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More ways to get Braingle...

Two Coins

Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.


Puzzle ID:#29228
Fun:*** (2.67)
Difficulty:** (1.77)
Submitted By:lips*****
Corrected By:Winner4600




Suppose I flip two coins without letting you see the outcome, and I tell you that at least one of the coins came up heads. What is the probability that the other coin is also heads?



For two coins, there are four possible outcomes: HH, HT, TH, and TT. Since we know that at least one was a head, we can eliminate TT. Of the remaining three possibilities, only 1 allows the second head: HH.


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Mar 21, 2006

I can't agree with this. The possible outcomes of two coin flips are HH, HT, TH, TT. Since the first flip is known to be H, the possible outcomes are now HT and HH. You can't claim TH to be valid any more. It turns into a single coin flip.
Mar 21, 2006

He didn't say that the first was heads, he said at least one was heads. There are indeed three equally-likely ways to get "at least one" with heads, and only one of these has two heads. The answer is 1/3.
Mar 21, 2006

This is analogous to the dogs question, so it shouldn't really be here.
If you don't know which of the coins is a head, the chance of getting 2 heads is 1/3.
I've tested it with a computer program, and the result just keeps getting closer to 0.33333 the more experiments are done.
You can use a tree driagram too.
Basically after there are 4 options HH,HT,TH,TT, each with a 1/4 chance of occuring, but since TT must be ruled out, we only have 3 equal options HH,HT,TH, each with a 1/3 chance.
Mar 22, 2006

i think the answer would have to be 1/4. as you have said, there are 4 possible combinations. and knowing that at least one coin turned up heads doesnt change the probability of getting HH in flipping 2 coins.

we cant just eliminate TT from the equation bcoz before you flipped the coins, TT has 1/4 chance of turning up. and it cant be 1/2 either (no, it doesnt turn into a single coin flip), bcoz its like saying that the first coin is sure to turned up heads.

but if you said right from the start that the coins are somehow biased that at least 1 coin will always turned up heads, then i would agree that the probability of getting HH is 1/3.

otherwise if getting HH, HT, TH or TT have equal chances, then it should be 1/4. (the 1st coin has 1/2 chance of turning up H, and the other have 1/2 chance also. so 1/2 x 1/2 = 1/4!)

(thats what i learned from joining math contests back in elementary and high school . if i remembered correctly, this is just a trick question! it makes you think of other answers when in fact, the answer is quite simple.)
Mar 24, 2006

wow, i didnt follow any of those comments! still good teaser
Mar 24, 2006

[1] This is a duplicate of the teaser "Bar Room Toss".
[2] It should be stated "What is the probability that they are both heads?" because once you say "The other one" it implies that you are indicating that a particular coin is a head. In that case the answer would be 1/2.
[3] Provided the wording is changed to "both heads" instead of "the other one is a head" then the answer is 1/3.
Mar 25, 2006

So much ado about a simple textbook case in simple probability 101; it is 1 in 3 chances.
Mar 27, 2006

I take back point [2]. In this context, "The other one is a head" and "both heads" are equivalent. Sorry!
Mar 27, 2006

why not 1/2? i think we can assume that there's only one coin which need to be tossed.. i dunt really undrstand (i'm a beginner).. can anybody explain? thx.
Mar 28, 2006

Yeh, because basically you don't know which of the 2 coins is a head to start with.
There are 3 equally likely options:
Mar 30, 2006

Mar 30, 2006

It would be 1/2 because the events are independent of eachother. One occurence does not influence the other
Apr 04, 2006

Okay, here is a clear explanation why 1/3 is correct.

Lets start from the beginning. The chance of getting two heads (HH) if you flip two coins is 1/4. This is true because you have four possible outcomes HH, HT, TH, TT. Mathematically, if you have a 1/2 chance of each coin landing H, then 1/2 * 1/2 = 1/4

The question posed is this: What is the chance of having two heads (HH) if I tell you that the outcome is NOT two tails (TT)?

[This is the EXACT same question as the teaser, worded differently]

This now means we have three possibilities HH, HT, TH. The chances of getting HH, of these three possibilities, is 1/3.

Basically what is happening here, is 1/4 of the data is being thrown out, that is all occurences of TT are not counted, bringing the total possibilities to 3.

This is NOT the same as the following question: If I have a coin that is heads, what is the chance of flipping another coin and having two heads?

In that case, the chance would be 1/2.

The difference here is simple, you are eliminating the chance of coin A landing tails, leaving the possibilities HH and HT.

The main point of confusion here is the fact that HT and TH are NOT equal. These are two distinct events, even though they appear the same.

Try this at home using a dime and a penny, keeping track of what each coin does independently.

Hope this helps.
(user deleted)
Apr 06, 2006

I totally agree with the comment given by PATH, since, earlier, i was also confused with the same stuff.

