Do This by Logic

Again I am trying to get you to think, not blindly use formulas. You reach into a bowl that you know contains three fair coins and one double headed coin. You randomly pick one. Without looking at it, you flip it two times and get two heads. What's the chances you picked the bad coin?

...I thought it said 2 fair coins, not 3. So my answer would be 25%.


Ryan

I dont think 2 flips of heads, would be a big enough sample to change any percentage. There are 3 regular coins and one bad (1 out of 4) thats why I say 25%. Now if it came up 12 times in a row, then your odds would be almost 100%.

The funny coin will be picked 1/4 of the time. A fair coin will be chosen 3/4 of the time and come up heads twice 1/4 of the time, which means this sequence occurs 3/4 * 1/4 = 3/16 of the time. Thus, HH will appear 3/16 + 1/4 = 7/16. So the probability that HH is caused by the funny coin is 1/4 / 7/16 = 4/7.

... is 4/7 !?

First try:


The chances you would initially pick a fair coin are 3/4. The chances that a fair coin would flip two heads in a row is 1/4. So the chances that you picked a fair coin AND flipped 2 heads in a row with it is 3/4 * 1/4 = 3/16. Therefore the chance that you picked the double-headed coin is 13/16, since 3/16 plus 13/16 = 1.0


Intuitively this seems high, so I am going to think about it a bit more while I go out for a bit of exercise.

25%

The fact that no tail came up only proves that it is possible that you chose the double headed coin.

How'd I do?

I was erroneously counting the ways that it might have not been two heads in a row as belonging to the double-headed's claim.


Out of a total of 16 possible picks and flip sequences, 4 give two heads with the double headed coin. 3 give two heads with the single headed coins. Therefore the odds are 4/16 vs. 3/16 which reduces to 4:3 in favor of the double-headed coin. (I believe we must count the double-headed coin as having two distinct sides even though both sides give the same result).

One in four that you chose the bad coin.


Jimbo

I thought it was funny that the early reponses decided to show their answers determined by formulas.


They must have skipped over the first sentence in David's post: Again I am trying to get you to think, not blindly use formulas.


They must have missed the point of the exercise. I'll go with 25% too.

If fair coins turn up HH 1/4 of the time that means the other 3/4 of the time it came up HH it would be the double headed coin.


Just a guess since DS asked for logic and not formulas.


Flame away.

4:3= 4/7 = the chances you picked the bad coin

another theory although most likely incorrect.


P(HH for fair coin) = .25

P(HH for unfair) = 1.00


P(unfair HH) divided by P(fair + unfair coins) = .80


1.00/1.25 = 0.80 or 80%


again flame away.

once you flip a coin and get a side, then if you flip it again theres a 50% chance you get the same side in a fair coin, or 100% in a double sided coin.


thats why i say its double fair chance of 1/4.


so 1/2.


but now i see it wasnt a double sided coin but a double headed coin. ugh. too much scotch i guess


brad

(3(.25) + 1(1) )/ 4


=


1.75/4


= 43.75%


?

i think there are two events here


picking


and flicking


(you can easily remember this "formula" by thinking of dynasty's nose)


picking is not affected by flicking


so the chances that you picked the double-headed coin are 1 in 4


once you've picked it it doesn't really matter if you flick ten heads in a row - although unusual, it's already on your finger

We pick the bad coin 1/4 of the time. We get 2 heads 1/4 of the time when we pick the 3 good coins, and all the time when we pick the bad coin, so the total probability that we get 2 heads is (1/4)(3/4) + (1/4)(1) = 7/16. So the fraction of the time we have the bad coin when we get 2 heads is (1/4)/(7/16) = 4/7.

there are four apples in a basket


one of them is laced with sleeping potion


a guy picks one at midnight and eats it


he falls asleep


another guy takes a bite and he falls asleep


what is the probability that this is the funny apple?

1/4 assuming they would fall asleep at midnight regardless.

bit slow as usual but i think i understand it now


the result that you see is HH


although there were two events leading up to this result the first event of picking has long gone under the bridge


the way you can have such a result is by the H/h coin falling HH, Hh, hh, or hH, or the three H/T coins falling HH


so there are seven different possible forms of the HH you can see, four of which could be made by the H/h coin and three by the H/T coins


hence 4 out of 7 times that you do see HH it is as the result of having picked the H/h coin

what are the chances you picked the bad coin, not what are the chances you picked the bad coin and flipped it twice and came up with two heads. I think no matter how many times you flip heads, until you flip tails the answer is still 1 in 4...i think

The fact that the coin came up HH twice is relevant.

For instance, if you knew that after flipping the coin twice that it came up TT, the probability that it is the bad coin is 0 (even though initially the probability of picking it was 1/4).

Since two flips won't indicate anything, probability-wise

3 fair coins each have 1/4*1/4

1 bad coin has 1/4 chance of being flipped 2 heads


3/16 chance that fair coin was flipped 2 heads

1/4 chance that bad coin was flipped 2 heads


thus prob of bad coin is 4/7


peter

Funny, i got the number 4/7, by a different route (I think) to most others...only read a few after I did it myself.


Fair coin gives HH 0.25 of time. Funny coin 1.0 of the time. Just compare the added probabilities for fair and funny coins (ie 0.75 cf 1.0 or 4/7).

cf ?

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.