On Friday I posted this puzzle…..

A box contains two coins. One is a double-headed coin, and the other is an ordinary coin ( heads on one side, and tails on the other). You take one of the coins out of the box and look at one of the sides. It’s a head. What’s the probability that the other side also contains a head?

If you have not tried to solve it, have a go now. For everyone else the answer is after the break.

When the coin is taken out of the box, there are four possibilities:

Coin Drawn Side Shown Other side

Double headed Heads Heads

Double headed (other side) Heads Heads

Normal coin Heads Tails

Normal coin Tail Heads

The last possibility didn’t occur, so there are three remaining possibilities. Of the three, two of them will show heads on the other side and only one will show tails on the other side. So the probability that the other side of the coin is heads is two thirds.

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called **PUZZLED** and is available for the **Kindle **(UK here and USA here) and on the **iBookstore** (UK here in the USA here). You can try 101 of the puzzles for free here.

Yes, found that.

Yes, the answer is 50/50. The other side of the coin is either a head (if it’s the double) or a tail (if it’s the normal coin)

The answer is 50/50. The other side of the coin is either a head (if it’s the double) or a tail (if it’s the normal coin).

You have to remember that it’s more likely that you draw the double-headed coin (since you’re looking at a head). Thus, if it’s more likely that you’re standing with the double-headed coin, then it’s also more likely that the other side is a head.

Formally: You draw a head, so remaining in the game are 2 heads and 1 tail. So that’s 66% chance to get a head when you flip the coin.

Well, I think thats wrong.

When you take out a coin and see a head, then the other side is either another head (you have the double-headed coin in your hand) or a tail (you have the normal coin in your hand).

So there’s a 1 in 2 chance of the other side being a head.

That’s what I got, Martin. I can’t see why 50/50 is wrong if you’ve already *got* a head. Doesn’t the probability start from that point onwards?

There are 3 ways you can have pulled out a head only one of which has a tail on the other side…therefore 2 out of 3 it will be a head on the other side.

You’ve forgotten that there’s not an equal chance of pulling out a head or a tail to begin with. To pull out a tail you first have to select the ordinary coin and then hold it the right way up. The shift in probability is due to the other coin not having a tail.

Hi Martin!

I’m not a mathematician, but I think Richard is right.

It might be easier if you imagine this experiment on a larger scale, say you do it 1000 times. Then you will get heads in roughly 666 trials and a tail in about 333. Not a 500/500.

This is because one does not know WHICH head one has got at the beginning and calculate for all possibilities.

The probability of getting another head is the sum of all probabilities of having one head (there are three for the taking, it does not matter wether they are on the same coin or not) and getting another. 1/3+1/3 therefore 2/3.

Another view:

Obviously it is 50/50 wether you grab the coin with the two heads or the normal one, but we only count the times when you see a head first. In this instance the double headed coin features twice as often, as all the tail-sightings are dismissed. Therefore the probability of picking the normal coin are halfed and getting another head is now twice as likely as getting a tail.

I hope this helps.

L

Martin,

There are 3 heads. For 2 of those heads, they have a head on the other side. 1 head has a tail on the other side.

The fourth scenario is that you see a tail first, which also has a head on the other side. But that did not happen.

Great puzzle. Glad I got it right.

That’s interesting. That’s the first answer I got and then I convinced myself I was wrong and I changed it to 50/50. Once you’ve already *got* a head (which is what the wording suggested to me), surely you can turn it over and it will either be a head (if the coin is the double-head) or a tail (if the coin is a normal coin). So, I thought 50%. (I thought I’d been tricked by the fact that there were two more heads and one tail ‘available’ when you turned over the coin.)

I don’t want to create a trouble but I’d be grateful if someone could explain why this is wrong.

Anne

Barry Goddard’s post at 7.39 on 18/1/14 on Friday’s puzzle thread may help.

AE

Anne, we are both right. Barry’s comment doesn’t apply to this case. There are NOT 4 coins. Only 2 coins.