A nice teaser
Apr 07, 2006

..... and if you don't believe it.. test it.
Apr 09, 2006

The answer is definately one half. The basic problem in the concept of the teaser is that HT and TH are the same exact thing. If one of the coins is head, then the other coin being heads or tails is unaffected. There is still, no matter what, a 1/2 chance of the other being heads. This teaser is not really s teaser but a paradox that seems to confuse people with misinformation such as the teaser with the three men going into a hotel and there being a missing dollar. Sorry...
Apr 10, 2006


Please take another moment to think about your logic. HT and TH are actually not the same thing. Ask yourself the following questions:

What is the chance of getting HH?
1/2 x 1/2 = 1/4

What is the chance of getting TT?
1/2 x 1/2 = 1/4

What is the chance of getting one heads and one tails? This is just 100% - 25% - 25% (from above)
1 - 1/4 - 1/4 = 1/2.

As you can see, you are twice as likely to get heads and tails combined, because there are two ways to accomplish it.

Use Excel to generate random 1s and 0s in two columns to model this situation. It is a great proof by experiment.
Apr 13, 2006

I disagree with the answer. The probability would be the same if flipping one coin which is 1/2...
Apr 24, 2006

The answer would be 1/2 for the coin to land on heads. (Two events)

It would be 1/3 for both coins to land on heads
(one event)

You need to reword your question
Apr 28, 2006

Thanks for the explanation, PatH... it was very helpful. I think I got it now!
May 02, 2006

1 out of 2
May 04, 2006


Your comment is snotty and ignorant, provide some explanation for your claim if you want to be taken seriously.
May 16, 2006

This is similar to the following:
A prize is behind 1 of 3 doors. After you pick a door, the host opens 1 of the other doors, which does NOT contain the prize. If you are allowed to change your choice to the other unopened door, should you do so? The answer is YES. By changing, you have a 1/2 chance of winning. If not, then it is 1/3.
(from Ask Marilyn in Parade.)
May 17, 2006

rock57 -- two comments. (1) In the games show example you use, the probability of winning if you switch is 2/3, and if you stay is 1/3. Otherwise, the sum of the probabilities would be less than 1, which makes no sense. (2) This is not really the same issue. In the game show, there is asymmetric information which allows the host to show you "the goat"; here, there is just the elimination of the TT case, but not the same shift from a priori probabilities.
May 17, 2006

Of course you are correct. (I had a brain cramp)
May 22, 2006

The solution 1/3 is correct. There are only two coins being tossed. Since there are only four possibilities, write them down on a piece of paper. Cross out the two tails possibility, since we are told at least one coin is heads. See what you have left. Walk through the logic and come up with the answer.
May 23, 2006

This conversation has been amusing to read. Both people on both sides are so sincerely addiment that they are correct.
Basically you both could be right if the wording was slightly different in several various ways in 3 areas, the tense of the questions and the detail of knowledge of the outcome and the level of detail needed to be seen in result given.

I challenge you to toss 2 coins (and repeat this exercise 20+ times). If at least one of the coins comes up Heads (as the riddle states has happened, we officially know at least one of the coins is a Head) then you are able to use the result in this experiment
Now, mark down if the face of 'THE OTHER' coin was heads or tails.

This is a practical example that does what the question states. You will find the result. The result will surprise half of you!

I haven't time to explain tonight why both arguments are so strong but one is incorrect, maybe later in the week if people still doubt this experiment.

Probability is so simple but can easily deceive with its presentation.

May 26, 2006

I agree with Sword of Fury, kind-of. HT and TH are the same thing, but they don't then comprise 1/2 of the remaining chances, but 2/3 of the remaining chances. So the 1/3 solution is still correct.
May 29, 2006

This seemed like such a nice, and correctly worded, teaser to create such controversy. For 2 coin tosses there are four outcomes. 3 of these include at least one head. We are told there is at least one head. Only one of the outcomes has a second head. 1 in 3 is the correct and only answer. The 2 tosses had already been made. We are not asked if a head has just been tossed what's the probablilty of a second head. What's the probablilty of everyone getting their heads around this? Very slim
Jun 04, 2006

Seemed like 1/2 after the fisrt read, but 1/3 is certainly the correct answer.
Jun 17, 2006

The problem with 1/3 is that one coin toss has already been decided. The prompot told us "that at least one of the coins came up heads". Thus either toss 1 or toss 2 is heads before we start the problem, so what is the probability that "the other coin is also heads". To me, the language suggests we're considering only 1 flip. The "also", especially, implies that one coin is already set as heads and we're dealing with the other coin.
Jun 20, 2006

By changing the wording to: "What is the probability that both coins came up heads" you would eliminate a lot of confusion.....however it is true that the answer is 1/3 because you're not looking at two separate events but rather one sample space that includes two events.
Jul 01, 2006

After reading this again I'm convinced that the way you worded it is wrong, and that it should be 1/2.
Jul 20, 2006

I disagree with this. I think it should be a 1 in 2 chance. The question is worded poorly.

