I am assembling a jigsaw puzzle. I have two candidate pieces A and B (among many others) to try and fit onto a corner of the partially solved puzzle. A and B are different kinds of pieces. There are two potential ways to fit B onto the corner and one potential way to fit A onto the corner.
Are A and B equally likely to fit?
It is twice as likely that $B$ will be the piece that fits there than $A$ ... but that does not mean that putting $B$ there as your next move is more likely to get you to solve the whole puzzle than putting in piece $A$: if you put $B$ in with the wrong orientation, you'll have to redo everything just as much as when $A$ turns out to be the wrong piece. Put differently: you have $3$ possible ways to proceed, each of which is equally likely to be the correct one.
To prove that trying piece $B$ as your next piece does not provide you with advantage over piece $A$, let's make the following assumptions:
If you correctly place your next piece (in the corner), it'll take $n$ moves to finish the puzzle.
If you incorrectly place your next piece, you'll end up wasting $m$ moves. Given that we don't know anything else about the pieces, we can assume this is the same for any of the wrong moves.
OK, so let's compute the expected number of moves to finish the puzzle when using piece $B$ as your next piece and compare this to the expected number of moves to finish the puzzle when using piece $B$ as your next piece. That is, let's compare $E(B)$ with $E(A)$
Now, again, there are three moves you can make: one move involving piece $A$, and two involving piece $B$, and again, since we don't know anything else about the rest of the pieces, we can assume that any of these three moves is equally likely to be correct, i.e. each has a chance of $\frac{1}{3}$ to be correct.
Let's first consider placing piece $A$ as our next move. There is a chance of $\frac{1}{3}$ for this to be the correct next move, in which case we end up making $n$ moves to complete the puzzle.
If, on the other hand, this turns out to be the incorrect move (and therefore incorrect piece), we have to try piece $B$. Now, there is a $\frac{2}{3}$ chance that piece $A$ was indeed the incorrect piece to use. However, just because we have mnow found thje ocrrect piece to use,m doesn;t mean that we know what is the correct move to make. Indeed, there are two possible moves we can make with $B$. There is a chance of $\frac{1}{2}$ that we make the correct move, and a chance of $\frac{1}{2}$ that we make the incorrect move. So if, after we wasted $m$ moves on trying piece $A$, we make the correct move with piece $B$, we end up finishing the puzzle in $m+n$ moves. However, if we make the incorrect move with piece $B$, we end up finishing the puzzle in $2m+n$ moves. In sum:
$E(A) = \frac{1}{3} \cdot n + \frac{2}{3} (\frac{1}{2} \cdot (m+n) + \frac{1}{2} \cdot (2m+n)$
$ = \frac{1}{3} \cdot n + \frac{1}{3} \cdot (m+n) +\frac{1}{3} \cdot (2m+n)$
$ = \frac{1}{3} \cdot n + \frac{1}{3} \cdot m +\frac{1}{3} \cdot n + \frac{1}{3} \cdot 2m+\frac{1}{3} \cdot n$
$ = n + m$
OK, now suppose we immediately try piece $B$. After all, piece $B$ is more likely to be the correct piece to use. However, even if $B$ is the correct piece, we could still make the incorrect move. So, let's see what happens:
First, piece $B$ is the correct piece with a probability of $\frac{2}{3}$. Even if $B$ is correct, however, we can immediately make the correct move with piece $B$ with a chance of $\frac{1}{2}$ (leading to $n$ moves to complete the puzzle), but there is also a chance of $\frac{1}{2}$ that we first use piece $B$ incorrectly (but, since $B$ was correct, that means that the next move with $B$ is correct, and thus we end up making $m+n$ moves for completion). Finally, there is of course the $\frac{1}{3}$ chance that $B$ is not correct, and that $A$ is correct. In that case we waste $2m$ moves trying piece $B$ borh ways, and thus we finish the puzzle in $2m_n$ moves. In sum:
$E(B) = \frac{2}{3} (\frac{1}{2} \cdot n + \frac{1}{2} \cdot (m+n)) + \frac{1}{3} \cdot (2m+n) $
$ = \frac{1}{3} \cdot n + \frac{1}{3} \cdot (m+n) +\frac{1}{3} \cdot (2m+n)$
$ = \frac{1}{3} \cdot n + \frac{1}{3} \cdot m +\frac{1}{3} \cdot n + \frac{1}{3} \cdot 2m+\frac{1}{3} \cdot n$
$ = n + m$
So, it doesn't matter which piece you try next!