By how much does the 50/50 Joker option increase the probability of getting the correct answer? I am trying to determine by exactly how much the probability of getting the correct answer increases if we chose the 50/50 Joker option in "Who wants to be a millionare?", as compared to taking an entirely blind guess at the question.
Put formally, I (the participant) have four options, and I have zero knowledge about the question or the options given. A blind guess has a 1/4th chance of getting it correct. What are my chances of getting it correct if two incorrect options are eliminated? Is it 1/2 or higher?
I came up with the following three disjoint subsets and probabilities of winning in each:


*

*I (a) chose the incorrect option and (b) it does not get wiped out in 50/50, then (c) I will switch with 100% probability of winning $=3/4\cdot1/3\cdot1$

*I (a) chose the incorrect option and (b) it does get wiped out in 50/50, then (c) I will chose the correct answer from the remaining two with 50% probability of winning $=3/4\cdot2/3\cdot1/2$

*I (a) chose the correct option and (b) it does not get wiped out in 50/50, then (c) I retain my answer with certainty of winning $=1/4\cdot1\cdot1$
Thus the probability of winning if you use 50/50 seems to be  $=1/4+1/4+1/4=3/4$ and not $1/2$ as earlier suggested.
However, if we try to calculate probability of losing using a similar approach, we land up in trouble. Take these disjoint subsets: 


*

*I (a) chose the incorrect option and (b) it does not get wiped out in 50/50, then (c) I will not switch with 100% probability of losing $=3/4\cdot1/3\cdot1$

*I (a) chose the incorrect option and (b) it does get wiped out in 50/50, then (c) I will chose the incorrect answer from the remaining two with 50% probability of winning $=3/4\cdot2/3\cdot1/2$

*I (a) chose the correct option and (b) it does not get wiped out in 50/50, then (c) I switch my answer with certainty of losing $=1/4\cdot1\cdot1$
The sum of these is also $3/4$! Therefore, the probability of winning + probability of losing appears to be $3/4+3/4=1.5$ which certainly cannot be true since they are disjoint events for the given situation.
Where then have I gone wrong in my logic exactly?

Clarifications:


*

*Do you choose an option before the two are eliminated? - yes

*Can they eliminate the one you have chosen? - yes, this is the case 2 in both lists above

*How are the two that are eliminated chosen? - randomly and uniformly. There are three incorrect options. So, you can form $^3C_2 = 3$ groups of size two. Thus, each group has a probability of $1/3$ of getting wiped out.

 A: You haven't included the probabilities for choosing whether to swap your answer in cases 1 or 3. Either you have already decided, and one will have probability 1 and the other 0. Or you toss a coin when given the choice (either physically or mentally, fair or unfair coin). You can even use probability $p$ for swapping and $(1-p)$ for not swapping, if you haven't decided how you will decide yet.
Once you include that, the numbers should add up.
A: If the options to be eliminated do not depend on your original choice the two remaining options are equally likely.  You might as well not have chosen before.  Now you have $\frac 12$ of getting the right one.
A: How do you know whether to switch or not?
In the first calculations you assuming that you can magically switch to or stay on the correct answer if your first choice was not eliminated. But how would you know whether your answer is correct, i.e. whether you were in case 1 or case 3? 
In the second set of calculations you similarly assume that you can magically switch to or stay on the incorrect answer if your first choice was not eliminated.
In those two sets of calculations you follow two different (magical) strategies, so it is not surprising that their resulting probabilities do not add up to 1.
To get the correct answer, first you must determine what your strategy will be. If your answer is not eliminated, will you always switch? Or will you always stay? Or will you switch with a certain fixed probability, like 50/50? For each of those strategies you can calculate the probability of winning or losing. Your strategy cannot be based on knowledge of whether your first guess is correct, because you do not have such knowledge when playing the game.