The question asked is, having removed a coin and see a head:

Quote ‘What’s the probability that the other side also contains a head?’ and the answer HAS to be 1 in 2 as described above.

coin1: Head1 – Head2

coin2: Head3 – Tails

take out:

coin1: 50% chance

coin2: 50% chance

See first:

Head1: 25% chance

Head2: 25% chance

Head3: 25% chance

Tails: 25% chance

Seeing tails first did not happen, leaving 25% chance out of 75% that you see head3 first. 50% out of 75% that you have head1 or head2.

Imagine the same experiment with 2 normal coins. Then the probability is 50/50.

Are you saying that the probabilities are the same?

Trust me, Richard is right.

Hi

I believe this is a variation of the monty hall problem; http://youtu.be/mhlc7peGlGg

I think in the Monty Hall problem the key is that the gameshow host knows which door the prize is behind.

I agree, surely there’s only 2 possibilities from the start? You’ve either picked the double-headed coin or not? Just because you can’t tell from looking at the first side doesn’t change that does it?

Correct you have two possibilities but they are not equally likely.The most probable reason you see a head on the first side is because you pulled out the double headed coin. So now knowing you have a head means you know its most likely you have the double headed coin and therefore that there is a head on the other side.

David, The flaw in your reasoning is that you are ignoring the additional information about the coin that was chosen. You are assuming that this info has no effect on the probability of the coin being the double-headed coin. Suppose you choose one of the coins at random. It is indeed true that the probability that you have picked the double-headed coin is 1/2. But you have more info to go on. With the additional info that one of the sides is a head, the probability that you have picked the double-headed coin has increased to 2/3. Richard illustrates one approach to arrive at 2/3. Another way is to apply Bayes’ Theorem.

I knew there would be some controversy and I can’t explain why you’re wrong but it may have something to do with needing to consider the whole operation rather than just starting with the premise of having a coin in your hand showing a head.

For those not convinced by the answer, try it out with real coins.

Use a pencil to write a big H on the tails side of one of the coins.

Select a coin randomly, looking at one side only.

If it is tails, discard that selection from the trial.

If it is heads, record what you see when you flip the coin over.

The more repetitions you do, the more you will see that 2/3rds of the result set will be heads.

Another way to see the right answer is to consider the case with 99 double headed coins and one normal one. You pick one and see it has a head on one side. What are the chances it is the normal coin. Still 50/50?

Those who are not convinced, here is another example:

I have two decks of 52 cards. One of them is 52 aces of spades, the other one is a normal 52 cards deck.

I pick one deck randomly, and from this deck I pick two cards. I look at the first one, it’s the ace of spades. What are the chances that the other one is the ace of spades ?

It is obvious that it is more than 1/2, because the fact that the first card I picked happened to be an ace of spades makes it much more probable that the deck I picked is the one with only aces of spades. Thinking that probability “starts from a point onwards” can be right, but you need to make it properly. Ok, I have an ace of spades in my hand, but it absolutely does not mean that the deck I picked can be either one with probability 1/2. One option is more probable than the other.

With the coins it is the same thing: if I draw heads, it is more probable that the coin I picked is the double headed coin. Three ways to pick heads, two of them with the double headed coin, so the probability that the one in my hand has two heads is 2/3.

Nice! That’s a good way of explaining it.

If we’re talking about the ‘overall’ probability – i.e. from *before* you have the first head) then I completely agree with the 2/3 answer (which is what I got in the first place and should have stuck to🙂 ). But how I read it afterwards was that you begin the probability from a point when you *already have* the head, after which there is only a 50/50 choice.

Good puzzle, though! It taught me (emphatically a non-expert at Maths) that many people took the wording to mean that probability must ‘start’ (for want of a better word) from the very beginning of the overall puzzle, and not from when you already have the head. Hopefully a lesson learned for next time if I can remember it then.

