I wrote a PARI program to compute the number of RPN strings of each length, and the probabilities of an operator at each position. I assumed that there are only two characters, one of which is an operator. If instead we use j operator characters and k other characters, this multiplies the numbers of strings of length $2n-1$ by $k^nj^{n-1}$ but doesn't change the probabilities.

The number of strings of each length turns out to be the Catalan numbers (see http://oeis.org/A000108.)

Here's my code, and the first few lines of its output:

LIMIT = 100;

count = vector(LIMIT); \count[i] = number of reverse polish strings of length 2i - 1

ops = vector(LIMIT); \ops[i] is a vector v of length 2i - 1 such that v[j] gives the number of strings with an operator in the jth position
count[1] = 1;

ops[1] = [0];

{

for (i = 2, LIMIT,

```
ops[i] = vector(2*i - 1);
\\A string of length 2i - 1 is (arg1)(arg2)op. Let 2k-1 be the length of arg1
for (k = 1, i - 1,
\\Number of strings of length 2i-1 in which arg1 has length 2k-1 is
```

count[k]*count[i - k]

```
c = count[k]*count[i - k];
count[i] += c;
for (j = 1, 2*k - 1,
ops[i][j] += ops[k][j]*count[i - k]
);
for (j = 2*k, 2*i - 2,
ops[i][j] += ops[i - k][j - (2*k - 1)]*count[k]
)
);
ops[i][2*i - 1] = count[i];
write("rpn.txt", "n = ", 2*i - 1, " number of strings = ", count[i], " prob of operator = ", ops[i]/count[i])
```

);

}

n = 3 number of strings = 1 prob of operator = [0, 0, 1]

n = 5 number of strings = 2 prob of operator = [0, 0, 1/2, 1/2, 1]

n = 7 number of strings = 5 prob of operator = [0, 0, 2/5, 2/5, 3/5, 3/5, 1]

n = 9 number of strings = 14 prob of operator = [0, 0, 5/14, 5/14, 1/2, 1/2, 9/14, 9/14, 1]

n = 11 number of strings = 42 prob of operator = [0, 0, 1/3, 1/3, 19/42, 19/42, 23/42, 23/42, 2/3, 2/3, 1]

n = 13 number of strings = 132 prob of operator = [0, 0, 7/22, 7/22, 14/33, 14/33, 1/2, 1/2, 19/33, 19/33, 15/22, 15/22, 1]