A simple problem with a strange answer

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I'm 'reading' this book called The Book of Brain Games by Ivan Moscovich. I'm sure you can guess what the book is about. Anyway, the problems are rated 1-10 in difficulty and this is about one problem in particular that was rated 7. I knew the answer to this problem immediately and my problem is not with the problem but a note in the answer to it in the back of the book. I'm going to type it word for word here and I would like you guys to answer it. Please don't google or search for the answer, just read it and give an answer. When (if) this gets to page 2 I will post the answer, word for word, and explain why it seemed strange to me. I'm just really curious how people on the spectrum will perceive this problem.

Well, I'm 100% certain that I'm wrong since it appears to be too easy but I'll just go with 2/3 chance. Can't see any other logical answer but than again probability was never my strongest when it came to math.

Nope, that was my instant response too (2/3). I was reluctant to answer because it feels like a trick question.

It also reminds me of a big probability debate I heard about a while ago which has had world experts at loggerheads, but I can't for the life of me remember the details. Could this be a similar scenario?

2/3

You have 2 coins out of 3 total with the same sides so I am gonna go with that answer.

Seems too easy though.

yeah, I get 2/3 also. I tried thinking in terms of there being six possible sides in total we are considering but you then just get 4/6 (=2/3).

I dunno.
Its partially a matter of wording.

If you radomonly grab a coin, and put it down- and DONT look at it.

Then what are the odds that WHATEVER is visible to a second person in the room is going to be the same as whats on the down side? That seems to be the question.

Thats gotta be 2/3-that its going to be the same thing.

But if you a grab a coin and SEE the visible upside on the table. And your asked what the odds are that whatever you see is the same as the hidden side-then that changes it.


If you see heads - then you know it cant be the coin with two tails.

So it has to be either: the normal coin, or the two headed coin.

The odds of either of it being either of the two possible coins are 50 50.

If its the normal coin- then the hidden side is tails, otherwise the hidden side is also heads.
So the odds of different vs the same thing on both sides is also 50-50.

So if its phrased that way- then the odds are 50-50.

One would think the answer is 2/3, but since this is a trick question, I have a few options:
1. One coin has heads and tails, one has two heads, and one has two tails. The question didn't state the heads or tails were on opposite sides of the coin. One would assume that, but since this is a trick question, it could very well be that both tails and heads were on the same side of the coin with nothing on the other side. What are the chances, then, that both sides are the same? 0%.
2. Another argument, albeit far fetched, would suggest the question did not apply to the extracted coin but to the coins left in the basket.
3. Even weirder would be to ask how can one side be visible since coins, by definition of their physical properties, cannot be invisible.
4. Where's Dr. Spencer Reid when you need him!

Of course it's two out of three. the number of flips doesn't even matter. The question isn't about heads or tails, or the chances of which. It's not a trick question at all. It's so simple is LOOKS like a trick question, but it's really quite simple

Three coins right? Two out of three have the same face on one side as the other. The other has both a head and a tail.

You withdraw ONE COIN from the hat. Only one. LAY it on the table. (There is no flipping!).

What is the chance it will be ONE of the TWO coins with matching faces?

Two out of three. Period. 66.66%

Done.

The only reason it might be confusing is that most of these problems involve flipping coins with both heads and tails on each coin. None of that logic applies here. Even the fact that you're placing it on the table is irrelevant.

All that matters is that two of three coins have matching faces, the other doesn't. What are the chances the one you pick from the hat is one of those two? Two out of three. Done.

According to Bayes' theorem, it is 2/3, which is pretty irrefutable unless there is some kind of word trick in the problem that I'm not seeing.

Okay.

If you know one side of the coin, then you can exclude one coin because it is clear that one coin absolutely has to still be in the hat. That leaves two coins and so it is 50/50 which of the two coins was selected.

If you do not know anything about either side of the selected coin, then you cannot exclude a coin and the odds are 1/3 that it is Heads/Tails, 1/3 that it is Heads/Heads, and 2/3 that it is Tails/Tails. Therefore, 2/3 of the time both sides will be the same.

You know this is painful right? I'm just gonna keep posting until page two finally shows up ya know!

BABOOM! Answer please?

*drumroll*