Well you do “already have” the head side of a coin showing. The probability that the other side of the coin is also heads is 2/3.

The only probability that is 1/2 is the probability of picking the head/head coin (or the head/tail coin) to begin with.

The trick in probability problems is foten to break the potential happenings out into equi-probable events. Richard shows that there are 4 equi-probable events. One of those is picking a tails so that one is out. Of the 3 remaining equi-probable events, 2 are the double-header adn one is the normal coin. If you don’t find this an appealing and intuitive explanation, you will continue to have problems with probability.

If I am looking at one sided of the double-headed coin, the other side is heads.

If I am looking at the other side of the double-headed coin, the other side is heads.

If I am looking at the head on the normal coin, the other side is tails.

There are no other options, so the answer is 2/3.

I don’t really see what the problem is.

If you are not convinced the answer is two thirds, it may help to ask the question in a different way. I think the question can also be stated as, “given that you have chosen a head, what is the chance that you chose the double headed coin?”. There are three ways to have chosen a head, two of which by choosing the double headed coin, so the probability you chose the double headed coin is two thirds.

For those of you who think the answer is 50/50, how about a slightly different question “given that you have chosen a tail, what is the chance you chose the double headed coin?”. If you followed the same reasoning, would say that is 50/50 as well?

I disagree.

2/3 is the probability, after sides of the coins have been examined at random until a head is found, that the other side will be found to be a head also. This is NOT what the question asks. What the question asks for is the probability that the other side will be a head, given that you’re already looking a a head, having drawn a coin at random without knowing what is on the other side. THAT answer is 1/2, as many have stated.

Probability depends on prior knowledge.

I thing you’ve got your answers the wrong way round. If you pick a coin at random, examine both sides and put the coin down with heads guaranteed to be showing, then the answer is 1/2.

If you pick a coin at random look at one side only AND it’s heads then the answer is 2/3

Thanks Mike

Keep it up

Agree with the 2/3s answers. Some lovely explanations for it in the above comments too!

Here’s a brute force proof I knocked up in Excel.

The coin is pulled out at random. The face looked at is selected at random.

https://drive.google.com/file/d/0B1OjAVr3nACuZkJIU1piOXQ4WW8/edit?usp=sharing

If you click the link you’ll only get one “state” as it were. If you download the file, you’ll be able to run the simulation over and over (F9 should do the trick). The numbers will tend towards the 2/3 1/3 split, and be well away from a 1/2 1/2 split. (And the design honours the wording of the puzzle.)

The answer is 1/2!

> You take one of the coins out of the box and look at one of the sides. It’s a head.

This is the pre-condition. It is the same as “given you see head”.

And now it could be either the be the single-headed or the double-headed coin.

=> chances are 1/3

If the question would have been: what is the chance you pick up the coin showing head and on the other side is also head. But that is another question!

Here the question was “given you see head, what is the chance of head on the other side”.

It isn’t a pre-condition that you must see a head. The question clearly states you only look at 1 side of the coin. You could have seen a tail – but didn’t.

The answer is 2/3

The probability you select the head/head coin AND see a head is 1/2*1 = 1/2

The probability you select the head/tail coin AND see a head is 1/2*1/2 = 1/4.

Therefore you are twice as likely to see a head from the head/head coin as you are likely to see a head from the head/tail coin, hence the answer is 2/3.

If the question had been: “what is the probability you selected the head/head coin?”, the answer would be 1/2, but that’s a different question.

Nope.

You don’t know which one of the three heads it is “given” you are looking at. Two of those three heads have a head on the other side. The other one has a tail.

If you are shown both coins with heads up and pick one then it is 50/50 that the other side is a head.

It’s pretty clear that IF you see a head, there’s a 2/3 chance that you’ve chosen the double-headed coin, hence a 2/3 chance that the other side will be a head.

Palmer has the best explanation I’ve read so far. Perfectly stated.

