3 added proof that the probabilities form a dense set

EDIT to add some details:Given a universal TM $M$ with tape alphabet $A$, and given a subinterval of $[0,1]$, choose an integer $n$ so large that your given interval includes one of the form $[k/|A|^n,(k+1)/|A|^n]$. Let $S$ be a set of $k$ words of length $n$ over $A$, and let $w$ be another such word that is not in $S$. Modify $M$ to $M'$ that works as follows. If the first $n$ symbols on the tape are a word from $S$, then march to the right forever, ignoring everything else. If they are the word $w$, then simulate $M$ on the remainder of the tape (the part after $w$), moving any final answer into the right location, as in my previous edit. Finally, if the word consisting of the tape's first $n$ letters is neither in $w$ nor in $S$, then halt immediately. Then the probability that $M'$ moves infinitely to the right will be at least $k/|A|^n$ (the probability that the initial $n$-word on the tape is in $S$) and at most $(k+1)/|A|^n$ (the probability that this $n$-word is either $w$ or in $S$) and therefore within the originally given interval.

2 added 1209 characters in body

EDIT to take into account the revision of the question:Given a universal TM, you can make trivial modifications that maintain universality but change the probability $p$ of going infinitely far to the right. For example, modify your original machine $M$ to an $M'$ that works like this: If the first symbol $x$ on the program tape is 0, then halt immediately; otherwise, move one step to the right and work like $M$ on the program minus the initial symbol $x$ (and, just to guarantee universality, if the computation halts, go back to $x$, erase it, and move $M$'s answer one step to the left so that it's located where answers should be). That modification decreases the probability $p$. You can increase $p$ by having an initial 0 in the program trigger a race to the right by $M'$ --- it just keeps marching to the right regardless of what symbols it sees. You can achieve some control over the amount by which $p$ increases or decreases by having the modification $M'$ begin by checking more than just one symbol at the beginning of the program. As far as I can tell, such modifications, carried out with enough care (which I don't have time for just now) should give you a dense set of $p$'s.

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As far as I can see, if you consider a single TM, then you get only one specific probability, not a dense set, whereas if you let the TM vary then $\Omega$ will vary also, and the set of probabilities will contain all rational numbers in $[0,1]$ (and some other numbers too). If you fix the number of symbols but let the TM vary, it's not so clear that you'll get all the rationals, but you'll still get a dense set.