Transfinite Guessing Game Presented here is a description of a game.
The game consists of two players: the Player and the GameController.
The object of the game is for the Player to guess a real number called the Number, that the GameController has somehow chosen. If at any point in the game the Player guesses that real number (the Number), the Player is declared the winner.
You decide to play the game.
Before the start of the game the GameController gives you the following details on how to play: 
This game is played with the help of three "transfinite machines" called, Generator1, Generator2, and Comparator1. The machine Generator1 is used by me to produce the Number which you will try to guess. The Number will be a random real between 0 and 1 and stay the same for the duration of each 
game played. The Number comes directly from transfinite machine Generator1 and, so far at least, I have never seen the machine produce a rational number in this range, although the manufacturer claims the machine is "capable" of producing any real number.
Generator2 is for your use throughout the game. Generator2 will produce a random real number when given an input range to select from. Generator2 is guaranteed to be "capable" of producing the Number I have selected.
You are not required to use Generator2 to produce a guess, any well defined real number you can describe in a reasonable amount of time I will accept as a guess, or you can use a combination of Generator2 and your own intelligence to produce the guesses. I would, however, strongly recommend you do not "go it all alone" since these transfinite generators are capable of producing what your mathematicians like to call "uncomputable numbers", and the Number I have might be one of them.
I have set up both Generator1 and Generator2 so that the real numbers are represented in binary form such as .0111010010001001..., so effectually you are trying to guess the infinite binary sequence that Generator1 has produced and is known by me but not you.
When you have a guess ready I will put that guess into transfinite machine Comparator1 which will quickly give an answer as to whether your guess and my Number match. If the numbers do not match I will tell you whether the Number is higher or lower than your guess, and you may guess again. If the numbers match, I will declare you the winner.
Question: 
If a player or players play the game repeatedly how often will they win the game assuming an unlimited number of guesses can be made for each game played, and optimal or near optimal play under the rules is achieved:
a) never
b) some of the time
c) every time
?
Now suppose the GameController modifies the game as follows:
You can play the game as before plus I have just enabled Generator2 so it can also produce a random countably infinite list of real numbers within the range you provide. So now you may use a generated list for any of your guesses. Comparator1 can quickly process the entire list and if a match
with the Number is found anywhere on the infinite list I will declare you the winner. If the Number is not on the list I will tell you to try again.
Similar to before, you can use your own scheme in conjunction with Generator2's list for the final list for Comparator1 to scan. If you choose to scheme up a list, it must be constructed with a "well defined simple process" or Comparator1 may reject it. I will let you know if that happens.  
How does the answer to the above question change ?
What is a good strategy to use for either variant of the game ?
 A: How many guesses is the player allowed?  The title of transfinite implies that the player is allowed any ordinal number of guesses, not just a finite number.  With each guess the player can determine one bit of the real.  The first guess should be $\frac 12$.  If this doesn't win, the player knows the first bit of the binary expansion.  Keep going.  Each guess gets the player another bit.  As long as the player is only allowed finitely many guesses the player can't know the number.  With $n$ guesses the player knows it within a range of $2^{-n}$.  But if the player gets $\aleph_0$ guesses it is known.
A: The machine picks some $a\in [0,1]$. We need to guess this number. If we were to pick any random value then the probability of it being $a$ is zero (assuming the machine uses a uniform distribution). Now if we are given a randomly chosen countable list of numbers $A \subset [0,1]$, the probability of $a$ being in $A$ is $$P(a\in A) = P(a=a_1) + \cdots + P(a=a_n) +\cdots$$ where $A = \{a_j\}_{j=0}^{\infty}$. By what we have said before, this is a sum of zeros and so is zero. This means that the probability of winning after one guess is zero. Notice, if we press the button on the machine and get another countable set $B$, this still doesn't help because we are in the same situation as before just with $A\cup B$ instead of $A$.
EDIT: above we have assumed the the comparator returns a YES/NO answer (in the second game). As commented by Xoff, if the comparator returns a list of numbers below the solution $a$, then let's assume we input a countable dense subset and the comparator returns the set $B$ of numbers below the solution. Our next guess can be $b = \sup B$. We have that $a$ is an upper bound for $B$ by definition of $B$ (so $a\geqq b$). Also, for each $n\in\mathbb N$ we can find a $b_n \in B$ such that $a-b_n< \frac{1}{n}$ and for all $c<a$ there exists a $b^\prime\in B$ such that $b^\prime > c$. This means $a=b$.
So if we make the additional assumption that the comparator returns a list of numbers below the solution, then it is possible. And not only is it possible, you can always guess the solution in two turns. Simply guess your favourite countable dense subset, it'll be wrong with probability 1, but then you can then input the correct answer by considering the supremum over the set that the comparator returns.
A: The two machines are guaranteed to be mutually "capable" of producing the same number thru a transfinite random selection process.
Multiple wins by a set of players do not conflict with Cantor's uncountability proof.
So the obvious answers are: 
Game version 1
c) every time
Game version 2
c) every time
The simplest strategy would probably be to let Generator2  produce all the numbers.