For people not accepting Richard’s answer:

Do you still think the answer would be 50% if the box contained 1 ordinary coin and 999 double-headed coins? Or 1 ordinary coin and 99999999 double-headed coins?

This is what helps me understand this problem and the Month Hall problem.

Of course not, but this doesn’t help. If I thought the original answer was 50/50 (I don’t). then in your 1000-coin case I would think you had a 99.90% chance of the other side being heads, since you have a 999/1000 chance of picking one of the double-headed coins. Actually the correct answer would be 99.95%, since we discard the cases where you pick the normal coin and see a tails. This isn’t exactly analagous to the Monty Hall problem..

Ken,

If you don’t think that 50/50 was the answer to Richard’s original question, what were the two “plausible” answers you referred to in your post on Friday?

Thanks

Another Elk: On Friday, I was referring to 50/50 (or 1/2) and 2/3 as the two plausible answers. But only one of those two is correct–2/3.

I think this shows that when you repeat this a thousand times, in the end it will be 666 against 333 that it’s a head.

But when you do it only once, as in the puzzle, it is 50/50.

Doing it once, or doing it a thousand times will not affect the probability.

Repeat the puzzle 1000 times, on average you will pick the double headed coin 500 times, and the ordinary coin the same number of times. Of the times you picked the ordinary coin, on average you will have picked it heads up 250 times.

On average, you will have picked a coin heads up 750 times, but only 250 times will it have been the ordinary coin. So, given you have picked a coin heads up, the probability the other side is heads is two thirds.

I think the question by itself is unclear.

Main problem why people argue with each other is because no clear point from which in the history probability starts.

If the question would be smth. like: What is the probability that second side is a head among all coins which we have taken with head on top?

Then it should be 2/3

If the question would be smth. like: What is the probability that second side is a head among all taken coins when first side is always head (two-headed coin)?

Then is should be 1/2

It’s can be simply illustrated by coin and it’s orientation in the space. Imagine left side is a top and right is a bottom. Two-headed coin has H1 & H2 (two head for each side) and normal coin has H & L. Possible variants of each coin when it lies in the box are:

H1H2

H2H1

HL

LH

The probability that we have selected a coin with a head on top is 3/4

H1H2 H1H2

H2H1 => H2H1

HL HL

LH

Now, the probability that second side is also head among rest is 2/3

H1H2 H1H2 H1H2

H2H1 => H2H1 => H2H1

HL HL

LH

So, if we will calculate probability starting from the moment of taking the coin from the box, then:

3/4 * 2/3 = 6/12 = 1/2

But, if we will calculate probability assuming that coin which we have taken always has a head on top (probability 1), then:

1*2/3 = 2/3

So, we have two possible solutions, I think author must change the question in his task to make it more clear. Because “You take one of the coins out of the box and look at one of the sides. It’s a head” can be interpreted as one of variants, but not like a “must-have” condition.

Sorry, ascii pictures corrupted, they should be like:

H1H2 | H1H2 | H1H2 |

H2H1 | => H2H1 | => H2H1 |

HL | HL |

LH |

No, it’s perfectly clear.

The question is perfectly clear imo.

Two coins have four sides between them. Consider 4 tokens instead, one black and three white.

You pick out two tokens and look at one. It is white (like looking at a “heads”). The probability of the other, as yet unseen, token being also white is 2/3. You picked the tokens together, as you do the two sides of a coin but can only see one token / side when calculating the probability.

Because there are three unseen tokens /sides, the probability must be x/3 and CANNOT be 1/2, as this would be “1.5/3”. Remember, there are 3 heads and 1 tails, not 2.5 heads and 1.5 tails on the two coins!

Yes. I see.

Essentially the way I analyzed it too. You look at one side of one coin, how many sides are unobserved? Three, two of which are heads.

I think that the Big-Endians are winning the argument at present (see Gulliver’s Travels).

