the probability for $c_{k+1}=1$, and which is the probability for $c_{k+1}=0$? There are $(n-w)$ white balls, and $(n-b)$ black balls still in the urn. In total, there are $(n-w)+(n-b)=2n-w-b=N-k$ balls still available. The probability to get a further white ball is $(n-w)/(N-k)$. The probability to get a further black ball is $(n-b)/(N-k)$. So we compute: $$ \begin{aligned} &\Bbb E[\ Y(k+1)\ |\ A(c')\ ] \\[2mm] &\qquad= \Bbb P [\ c_{k+1}=1\ |\ A(c')\ ]\cdot Y(k+1)(c'1) \\ &\qquad+ \Bbb P [\ c_{k+1}=0\ |\ A(c')\ ]\cdot Y(k+1)(c'0) \\[2mm] &\qquad= \frac{n-w}{N-k}\cdot\frac{X(k+1)(c'1)}{N-(k+1)} + \frac{n-b}{N-k}\cdot\frac{X(k+1)(c'0)}{N-(k+1)} \\[2mm] &\qquad= \frac{n-w}{N-k}\cdot\frac{w+1-b}{N-(k+1)} + \frac{n-b}{N-k}\cdot\frac{w-1-b}{N-(k+1)} \\[2mm] & \qquad=\frac 1{(N-k)(N-k-1)} \underbrace{\Big((n-w)(w-b+1)+(n-b)(w-b-1)\Big)} _{(n-w+n-b)(w-b)+(n-w)-(n-b)} \\[2mm] & \qquad=\frac {(N-k)-1}{(N-k)(N-k-1)}(w-b) =\frac{w-b}{N-k} \\[2mm] & \qquad=\Bbb E[\ Y(k)\ | \ A(c')\ ]\ . \end{aligned} $$ So $Y$ is a martingale.
A similar computation can be done for $Z$. Constrained to $A(c')$ we have $Z(k)=\frac{X(k)^2 - (N-k)}{(N-k)(N-k-1)} =\frac{(w-b)^2 - (N-k)}{(N-k)(N-k-1)}$. From $c'$ the world can see only $c_{k+1}=1$ or $c_{k+1}=0$ with the same probabilities as above, they determine $Z(k+1)$, and the win is from the point of view of the information in $A(c')$ equal to... $$ \begin{aligned} &\Bbb E[\ Z(k+1)\ |\ A(c')\ ] \\[2mm] &\qquad= \Bbb P [\ c_{k+1}=1\ |\ A(c')\ ]\cdot Z(k+1)(c'1) \\ &\qquad+ \Bbb P [\ c_{k+1}=0\ |\ A(c')\ ]\cdot Z(k+1)(c'0) \\[2mm] &\qquad= \frac{n-w}{N-k}\cdot\frac{X(k+1)(c'1)^2-(N-(k+1))}{(N-(k+1))(N-(k+1)-1)} + \frac{n-b}{N-k}\cdot\frac{X(k+1)(c'0)^2-(N-(k+1))}{(N-(k+1))(N-(k+1)-1)} \\[2mm] &\qquad= \frac{n-w}{N-k}\cdot\frac{(w+1-b)^2-(N-k-1)}{(N-k-1)(N-k-2)} + \frac{n-b}{N-k}\cdot\frac{(w-1-b)^2-(N-k-1)}{(N-k-1)(N-k-2)} \\[2mm] & \qquad=\frac {(n-w)(w-b\color{red}{+}1)^2+(n-b)(w-b\color{red}{-}1)^2-(n-w+n-b)(N-k-1)} {(N-k)(N-k-1)(N-k-2)} \\[2mm] & \qquad =\frac {\scriptstyle(n-w+n-b)(w-b)^2 \color{red}{+ 2(n-w)(w-b)-2(n-w)(w-b)}+(n-w)+(n-b) -(N-k)(N-k-1)} {(N-k)(N-k-1)(N-k-2)} \\[2mm] &\qquad =\frac {(N-k)(w-b)^2 \color{red}{- 2(w-b)^2} + (N-k)-(N-k)(N-k-1)} {(N-k)(N-k-1)(N-k-2)} \\[2mm] &\qquad =\frac {(N-k-2)(w-b)^2-(N-k)(N-k-2)} {(N-k)(N-k-1)(N-k-2)} \\[2mm] &\qquad =\frac {(w-b)^2-(N-k)} {(N-k)(N-k-1)} \\[2mm] & \qquad=\Bbb E[\ Z(k)\ | \ A(c')\ ]\ . \end{aligned} $$ So $Z$ is a martingale, too.