Pólya's Urn Long-Term Probability I am tasked with the following problem:
Let $M_n$ be the fraction of white balls in Pólya's urn after $n$ draws, where you start with one white and one black ball, and after every draw, you add another one of the color you drew to the urn. Suppose that we know that $M_{20}=\frac{7}{22}$. Prove that:
$$P\Big\{M_\infty>\frac{3}{4}\Big|M_{20}=\frac{7}{22}\Big\}<\frac{4}{9}$$
I already know that $M_\infty=\lim_{n\rightarrow\infty}M_n$, i.e. after some arbitrary time $n$, the proportion stabilizes around some fraction between $[0,1]$, and that this is uniformly distributed. I was thinking about using the Martingale Convergence Theorem, since this proportion is bounded by $1$, but I wasn't certain how to apply it to this problem, or if it is even appropriate to use in this circumstance. Any suggestions as to how to start would be sincerely appreciated. Cheers. 
 A: Try using Markov's inequality. Realistically speaking, your probability should be very, very close to $0$. But this inequality provides an (rather weak) upper bound that is just slightly less than $\frac{4}{9}$, which I believe is what you are looking for. You should also use the Optional Sampling Theorem or the Martingale Property along the way.
A: The existing answer already shows how to obtain the weak bound of $\frac49$ using Markov’s inequality, which only requires the mean $\frac7{22}$. (At least I think that’s what is meant, since the link, despite saying “Chebyshev’s inequality”, leads to Markov’s inequality, and there’s no variance calculation.) A much sharper bound can be obtained using Chebyshev inequality, which requires the variance.
Pólya’s urn is equivalent to drawing a probability $p$ from a uniform distribution on $[0,1]$ and then drawing white and black balls with each ball having independent probability $p$ to be white (see How to prove the rule of succession without calculus?). From this perspective, the $6$ white balls and $14$ black balls already drawn are data on $p$ that yield a likelihood proportional to $p^6(1-p)^{14}$; that is, conditional on $M_{20}=\frac7{22}$, $p$ has a beta distribution with $\alpha=7$ and $\beta=15$, the numbers of white and black balls in the urn after the $20$-th draw.
The mean of this beta distribution is $\frac\alpha{\alpha+\beta}=\frac7{7+15}=\frac7{22}$, and the variance is
$$
\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}=\frac{7\cdot15}{(7+15)^2(7+15+1)}=\frac{105}{11132}\;,
$$
so by Chebyshev’s inequality we have
\begin{eqnarray}
\mathsf P\left(M_\infty\gt\frac34\mid M_{20}=\frac7{22}\right)
&\le&
\mathsf P\left(\left|M_\infty-\frac7{22}\right|\gt\frac34-\frac7{22}\right)
\\
&\le&
\frac{105}{11132}\left(\frac34-\frac7{22}\right)^{-2}
\\
&=&
\frac{420}{8303}
\\
&\approx&
0.05\;.
\end{eqnarray}
Of course we can also integrate the distribution exactly to obtain
$$
\mathsf P\left(M_\infty\gt\frac34\mid M_{20}=\frac7{22}\right)=\frac{\int_\frac34^1p^6(1-p)^{14}\mathrm dp}{\int_0^1p^6(1-p)^{14}\mathrm dp}=\frac{5628239}{549755813888}\approx10^{-5}\;.
$$