I appreciate that this is not a great parallel, in a way, in that there is a “correct” answer to this puzzle, being 50% (just kidding)

TMT

Intuition says there are 2 possibilities: either heads or tails. Then in school you get taught to count the number of options for each, resulting in 2 heads out of 3 options. It is so counter-intuitive that people will get into very heated debates about it.

At the risk of being on the cover of “Psychology Today”, Richard has already been here with the “2nd Daughter Problem”. Both involve “non-conditional” probabilities: “Given a couple with two children, one of whom is a girl, what is the probability both are girls?”

2/3! Richard is right!

If are were told that the first child (or the one with freckles) is a girl, the probability that the second child is also a girl is 1/2. Likewise, if Richard said that the head you looked at belonged to the two-headed coin the probability is 1 that the other side is also a head. If he said the head you looked at belonged to the fair coin, the probability that the other side is also a head is zero.

Knew this one. It’s a variation of the 4 door game were a prize is begin one door. You select a door and then are shown that the prize is not behind another door. You are then offered to change your choice. You would think it shouldn’t make a difference but probability wise you re better off changing choices.

I knew this would generate a lot of discussion. Those who still think the answer is 50/50, try it. Put two pennies in a bag and mark one of the tails (with a little scratch or magic marker) indicating it should be a heads. Shake the bag and remove a coin and observe one side. If it’s tails (the one you didn’t mark), don’t count that trial–we’re only interested in cases where you see a head. Otherwise, turn the coin over and record whether it’s heads or tails. Repeat 100 times, and see if the number of heads you get is closer to 50 (1/2) or 67 (2/3).

Thank you Ken! This explanation convinces me that the answer is indeed 2/3. The point is, that you eliminate the cases where you see Tail. So, you look only at 75% of the potential outcomes. And of those you see 2 in 3 times a Head.

If you are not convinced that the answer is 2/3 (like I was, see my comments above!), run the experiment that Ken suggested yourself:

https://github.com/scharf/friday-puzzle/tree/master/2014-01-17

Interesting followup, Michael. Thanks for posting. Now I’m distracted by your github site and the past Friday puzzles…🙂

…Ken

Another variation of the same puzzle:

You have a bag, it has either a white or a black ball in it (you don’t know which). You add a white ball to the bag, you then remove a ball at random. It’s a white ball. What’s the probability that the ball left in the bag is white?

I think Richard posted this puzzle a couple of years ago, or something like it.

Much debate ensued – not all of it civil.

Anon: If memory serves, this is the first *civil* exchange of opinions in the last six months or so.

Those who believe the answer to Richard’s question is 2/3 are answering the question:

What is the possibility of

1. drawing and seeing a head on one coin

AND

2. there being a head on the other side.

In this case I agree that the answer is 2/3. No problem, but that is not the question that was asked.

BUT Richard said:

You have drawn a coin and looked at one side.

It is a head.

(If you saw a tail there is no question)

His question, at this point, is

What is the possibility of the other side being a head?

You have in your hand either the double-headed coin or the normal coin, so the answer is 1/2.

I agree with this completely (as I mention above).

The answer to both your questions is 2/3.

In your 1st question you’ve actually asked 2 questions:

1) What is the possibility of drawing a coin and seeing a head? Answer is 3/4

2) Given 1) what is the possibility the other side of the coin is a head?. Answer is (1/2)/(3/4) = 2/3

And the answer to your 2nd question is also 2/3.

The probability you selected the head/head coin AND see a head is 1/2*1 = 1/2

The probability you selected the head/tail coin AND see a head is 1/2*1/2 = 1/4.

Since you are twice as likely to see a head from the head/head coin as you are likely to see a head from the head/tail coin, you are twice as likely to be looking at the head/head coin. The answer is 2/3.

Your solution of 1/2 doesn’t logically or mathematically follow from the premise.

Palmer: This line in your message:

<>

is not valid because you already have a head, according to the wording.

But I suspect we’re dragging this on now, so let’s leave it.

Anne Elk.