'Suppose I flip two coins without letting you see the outcome, and I tell you that at least one of the coins came up heads. What is the probability that the other coin is also heads?'

It doesn't matter that the first coin was heads. That doesn't have anything to do with the equation. The question is: What is the probability that the other coin will land on heads.

I understand what the author means with the answer explanation, but I think you should rethink how you wrote that.
Jul 20, 2006

It is a good teaser, but it has multiple solutions. I looked at it again and I could see how you said that it was 1 in 3.

By saying that at least one of the coins came up heads, does that mean there is a 1 in 2 chance that it also came up both heads?

If this applies, then I think it would actually be a 1 in 1.5 chance of getting both heads. I'm not too sure about this strategy, though.
Aug 07, 2006

There really is not controversy here. This one is easy to do on paper because there are not many possible out comes. So lets just go through the answer with a little more detail.

There are four possibilities when flipping both coins. They are:

Coin 1 | Coin 2

Heads | Heads
Heads | Tails
Tails | Heads
Tails | Tails

Since at least one of the coins is heads, Tails|Tails can be eliminated leaving you with three possible coin flips that fit this scenario:

Coin 1 | Coin 2

Heads | Heads
Heads | Tails
Tails | Heads

Now we need to find out how many of the three above possible coin flips where "the other coin is also heads." In other words, where both coins come up heads. When we look at the possible coin flips, there is only one where both are heads. So the answer is one out of three, or 1/3. This may seem counterintuitive, but it is the correct answer nonetheless.
Aug 09, 2006

Okay here is a nice sample space, of tosses for a pair of coins, with 78 entrants


19 of these are TT so they cannot be used. That leaves us with 59
Of those 59, 21 of them are double heads. 21/59 is almost 36% very close to 1/3 not so close to 1/2

If you don't trust me try it yourself.
Aug 26, 2006

I don't see the difference between the combination HT and TH. It doesn't matter. This means the only other chances are HH or HT. I don't think it is 1/3.

Plus this is a duplicate of #3230 (Dogs).
Aug 27, 2006

grungy, there are TWO coins. Say Coin A is Heads and Coin B is Tails. That is different from Coin A being Tails and Coin B being Heads.
Oct 21, 2006

I knew when I read this teaser that there would be controversy. Probablility teasers can do that sometimes. I knew it would be more fun to read everyone's comments than to actually solve it.

Consider these different ways it could have been worded:

A. Suppose I flip a coin and it came up heads. Then I flip another coint. What is the probability that this 2nd coin is also heads?

B. Suppose I flip two coins without letting you see the outcome, and then I tell you that the first coin came up heads. What is the probability that the other coin is also heads?

C. Suppose I flip two coins without letting you see the outcome, and I tell you that at least one of the coins came up heads. What is the probability that the other coin is also heads?

D. Suppose I flip a dime and a penny without letting you see the outcome, and I tell you that the dime and the penny were not both tails. What is the probability that both the dime and the penny turned up heads?

The answer to A and B is 1/2. The answer to C and D is 1/3. Take a look at which one was the actual wording.

Does anyone still think that saying "the other coin is also heads" if different than saying "both coins are heads"? They amount to the same thing, right?
Oct 21, 2006

You could also replace "at least one of the coins" with "either the first or second coin" on C.
Feb 06, 2007

For any of you still working on this one. The main thing to remember is that when you flip two coins, you are twice as likely to get a heads and a tales, than you are to get two of a kind.

This is true because the heads and tales combination can be accomplished two ways, while heads/heads and tales/tales can only be accomplished one way.

This is the FIRST part that has to be understood in solving this puzzle.

Also, there is no excuse for not TRYING this at home, it takes only moments with a couple of coins to prove out this concept (flip at least 25 times).

(user deleted)
Jun 21, 2007

Come on now. The first coin is already chosen, there are only two sides to the second coin. It is without a doubt 50%
Jun 23, 2007

at first, I thought the answer was 1/4, and reading the answer, I thought the answer was wrong, but having it argued in here, I am convinced that not only is the answer correct, but the question is phrased properly. the people who believe it is 1/2 are reading too much into the question. I'll explain it one event at a time, listing ALL possible results from each event
EVENT 1: Suppose I flip two coins without letting you see the outcome,
**results: HH, HT, TH, TT
EVENT 2: and I tell you that at least one of the coins came up heads.
**results: HH, HT, TH
**alternately stated: for a fact, one coin is heads, the other coin is unknown at this time.
**it is not stated that the coin in this event is the first coin, or the second coin. this is the inverse-parallel of the "I have two coins that equal $0.30 in my pocket. one of them is not a nickel, what are the two coins?" question
EVENT 3: What is the probability that the other coin is also heads?
**results: HH.
**this is the only answer that satisfies all three events. I originally thought it was 1/4 because I had included TT as a possible result, but it is explicitly excluded in event 2
Aug 28, 2007

