The toughest probability question i've ever heard, does anybody know the answer?

Question

You have 3 coins, one has heads on both sides, one has tails on both sides, and one is a normal coin. You put them in a bag and shake it. You draw one without looking and look at it in your hand. you see heads. AT THIS POINT what is the probability that you drew the double headed coin?
there are 2 possible answers for this:
1: 50% as you have 2 possible coins one is the double sided one the other has tails on the back, therefore 50%.
or
2. 2/3 as since it's heads you had a higher probability of it being the double headed one because it has 2 possible ways to draw heads.

I maintain that it's 50% as you had no possible way of knowing you were going to draw heads until you drew it therefore it was tottally random until you looked at it at which point there is 2 possible answers therefore 50%. It would not be 2/3 unless you TRIED TO DRAW HEADS, which is a complicated logical thought. what do you think?

I think it has a lot to do with how you look at the problem. yes if you start from the beginning and draw a coin then the probability would be 2/3, but if you start at the at the point where you have the coin in your hands, everything before that is irrelevent. Also since you can't draw both sides of the two headed coin at the same time there is really only 2 options that could have happened, one, you drew the regular coin or two you drew the double headed coin, which SIDE of the coin you drew does not matter as they are both the same. I think you can get 2 different answers for this problem if you look at it from a mathmatical point of view as opposed to a logical point of view.

If you KNEW you were going to get heads (impossible) than you it would be 3 out of 6 possibilities and a 2 out of 3 chance of picking the double one. BUT as you drew randomly all you "know" is that you have heads which means the side you are looking at can't also be on the other side, therfore 1/2

I think i figured out where everybody is getting confused. People are trying to seperate the sides from the coins. look at it in this very simple way:

In the beginning you have a 1/3 possibility of drawing each of the 3 coins. When you draw one and it eliminates the tail/tail coin then you are down to only 2 posibilities, either it's coin "a" or coin "b". what is actually on that coin is meaningless other than for the fact that it eliminates all coins without heads on them. so now you have 2 possiblities, heads, or tails, os 50%. what people are trying to do is to seperate the 2 sides of the coins, for instance saying that you can draw heads on either side of coin "b" so there's a higher probability of that one. this is invalid as they are both the SAME option. we are drawing coins and thereby drawing sides. You don't have 6 options you only have 3 and THEN you would have 2. There is a break in the problem, you have to DRAW the COIN, THEN you check the probability. Agree, Disagree??

One more thing, there are a few of you who don't understand the problem. The problem is NOT "what is the probability of drawing a 2 headed coin?", it's "given that you have ALREADY drawn heads what is the probability that the other side is heads?". It's more logic and critical thinking than math i think.

Answer 1

It's 2/3.

Use Bayes Theorem, which states:

P(A|B) = P(B|A) P(A) / P(B)

or in this discussion:

P(HH|head showing)

= P(head showing|HH) P(HH) / P(head showing)

The probability of a head showing, given you have the HH coin, is 1.

The probability of drawing a HH is 1/3,

The probability of a head showing (considering all three coins), is 1/2.

So, P(HH|head showing) = (1) (1/3) / (1/2)

or 2/3.

You maintained "that it's 50% as you had no possible way of knowing you were going to draw heads until you drew it therefore it was totally random until you looked at it at which point there is 2 possible answers therefore 50%."

The mistake that is made when concluding that the probability might be 1/2 is that, because you see a head, the chances of having the head/tail or the two headed coin are equal. They are not equally likely probabilities. You had a 1/3 chance of drawing the two headed, and a 1/3 chance of drawing the head/tail. But half of the head/tail draws are elimintated from your consideration, because when you look at the coin, you might see a tail.

So ask yourself this; when you see a head, what percentage of the time is that head caused by a two headed coin? What percentage of the time is is caused by a head/tail coin?

The answer is, 2/3 of the time it would be caused by a two headed coin, 1/3 of the time it would be caused by a head/tail. The mistake you make in thinking it's 50-50 is thinking that the head is caused in equal proportion by the two-headed or head/tail coins.

The conditions that would cause you to decide the probability was 1/2 might be if *I* had the bag, looked inside it, saw all the coins, knew which was which, and chose to show you a head. Then, the two tailed coin would not be considered, and it would be an equal probability that I showed you a two headed or a head/tail coin.