The answer is only 1/2 if you were to pick a coin at random, examine both sides, and place the coin on a table with the heads side showing, and then ask the question But that’s not the question being asked.

Just because you see a heads doesn’t mean you couldn’t have seen a tails. There is nothing invalid in my last comment.

This problem is an example of Bertrand’s Box Paradox (but without having a coin with tails on both sides) – as is the Monty Hall Problem, the 3 Prisoner Problem, and numerous others.

Oddly, my previous message did not quote your line. I’ll try again. It’s this particular line –

“The probability you selected the head/head coin AND see a head is 1/2*1 = 1/2”

that is not relevant to this puzzle. (In case it doesn’t come out again, I’m quoting the second sentence of your second paragraph.)

Still, no harm done either way. I think we’re probably boring other people now so let’s call it a day, shall we? You’re entitled to your opinion and so am I.

It’s not my opinion, it’s the mathematically correct answer.

Let’s meet up and play a little game. We’ll do this 100 trials, if heads is exposed on first choice, each time the other side comes up tails, I give you ten dollars, each time the other side is heads, you give me six dollars. Deal?

this is probably too late for anyone to notice, but the fact that you’re looking at a head gives you some indication of what coin you’ve picked.

The fact that you have get a head do not give you any information on the last coin.

So the last coin could be EQUALLY either single headed or two-headed.

So the possible events are :

You get the face A of the double-headed coin which is Head

You get the face B of the double headed coin which is Head

You get the face A of the single headed coin which is Head

You get the face B of the single headed coin which is Tail

So the probability is 3/4 !

Do am I wrong ?

yes.

well looks like this time richard explained in detail

The problem statement is not clear. If it’s a repeatable experiment, it’s 2/3 because you will pick up coins, and each time the coin you pick is “tails”, you discard it, so yeah, you don’t count that fourth possibility, every time you pick a “heads” coin, 2 out of 3 will be the 2 headed.

But if it’s a declaration, a non repeatable event (You pick a coin, and Voila! coincidence! it’s heads), the chances are 50%, the 4th possibility was still there, but this time it didn’t occur. Because what do you see doesn’t affect the final result, the coin that you picked up was chosen without any preference over the other one.

You have a coin, you see that it’s heads, but you don’t know wich coin do you have, it could be any. The chances are 50% because when the “random” event happened, you knew the same that yoy know now, “it could be any”. And the original random event was 50%

The problem statement is perfectly clear imo : you pick a coin at random, see 1 side only, it’s heads. What’s unclear about it?

However, your answer is confusing. You say that over repeated trials the answer is 2/3, but for any individual trial it’s 1/2. How does that work?

In say 400 trials, you’ll see tails 100 times. Of the 300 times you see heads 200 will be the heads/heads coin. In a single trial, given that you see a head the probability you’re looking at the heads/heads coin is 2/3 – it’s definitely not 1/2

The answer could only be 1/2 if you looked at both sides of the coin and always placed it heads side up, then in 400 trials there would always be a heads showing, 200 from the heads/heads coin and 200 from the heads/tails coin – but that’s not the problem statement.

The answer to the problem statement is clearly 2/3.

Way behing the curve here, but I thought a comment would be helpful. What is probability? It is how LIKELY something is to happen which is determined by considering how many ways an event can happen divided by the total number of outcomes. With one head showing, there are three possible outcomes (two heads and one tails). We are looking for heads, so the probability is 2/3. Again, we are talking about PROBABILITY. If you are discussing probability as we are here, there is no other option.

The confusion with this problem is a lack of understanding the nature of probability. This is a specific question with a specific answer.

This problem can be viewed as if we take out a coin and see that ite one face contains head then what is probability that we have choosen double headed coin(because then only other face will be head)

so if one face is head then probability that we have choosen double headed coin is 2/3 Because double headed coin has 2 Heads while normal coin has 1 Head.

hence the probability is 2/3 Surly.

i think this will help.