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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?

----- Edit -----
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.

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    $\begingroup$ There isn't a 50-50 chance that the third flip will be the same as the two that are the same. Just go through the casework and you'll see that isn't true. Conceptually the point is that you didn't select the two that are the same by chance; you selected them because they're the same, so the remaining flip is more likely to be different. $\endgroup$ Sep 18, 2022 at 7:36
  • $\begingroup$ @QiaochuYuan I guess you must somehow be right since 50-50 is pretty obviously wrong. But it sounds like you're saying the way I epistemically look at the flips somehow affects their ontological outcome probabilities... or something like that, if I'm saying it clearly. And that doesn't seem right. $\endgroup$ Sep 18, 2022 at 7:42
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    $\begingroup$ All I'm saying is that if you decide to look at a specific set of coins because they're more likely to be heads, the remaining coins are more likely to be tails, and vice versa. I'm not saying that the first coin, or the second coin, or the third coin, is in any way affected by your choices. I am saying that if you choose to distinguish some of those coins via some property, the behavior of the remaining coins is affected, because you decided which ones "the remaining coins" are. Does that make sense? $\endgroup$ Sep 18, 2022 at 8:10
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    $\begingroup$ An analogous situation with bigger numbers might make the point clearer: suppose I flip $100$ coins, then direct you to pick out a subset of $99$ of them containing as many heads as possible. What's the probability that the remaining coin is heads? The answer is that the remaining coin is guaranteed to be tails unless I flipped a hundred heads, so the probability is $\frac{1}{2^{100}}$, and certainly not $\frac{1}{2}$. $\endgroup$ Sep 18, 2022 at 8:13
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    $\begingroup$ In case you haven't made the connection in relation to your edit, if the person who knows the flips doesn't make any choices, then you cannot change the probability from $1/2$. If the person who knows the flips makes some choices, then the implementation of that knowledge can affect the probability from the perspective of the one with limited information. As you have currently written the scenario in the edit, the probability would still be $1/2$. $\endgroup$ Sep 18, 2022 at 8:20

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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.

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  • $\begingroup$ Thanks again. Yeah, as per your answer and Brian's comment, I guess I was incorrectly seeing a Monty-Hall analogy in more or less the same way I was incorrectly seeing the original problem. But (as per reply to Brian), I'll leave that incorrect edit in place, just so that it's clear what your remarks are referring to. And thanks again, ditto Brian, for clarifying my misunderstanding(s). $\endgroup$ Sep 18, 2022 at 8:33
  • $\begingroup$ You're very welcome. FWIW, I think this is a great question and would make a good brain-teaser or exercise in an introduction to probability textbook or class. $\endgroup$ Sep 18, 2022 at 8:35

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