To answer your original question, indeed probabilistic algorithms are an extension of normal algorithms. One way to make the extension (and retain the property that algorithms give the same output on the same input) is to add a 'random oracle'.
[1] https://en.wikipedia.org/wiki/BPP_(complexity) [2] https://en.wikipedia.org/wiki/Church%E2%80%93Turing_thesis#V...
The output is a random variable.
Randomized algorithms are often used to avoid falling into worst-case scenarios and improve run-time, like with quicksort, or to avoid settling in local minima like with simulated annealing.
"Starting from an initial state and initial input (perhaps empty),[4] the instructions describe a computation that, when executed, proceeds through a finite[5] number of well-defined successive states, eventually producing "output"[6] and terminating at a final ending state"
The "finite" bit I guess is the key thing, that I think was what my lecturer would have been getting at - the canonical bozo sort is not - you could just run for ever, as you might not /ever/ produce a correctly sorted version of the input. That article also points to "randomized algorithms", taking some random input (you could image random input being an RNG seed I guess?).
That article does include monte carlo algorithms, which technically could not terminate (Montecarlo simulation of a bouncing photon could technically go for ever) so I am left confused.
The lecturer was Prof. Tadao Takaoka, who was great, and apparently (per google) was on the editorial board of the journal Algorithms (http://www.mdpi.org/algorithms/editors.htm)
I think for something like this, you could do asymptotic analysis of the expected run-time. Assuming all elements are distinct, there are n! permutations, so each would have a 1/n! chance of being selected each iteration. (But it's rather late, and this is a poor medium for discussing math, and someone has probably already done this, so I'm not going to go any further.)
Perhaps-Algorithm A:
step 1: x := sample a value from a roll of a fair D6
step 2: print x
Can we argue that we're in a well defined state after step 1 has executed? We have some variable x bound to one value from {1, ..., 6}, with a known probability distribution, although we don't know which integer we've got.
edit: on the other hand perhaps we're just haggling over definitions here. it is definitely the case that non-terminating algorithms exist and can be useful, and non-deterministic algorithms exist and can be useful. it might be the case that some powerful theory and analysis restricts attention to terminating, deterministic algorithms, it doesnt meant that that's all there is.
We have? I believe the probability is ~34%. http://mathworld.wolfram.com/PolyasRandomWalkConstants.html
Basically the distinction can be best illustrated with bozo/bogo sort. The definition many people use is:
1. Randomly shuffle the input
2. See if it is sorted, if not go to 1.
The problem, as explained to me by an algorithms lecturer a long time ago, is that the algorithm is not guaranteed to finish, and more over the same input may or may not result in it finishing on different runs.
Per that lecturer a “correct” bozo sort is
1. Generate a list of every permutation of the input
2. Iterate the list until you find a sorted entry and return that.
It has computable best case, worst case, and average case (all terrible). It is guaranteed that the same input will always produce the same output (terminate vs not terminate).
Obviously you could optimize this if you had an algorithm that could shuffle an input such that you were guaranteed that if you shuffled it’s own output you were guaranteed to produce every ordering of the original input. Then you would get closer to: shuffle; check; repeat. But still note that there is no random involved :)