A simple problem with a strange answer

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if it's 2/3 what the fuzz was the point of this thread????

I know I'm probably retreading what has been said before a bit, but I wanted to share the stream of consciousness I had when solving this problem.

Okay, so If you pull one of the coins out of the hat and set it down, without looking at it, the probability that both sides are the same is 2/3, because 2/3 of the coins in the hat had the property of having both sides the same.

However, if the question says that you pull out the coin, set it down, and look at it, then that actually changes things quite a bit. The coins themselves haven't changed, there are still two single-sided(head-head/tails-tails) ones, and when pulling one out of the hat there is still a 2/3 possibility of getting a single-sided one. No probabilities have been magically changed, but the question is now asking for a different probability.

In the version of the question where you don't look at the coin, it's asking for the probability that both sides of the coin will be the same, but whether or not both sides will be heads or tails is ambiguous. In the version of the question where you do look at the coin, you are now looking for the probability that both sides of the coin will be the same, given that you already know what one of the sides is(because you're now allowed to look at the coin). That extra "given" on the end subtly changes what you're looking for, although this wouldn't be immediately apparent just going by the question.

So, since you'll be able to know the symbol on one of the coin's sides, that means you immediately know that you can't have drawn the one coin that lacks that symbol. If the coin you draw has a heads on one side, then there is a 0% chance that you've drawn the two-tailed coin, and vice-versa. So now you know your coin must be one of two coins: either the single-sided one that has the same symbol that you saw on the top side of your coin, or the standard double-sided one. That might lead you to conclude that there's a 50/50 chance of having a single-sided coin.

However, you have to consider that half the time you draw the standard coin, you will see the heads side, and half the time you will see the tails side. You are twice as likely to draw the two-headed coin and set it down heads-up as you are to draw the standard coin and set it down heads-up(and same for tails). Given this knowledge, you can conclude that there is a 2/3 chance that the bottom hidden side of the coin is the same as the top side.

Sorry, that ended up too long.

I just looked for the actual question, said to myself, "3 coins, 2 of them have the same thing on both sides," so two out of three. No intuition needed. Intuition never helped me much with math anyway.

pick a coin out.
1/2 chance that it'd be A or B if you took the AB coin out, regardless
2/3 chance that it'd be the A or the B one if you took the AA or the BB one out

(this covers all ranges), the total ways to do the 2 above scenarios is.... 3 in this way of looking at it

however specifics wise

I shall say A-B or B-A, what this represents is the heads/tails coin, and the first letter shall be what faces up

A-B out -->A (coin 1: pick A or A,coin 2: pick B or B)
4 possible events, 2 are "right", 2/4 or 1/2
B-A out -->A (coin 1: pick A or A,coin 2: pick B or B)
same thing here

A-B out -->B (coin 1: pick A or A,coin 2: pick B or B)
same thing here
B-A out -->B (coin 1: pick A or A,coin 2: pick B or B)
same thing here

AA out -->A (4 choices, pick A or B, pick B or B)
(1/4 chance, 2 coins each with half a chance to be one or the other side, thus 4 options 1 right one)

BB out -->B (4 choices, pick A or B, pick A or A)
same thing here, 1/4

4 scenarios/chances to match up! from picking the A-B/B-A coin... each scenario having 4 possibilities...

16 possibilities... 8 are correct

1/4 from getting the AA coin out, 1/4 from getting the BB coin out
8 possibilities.... 2 are correct


24 choice types, 10 are right

10/24

5/12

house wins. you're led on

66% win chance in your favour OMG GAMBLE YESSSSS
nope.

each thing unit on the top of this fraction is worth about 8.3%, so 5/12 is approximately 41.5% you're off by 50%, you're guessing the chance is 150% of what it is, the actual chance is approximately 66% of your 66% guess, slightly less, actually (maybe there's a blindspot for repetitive thinking, fractals and the like in people's minds?)


I almost failed math 11, and did fail math 12 for a while

I guess it's a matter of how you frame what you're thinking about, I was in denial over autism back then... so.... that might be why I did so poorly, was trying not to think the way I think... lol

ALWAYS SUSPECT THE ODDS, GAMBLING BIAS IS EVERYWHERE

seriously, I just wanted to solve the problem, curiousity strikes again. feels good man.

I'm getting the feeling this may end up being one of those eternal threads that won't go away.

oops. wrong problem. yeah 2/3

[Moved from GA to Off the Wall]

Unfortunately, the OP's original intent for this thread has gone largely unnoticed.

The "real" question isn't in the first post of the thread. The OP's point here is actually being missed by most of the posts.

The solution to the problem, taken straight from the book the OP got the problem from in the first place, has already been posted by the OP.

The problem in this thread though, is that people are missing the OP's purpose for posting it in the first place. The solution isn't what he was looking for. He already had the solution.

What the OP was really looking for were the answers to the questions he posted on page two of the thread, where he not only posted the solution to the problem, but the purpose of the thread.

http://www.wrongplanet.net/postp5046052.html#5046052