Answer 2

Use H to represent the double-sided head coin, T for the double-sided tail, and N for the normal coin. Use x to represent the event that you see heads.

x occurs if you draw H, (prob = 1/3) or if you draw N and it turns up showing heads (prob = (1/3)*(1/2))
Thus P(x) = (1/3)*1 + (1/3)*(1/2) = 1/2 as expected.

Now use conditional probability:

P(xH) = P(x given H)*P(H) or P(H given x)*P(x)

The first of these gives P(xH) = 1/3, which is just common sense because P(H) = (1/3) and when H happens, x always does, so the probability of them happening together is 1/3.

But applying the second expression to this gives
P(H given x)*P(x) = 1/3, and since we already know that
P(x) = 1/2, then

P(H given x) = 2/3, so your second answer is correct.

I suppose another way of looking at it is that there are three head faces altogether, of which two are on the double coin, so if you see a head the prob that it's on the double-head coin is 2/3, which is another way of saying what you said.

What's wrong with the 50% argument? It assumes both coins are equally likely, which isn't the case if a head is already showing. All three head faces are equally likely.

Answer 3

The answer is 2/3. It's actually quite simple to see if you think in terms of a repeated experiment, which is often the best way to resolve knotty probability questions.

Draw the coin out of the bag a large number of times, for the sake of arithmetic say 6000 times.

You are equally likely to see any of the six sides of the three coins in your plam, so each side would show in 1000 out of the 6000 trials.

But, half, 3000, of the trials would be discarded as they would not meet your criterion of seeing a head.

Of the remaining 3000 valid trials, there are 1000 showings for each of the three heads, the head on the normal coin and the two heads on the double-headed coin, so in 2000 of the 3000 valid trials you are holding the double-headed coin, the probability therefore being 2/3.

So long as you can't distinguish the coins when you reach into the bag, it makes absolutely no difference what you are thinking about when you go through the process (or, when you try it, if it does make a difference then you have mind-over-matter powers that could make your fortune!)

The same approach solves the Monty Hall problem others have referred to.

PS If you don't believe me, try it!

Answer 4

Use H to represent the double-sided head coin, T for the double-sided tail, and N for the normal coin. Use x to represent the event that you see heads.

x occurs if you draw H, (prob = 1/3) or if you draw N and it turns up showing heads (prob = (1/3)*(1/2))
Thus P(x) = (1/3)*1 + (1/3)*(1/2) = 1/2 as expected.

Now use conditional probability:

P(xH) = P(x given H)*P(H) or P(H given x)*P(x)

The first of these gives P(xH) = 1/3, which is just common sense because P(H) = (1/3) and when H happens, x always does, so the probability of them happening together is 1/3.

But applying the second expression to this gives
P(H given x)*P(x) = 1/3, and since we already know that
P(x) = 1/2, then

P(H given x) = 2/3, so your second answer is correct.

I suppose another way of looking at it is that there are three head faces altogether, of which two are on the double coin, so if you see a head the prob that it's on the double-head coin is 2/3, which is another way of saying what you said.

What's wrong with the 50% argument? It assumes both coins are equally likely, which isn't the case if a head is already showing. All three head faces are equally likely.

Answer 5

Okay, I am one of those been there unkempt women that now takes a lot of pride in their appearance. First and foremost ANYBODY is going to feel better and less depressed even if they already are depressed if they will do the best with what they have and practice at least good hygiene. Then the more that they WILL and CAN do with their appearance will improve their self esteem by 110%. I am living proof of that and up until 3 months ago after 22 years of marriage I could go 6 months not put makeup and lied to the world and myself that I was comfortable with it. Now I won't so much as go for a jog without at least eyeliner and lipstick. My marriage has gotten a much needed boost. I probably do, no, I know that I do have a husband that loves me unconditionally and will stay with me through thick and thin but that is no good excuse to let him look at a hag because of it. I like myself a hell of a lot more and if people are going to be haten on you for your question don't worry about it because there maybe a few that might realize that they resemble your remark and do something about it and be thankful you posted it. If there is any correct place to ask something of this nature, this forum is most certainly it. More of my 2 cents....This has just gone over some of these peoples heads, this is not about hating or judging for most it is actual concern. I wished I had woke up years ago.

