February 7th
Today I learned that random play in two-pile Nim give each player a $\frac12$ probability of winning if one pile has at least $2$ stones. Formally speaking, "random play'' means that a player on her turn lists all possible moves and then makes a random move. This is different from picking a legal pile randomly and then taking a random number of stones, but I don't think the result changes.
We prove this by brute-force tabulation. Let $p(m,n)$ be the probability that player $1$ if the two piles have $m$ and $n$ stones. By convention, we set $p(0,0)=0$ because player $1$ is the first player to be unable to make a move. From here, we note that we have the recurrence\[p(m,n)=1-\frac1{m+n}\left(\sum_{k=1}^mp(m-k,n)+\sum_{\ell=1}^np(m,n-\ell)\right)\]for $m,n,m+n\ge0.$ To prove this, we note that player $1$ must make a move, which consists of either taking $k\le m$ stones from the first pile or $\ell\le n$ stones from the second pile. Each of these moves occur with probability $\frac1{m+n},$ and afterwards, player $1$ is effectively playing as player $2$ in the $p(m-k,n)$ or $p(m,n-\ell)$ game, so we subtract the probabilities from $1.$
Quickly, we simplify our recurrence to\[p(m,n)=1-\frac1{m+n}\left(\sum_{k=0}^{m-1}p(k,n)+\sum_{\ell=0}^{n-1}p(m,\ell)\right)\]by flipping the sums. As some motivation, we remark that we could use the recurrence relation to build a table of the various values of $p(m,n).$ New values are computed by one minus the average of all terms above or to the left. Here is the start of the table.\[\begin{array}{r|c|c|c|c} & 0 & 1 & 2 & \cdots \\\hline 0 & 0 & 1 & 1/2 & \cdots \\\hline 1 & 1 & 0 & 1/2 & \cdots \\\hline 2 & 1/2 & 1/2 & 1/2 & \cdots \\\hline \vdots & \vdots & \vdots & \vdots & \ddots\end{array}\]However, the fast way to finish the proof is to claim directly that $p(m,n)$ is\[p(m,n)=\begin{cases} 0 & (m,n)\in\{(0,0),(1,1)\}, \\ 1 & (m,n)\in\{(0,1),(1,0)\}, \\ 1/2 & \text{else}.\end{cases}\]Note that this is the main claim. To prove, we see this matches our initial condition $p(0,0)=0,$ so it suffices to show that we satisfy the recurrence relation. We quickly check that $p(0,1)=p(1,0)=1-0$ and that $p(1,1)=1-\frac12(1+1)=0.$ It follows that, for $m\ge2,$ we see\[\sum_{k=0}^{m-1}p(k,n)=p(0,n)+p(1,n)+\sum_{k=2}^{m-1}p(k,m)=1+\frac{m-2}2=\frac m2.\]Here the trick is that $p(0,n)+p(1,n)$ is one of $1+0,\,0+1,\,\frac12+\frac12,$ all of which are equal to $1.$ Similarly, for $n\ge2,$ we see\[\sum_{\ell=0}^{n-1}p(m,\ell)=p(m,0)+p(m,1)+\sum_{\ell=2}^{n-1}p(m,\ell)=1+\frac{n-2}2=\frac n2.\]Again, the trick is that $p(0,m)+p(1,m)$ always evaluates to $1.$ It follows that\[1-\frac1{m+n}\left(\sum_{k=0}^{m-1}p(k,n)+\sum_{\ell=0}^{m-1}p(m,\ell)\right)=1-\frac1{m+n}\cdot\frac{m+n}2=\frac12,\]which completes the proof.
Quickly, we talk about the alternate method of "random play'' mentioned at the beginning: a player picks a random pile and then picks up a random number of stones. Let the probability of player $1$ winning in this game with $m$ and $n$ stones be $q(m,n).$ We still have $q(0,0)=0,$ and a similar computation as before gives the recurrence relation\[q(m,n)=1-\frac12\left(\frac1m\sum_{k=0}^{m-1}q(k,n)+\frac1n\sum_{\ell=0}^{n-1}q(m,\ell)\right)\]for $m,n \gt 0,$ and\[q(m,0)=1-\frac1m\sum_{k=0}^{m-1}q(k,0),\qquad q(0,n)=1-\frac1n\sum_{\ell=0}^{n-1}q(0,\ell).\]With these in hand, it's not difficult to actually show $q=p$ always. Indeed, our recurrence relations matches $p$ for the $m=0$ and $n=0$ cases, so we match there. We can also check that $q(1,1)=1-\frac12(1+1)=0$ still. Then for $m,n \gt 0,$ the internal sums will match what we evaluates for $p$ in the above proof, so we see\[1-\frac12\left(\frac1m\cdot\frac m2+\frac1n\cdot\frac n2\right)=\frac12.\]This finishes the proof that $p$ satisfies the recurrence relation of $q,$ so they are equal.