Do This by Logic

Ok, i figure it this way; The chance of the specified outcome for each coin is 25% for the fair coins and 100% for the cheat coin. Since you have an equal chance of pulling out any of the coins, this leaves a 4/175 chance of this happening, therefore the chance is 43.75%. Seems right to me, but I"m sure DS has something up his sleeve here.

David, are you going to reveal the answer. It seems that the consensious is split over 1/4 and 4/7. Both answers seem logical, just wanted to know your thoughts.


Ryan

Without formulas? If there were a thousand coins and 999 were straight I could pick one, flip it two heads in a row and still feel safe in saying I had a legit coin. But with only 4 coins and one is a trick it's different. Two heads has more weight. If there were only two coins and 1 was rigged if I picked one and flipped two heads I would now be inclined to take the side of the bet where I say I have the rigged coin. Without math it is hard to say but I figure 50 50 may look good now. I don't feel to happy about choosing either side and that is how I look for 50 50.

A classic Bayes Theorem question.


Each of the fair coins has a 1/4 chance of flipping heads twice and the double headed coin has a 100% chance of flipping heads twice. a priori each coin had an equal chance of being taken out. Now we have three possibilities whose a priori probilities are 1/4*(1/3), 1/4*(1/3), and 1*(1/3). To get the probability of any one of the possibilities, take its probability and divide it by the probability of all of them together. The 1/3's cancel and you have


1/(1/4+1/4+1) = 2/3.


Intuitively you could think "3 equal parts that are 1/4, 1/4, and 1".

I misread the problem. 2/3 would be right if there were 2 fair and one double headed coin. The correct answer is:


1/(1/4+1/4+1/4+1)=4/7.

Maybe I'm a little late.


Think of each fair coin as a package containing 4 objects :{ HH, TH, HT, TT}, adn the unfair coin as a package containing {HH,HH,HH,HH}.


So we think of this as really 16 objects w/ 7HH's. 4 of these pertain to the unfair coin.


SO the probability we have the unfair coin after flipping two heads is 4/7.


I think this way can generalize easily. Eg say we flipped the coin 3 times (and got 3 heads) . Then we have 3+8 objects pertaining to {HHH}, and 8 of them correspond to the unfair coin.


So the probability we have the unfair coin here is 8/11.

no sample size is sufficient enough to change the effective odds of which coin you pulled out. you could flip it a zillion times, get heads every time, and you could not be mathematically certain that you had pulled the double-headed coin. likely, but not certain.

we don't get HH 1/4 of the time when we pick a H/T coin. there is no law that says when flipping a coin it will alternate sides that it lands on. however, if you flip a coin twice, record the results, and then repeat this 2-flip procedure 1million times, HH will be very close to representing 1/4 of the results.

this hearkens back to the 'Lets Make A Deal' problem. the answer there was that you should switch because the probability that you picked the right door was 1 in 3, or 2-1 against. meaning that without further information (which revealing an empty door does NOT give you) you should still switch, because switching increases your odds to 2-1 in favor.


now, to apply this to the current problem:


your chances of picking the HH coin are 1/4 or 3-1 against. you flip the coin and flip H and flip again and get H. this does not preclude you from having picked a specific coin. you may have been falsely duped into thinking you have more reliable clues as to which coin you pulled out. however, you could flip heads a million times, it still wouldn't change the fact that this is possible with any of the coins. therefore it is still equally likely that you picked any one of them. the 'chances you picked the bad coin' are still 3-1 against, or 1 in 4.


anyone who doesn't get this should watch the movie 'Rosencrantz and Gildenstern are Dead'. in fact, you should all watch it. its pretty cool.