Answer 6

The correct answer as already explained by several people is 2/3. It's clear from Bayes theorem which computes exactly what you want: the probability of the double headed coin GIVEN that you see a head.

If you're not convinced, two possibilities:
1 - Try it yourself! Nothing more convincing than that ;)
2 - Imaging the following extreme example. You have 3 dices and each of them has 1000 faces. The first one has numbers from 1 to 1000, the second one has all its faces numbered 1 and the last one has all its faces numbered 1000. You pick up at random at dice and at random a face. You see the number 1000. Isn't clear intuitively that there is much higher likelihood that you have in your hands the dice all numbered 1000 rather than the dice with all numbers from 1 to 1000?

Answer 7

If you don't believe that the answer is 50%... try it at home!

- Draw a head on a sticker and stick it on the tails side of a coin. You now have a heads-heads coin.
- Draw a "T" on a sticker and stick it on the heads side of a coin. You now have a tails-tails coin.
- Add an ordinary coin.
- Put all 3 coins in a bag. Draw a coin at random as follows:
- Close your eyes and wear gloves so you won't feel the stickers.
- Pull a coin out of the bag.
- Place the coin on a table.
- Open your eyes and record what you see, heads or tails.
- Look at the other side and record what coin you picked, HH TT or HT.
- Repeat.

You will end up with a string of results like this:
H, HH (see heads up, heads-heads coin)
T, HT (see tails up, heads-tails coin)
T, TT
H, HT
T, TT
etc...

If you did this right, you will see the following happen:
H, HH will happen about 1/3 of the time.
H, HT will happen about 1/6 of the time.
T, TT about 1/3 of the time.
T, HT about 1/6 of the time.

So if the first side you see is heads, it is twice as likely to be the HH coin.

Answer 8

This is a great question and harkens back to the Monty Hall problem solved by Marilyn.

AT THIS POINT there are 3 ways it could happen: you drew the 2-headed coin and looked at the obverse side, you drew the 2-headed coin and looked at the reverse side, or you drew the normal coin and looked at the head. The probability that you drew the 2-headed coin is two out of those three ways, or 2/3.

The probability might become more intuitive if you have 100 normal coins, 100 2-tailed coins, and 1 2-headed coin. What is the probability of drawing the 2-headed coin? 1/201. But if you draw a coin and look at it and see a head, what is the probability it is the 2-headed coin? 1/101.

The probability is not affected by what you "try" to do. I was playing roulette once and the chap next to me bet on black, and when the wheel came up red he said, "I KNEW it was going to come up red!" I asked him why, then, if he knew that, had he bet on black. He ignored my question.

Answer 9

I agree that the correct probability is 2/3. A formal way of proving this is as follows:

Let "heads" and "tails" represent the event that you see a head or a tail respectively and "HH", "HT" and "TT" represent the three coins (I trust you can work out which is which).

There are two actions involved:selecting a coin, with 1/3 probability for each outcome, and selecting a side from that coin, with 1/2 probability for each side of the selected coin.

P(HH) = P(HT) = P(TT) = 1/3.

P(heads | HH) = 1
P(heads | HT) = 1/2
P(heads | TT) = 0

P(HH and heads) = P(HH) . P(heads | HH) = (1/3) (1) = 1/3.
P(HT and heads) = P(HT) . P(heads | HT) = (1/3) (1/2) = 1/6.
P(TT and heads) = P(TT) . P(heads | TT) = (1/3) (0) = 0.
So P(heads) = 1/2 (which we could have observed from symmetry, of course).

P(HH | heads) = P(HH and heads) / P(heads)
= (1/3) / (1/2) = 2/3.

Note that just because there are two possibilities there is no reason that there should be a 50% chance for each to be correct.

Answer 10

I get 2/3

First you have 3 branches representing the 3 coins, 1/3 each

HH
TT
HT

then each of these has two branches

HH has 2 branches, both H with p=1/2
TT has 2 branches, both T with p=1/2
HT has 2 branches, 1 H at p=1/2, 1 T at p=1/2

Now the conditional probability that you want to calculate is

P(coin HH | a head is shown) = P(coin HH and a head)/P(head)

the numerator is 1/3, the denominator is 1/2

So, the ratio is 1/3 divided by 1/2 or 2/3