A simple problem with a strange answer

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Marybird, I understand what you're saying. The issue isn't whether the odds of guessing the right answer changes. That isn't what the question asks as the OP wrote it.

The question asks ONLY what the odds are. Not what they are in your mind. That's what doesn't change.

The Monty Hall problem is different because the goal isn't to determine what the odds are. The goal is to win. To choose according to the best odds. What we know in the Monty Hall problem matters because we're trying to win the game. With this problem we're only trying to figure odds. Very specific odds, that don't change.

The odds of pulling a head/head or tail/tail out of the hat. Those are the odds the question asks for. All else is not relevant to the answer to that specific question.

EDIT: I just had a little epiphany.

I think the OP has a valid point. That autistic thinking may affect how we each approach the problem. I think I've got the answer to the OP's question posted on page 2 of this thread:



This problem does relate to the Monty Hall problem, but the goal isn't the same.

It is true that if you know what the face of the coin on the table is, and you take that into consideration, you can then THINK that the other coin in the hat with the opposite matching faces should be discounted. If you do that, and the goal is to guess whether the coin on the table has matching faces or not, your odds of being correct drop to 50/50.

In other words, let's say the table coin shows a head. You may THINK that over many times doing this, you should guess "Yes, it's match" 50% of the times you pull a coin out, and that would improve your chances of being correct, but the opposite is actually true.

You should guess "Yes" 2 out of three times. Not every other time. If you guess yes, two out of three times over 1000 times pulling one coin out, on average you will be correct almost every time.

If you do this 1000 times and guess "yes" every other time, and "no" every other time, you will be wrong more often.

I think we've actually been in agreement, and arguing over misunderstanding exactly what the other of us has been actually trying to say.

And that, I think, demonstrates the OP's point very well indeed!

Last edited by MrXxx on 22 Nov 2012, 11:46 am, edited 3 times in total.

I really tried to explain it as clear and step by step as I could but I'm not sure we're on the same page.


If we had a hat with one double-heads coin, one regular heads-tails coin and 98 double-tails coins and then selected one coin at random, what do you think the probability of that coin being double-sided is?

Now imagine putting that coin down on the table and looking at the top side of the coin only. You see it is heads. Now what is the probability of that coin being double-sided?

If you look at the WORDED problem on the OPs first post, its 2/3.
If you look at the other problem listed, that states to look at the visible top side of the drawn coin, then its 1/3, if the top is a head.
If it doesnt state what is on the coin, thats a diff kettle of fish.

i am on page 179 of a yet to be written manuscript.


99-1.


it is 50% considering that there are only only 2 states of every coin.

Last edited by b9 on 22 Nov 2012, 12:32 pm, edited 1 time in total.

Huh?

I'm sorry MrXxx but I still disagree with you. I agree that the intuitive answer is 1/2 and the correct answer is 2/3, but for different reasons.
I agree that the intuitive answer is 1/2 because after seeing heads on the top side of the chosen coin we intuitively want to eliminate the double tails coin, leaving two possible coins, one of which is double sided: 1/2.

But I think we do need to eliminate the double tails coin. After we see the heads on our chosen coin we know that the probability of that coin being double tails is 0. We can and must eliminate the double tails coin.

The denominator '3' in the answer DOES NOT refer to the 3 coins. It refers to the 3 possible positions of our 2 possible coins. As we know that the top side is heads, our two remaining possible coins can be in the following positions:

1. The coin is heads-heads, with heads-a facing up
2. The coin is heads-heads, with heads-b facing up
3. The coin is heads-tails, with heads facing up

Option 1 and 2 give us a double sided coin (the same double sided coin) while option 3 does not. Hence 2/3.

The intuitive error isn't eliminating the double tails coin. It is correct to eliminate that coin. The intuitive error is assuming that after eliminating the double tails coin, the two remaining coins must be equally likely. In fact they are not. When we see the heads on the top side of our chosen coin we eliminate not only the double tails coin, but we also eliminate the possibility of the regular coin being chosen and placed tails up. This makes the coin half as likely to be a regular heads-tails coin. It is this step that, to quote the official solution "is so counter-intuitive that many people refuse to believe it"

Last edited by Ellingtonia on 22 Nov 2012, 12:29 pm, edited 1 time in total.

Ok now we can extend that to the original problem with only 1 double tails coin. With only one double tails coin the answer to both questions is 2/3, but for different reasons.

Can you also see how the intuitive answer (for some people at least) is 1/2? Because after eliminating the double tails coins, whether there are 98 of them or only 1 of them, we are left with two possible coins. One of these two possible coins is double sided, and so the probability of our chosen coin being double sided seems to be 1/2. Can you see why many people would intuitively approach the question this way?

