Is the following coin-toss problem analogous to Monty-Hall? Suppose you flip a coin three times. What's the probability that all three flips result in the same outcome, i.e., all three heads or all three tails? There seem to be two solutions...
There are the obvious eight possible sequences: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT, and two of those eight satisfy our requirement. So the obvious answer is $\mbox{probability}=\frac14$.
But alternatively, if you flip a coin three times, then two of the three outcomes must be the same, i.e., each of the eight sequences enumerated above either have two heads or two tails. So then there's a $\ 50-50\ $ chance that the third flip will be the same as those two, whereby $\mbox{probability}=\frac12$.
So, I think $\mbox{probability}=\frac14$ is the obviously correct answer, but I'm not immediately seeing what's wrong with that $\frac12$ argument. What's wrong with it (seems to be a sets versus sequences kind of distinction)? And is it somehow analogous to the Monty-Hall problem?
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First, thanks to @QiaochuYuan for his comment/answer. And I think his remarks suggest that, yes, it's a bit analogous to Monty-Hall in the following way...
Suppose a coin is flipped three times, but you don't see the results. And now you're asked to choose one of those flips and guess it's outcome. So you have a 50-50 chance of guessing correctly. But if you're now told that the other two flips were both H, then you should change your guess to T (if it wasn't T already), and you'll now have a 75-25 chance of being correct. And conversely if the other two flips were both T. So that's not quite the same as Monty-Hall, but maybe bears somewhat of a resemblance.
 A: 
Suppose a coin is flipped three times, but you don't see the results. And now you're asked to choose one of those flips and guess it's outcome. So you have a 50-50 chance of guessing correctly. But if you're now told that the other two flips were both H, then you should change your guess to T (if it wasn't T already), and you'll now have a 75-25 chance of being correct. And conversely if the other two flips were both T. So that's not quite the same as Monty-Hall, but maybe bears somewhat of a resemblance.

This does not follow! Let's say I chose the first coin, so that I'm being told that the second and third coins flipped H. If you just look at the $8$ possible sequences, this means there are two possibilities, THH and HHH, and each of those occurs with probability $\frac{1}{2}$. Formally, assuming that the coin flips are independent, conditioning on the second and third coin flips gives me no information about the first coin flip. The same would be true if I picked the second or third coin by symmetry.
The difference between this situation and Monty Hall is that in Monty Hall there's no independence: the prize is behind exactly one door, so every time I learn the prize isn't behind a door it's now more likely to be behind another door.
The difference between this situation and your original question is that in the $\frac{1}{4}$ argument you were asking about, you chose to find two of the flips that are the same. In this guessing game neither you nor the "game show host" is making such a choice. You are making a choice of a single coin based on no information and then there are some possible worlds in which the remaining coins are the same and some other possible worlds in which the remaining coins are different, and you learn that you're in the first set of possible worlds.
In other words, there's a big difference between selecting some of the coins because they are the same, and conditioning on the event that some of the coins happen to be the same.