Either way works. And I think that's the OP's real point.

You went through some extra steps I found unnecessary, yet we both get the same result. You think it is necessary to eliminate the tail/tail coin. I don't.

The intuitive error occurs if the tail/tail coin is eliminated, and you DON'T take the extra steps of considering the three alternatives you listed, assuming then that the chances are now 50/50, but they are not. The same mistake most people make with the Monty Hall problem by the way.

IF you do eliminate the tail/tail coin, you MUST consider those possibilities, but if you do not eliminate it, you get the correct answer, correctly, in fewer steps.

no.

I think your method of not eliminating the double tails coins is actually incorrect, and gets the same answer only coincidentally. What if we repeated the problem with one double heads coin, one heads-tails coin and TWO tails-tails coins? How would this change your answer?

...

Why not?

You know what, never mind. I give up. It's nearly 5am and I'm going to bed.

Still didn't get it when I read this. I had to get 30 pieces of paper and move the money around. But now I get it.
When the sailors pay 30, but get back 3, they did pay 27 (which they don't have on them, it's the 2+25 the hotel has) thus you have to add the 3 they were returned, not the 2.

Back to the OP, after reading many of the very extensive explanations and reply, I thought I'd add my own... I interpreted the problem as.... you take a coin, put it on the table, look at it. And then what are the chances that the coin that you drew is double sided. Since the sentence is all together "If you withdraw 1 coin from the hat and lay it flat on a table without looking at it, what are the chances that the hidden side is the same as the visible side?" It's not clear whether it's asking the chances "all together" or the chances of that single coin of being double sided. But as many have explained the chance in both occasions is still 2/3 (the double sided coin increases the chances of that coin).


It's counter-intuitive because if from those two coins you were asked to pick between them, than the chances would be 50/50 because of the knowledge you have. But you don't get to pick between 2 coins. You picked from 3, and you can't change which you picked.

I took 3 pieces of paper and did it in real life. If I picked from the 3 and marked whether it was double sided or not like someone else did I got 11/16 which is close to 2/3.
BUT if I scrambled the pieces of paper and picked from the 2 that had come up the same, the chances were 5/5 or 50/50. It's the illusion of picking that makes it seem counter-intuitive.

Now for the real question... We need to ask this to some NTs and see what they reply. I see the majority here immediately said 2/3, only upon going in depth did everyone get confused. I think NTs will most likely say 1/2 but people with AS who interpreted it more literally (many not even thinking of looking, or thinking looking was not important) got the right answer (This is the right answer, right? What did the book say in the end?)

Radically most likely. Because then you've just constructed a completely different problem.

Let me counter with this:

Make it two double-headed coins. And one head-tail coin. Now do you see why it isn't really necessary to eliminate the other coin?

Not in this particular problem at least.

I think what you're doing is constructing a method that works for all scenarios. And there's nothing wrong with that, but it doesn't mean that it's the only correct way to solve the problem.

I approach all problems looking for the quickest way to get to the answer. As long as the answer is consistently correct that works. It may mean that if we change some variables in the problem that my method no longer works, but that doesn't make the method incorrect.

Your way creates a method that works for every variable, and that's fine, but it is a formulated method. My head hates formulated methods because they require memorization, and I'm no good at that. I'm much better at creating methods for each particular problem. Doing things this way does cause some occasional mistakes (see the Monty Hall problem thread, where I did just that), but practicing creating a method for each new problem can also create the ability to adapt to changing circumstance very quickly, and the ability to solve problems more quickly.

It means I have to keep up with it though. When I am regularly involved with solving problems like this, I can get to the point of clicking off correct answers very quickly with little trouble. When I am not in practice, I make more mistakes because I've not been in the groove so to speak. The Monty Hall problem is a pretty good example of that.

My istinctive answer before i read the actual answer: 50%.

Explanation: say you extract the coin and see the visible face shows heads. At that point, either the other face is also heads or it's tails, thus there is 50% chance it will be the same.

Aha I see I definitely overthought it. Immediately I thought 2/3 but it was supposed to be counterintuitive so I looked for an explanation and found 1/2. And it took me a long time to re-convince myself that it was 2/3! My head is spinning.

Yes. You are right. I was only explaining why, when you see the face of the coin you selected, your knowledge at that time gives you a 50% chance of guessing the outcome. Of course, when you turn the coin over you know 100% what the outcome is. The original probability hasn't changed but you can ask what the probability is each time you gain more knowledge.

Last edited by Marybird on 22 Nov 2012, 10:01 pm, edited 1 time in total.