I also don't agree with this teaser because although 1 coin has flipped heads, the other coin doesn't depend on the first. THE SECOND COIN IS BASED ON AN INDEPENDENT PROBABILITY. If it is 50% to get heads, then the answer is 1/2.
Feb 06, 2008

This question is worded wrong. The results of the first coin is TOTALLY irrelevant. If P=O/E, then P= 1 possible outcome out of 2 expected (it can only be Heads or Tails) 1/2 or 50% is the correct answer. Even if the first coin was tails, the answer on the second coin would still be 1/2.
Feb 06, 2008

srost24 -- The question is worded correctly, and the answer given is the right answer. Think instead of the case where I have 10,000 fair coins, and I flip them all, and I tell you NOT that the first 9,999 came up heads, but that I have "at least 9,999 heads". What is the probability that they are all heads? By Bayes Theorem, it is 1/10,001. This is that same problem, only severely reduced.
Nov 24, 2008

I'm sorry but you got this one wrong. Your could fix it by changing it just a little. You could say, "I toss the two coins and then you ask me, is at least one a head, to which I truthfully reply, yes".

Now the probability for two heads is 1/3 the way you state it 1/2 is the better answer.
Nov 26, 2008

There is a similar puzzle about a nearsighted gambler. He tosses the dice to the other end of the crap table but he can't see that far. So he asks, is there at least one 5 down there?

Siomeone answeres, "yes".

The probability for two 5's is now 1/11

But, if he asked instead, "What do do see down there"? And someone answered, "I see at least one 5".

Then the probability for two 5's is 1/6
Nov 28, 2008

I hate to be just one more drop in the 1/2 bucket, but there is a severe flaw in the 1/3 logic. The users who created a computer simulation must have built this flaw into their programs.

The flaw is in assuming that the player of this game knows that the possibility of TT is 0%. That is, you assume the person who flips the coin has somehow rigged the game so that TT would not occur.

Here is another way to explain this logic flaw. Say the coin flipping person decided to play this game with several different people. He played only one time with each person, and the players were not aware of his previous rounds with other players. Every round he plays this game, he announces that one of the coins is heads or that one is tails. He can't rig the game, so TT is a valid possibility. Therefore, he is sometimes forced to tell the player "at least one of the coins came up tails." By the "1/3" logic, each of the players should choose the answer opposite to what the coin flipper says. If he says "there's at least one heads" then the player should choose tails by eliminating TT from the four possibilities. If the flipper says "there's at least one tails" then vice-versa.

Here is the interesting result. The "1/3" logic tells us that 2/3 of the time, the player will be safe with the opposite choice. In other words, 2/3 of the time, the flipper (by chance) flipped both coins opposite (a TH or HT), and only 1/3 of the time flipped both coins the same (a TT or HH). But we know that that can't be. Despite the fact that the flipper announces one of the coins, he can't change the fact that 1/2 of the time both coins will come up the same.

Since we can't accept the outcome of this "1/3" logic, we have to look at the problem differently.

Some users have stated that there are 4 equal possibilities, HH, TT, TH, and HT, and that saying "at least one is heads" eliminates only TT. However, it is correct to eliminate both TT and TH. The reason is that saying "one is heads" is the same as labeling the coins as "one" and "the other". In fact, the coin flipper may as well show you the coin that is the "heads" coin to clearly demonstrate that the two coin flips are independent events. Then you can label the coins "left and right" "one and two" or whatever. If you've labeled them, then I think everyone can agree that TT and TH must both be eliminated.
Nov 28, 2008

M/duke got it just right. The proponents of the 1/3 answer assume the flipper of the coins simply ignores the instances when TT occurs. He simply skips right past those TT tosses and flips again and again until he in fact can announce that he has indeed tossed at least one head.
Nov 28, 2008

Sorry I got the name wrong, it's mrmanuke
Dec 21, 2008

The problem with the wording is that we have to assume that what we've been told about the outcome is arbitrary.

If we are always told there is at least on tails when there are any tails, then the second coin will always be heads.

If what we are told always depends on the first coin flipped, then the answer is 1/2.

If which coin we are given information about is chosen randomly, then the answer is 1/3. The problem doesn't specify how the information were given is chosen, so there isn't enough information to answer the question.
Dec 22, 2008


It's just like the puzzle where you meet a woman at the airport and learn through conversation that she is the mother of exactly two children.

In the firsst instance, you might ask her if at least one of her children is a boy.

Or, in the second instance, you might ask her to disclose the sex of one of her children, chosen at random, from the two.

You learn she is the mother of at least one boy.

Now the answer is 1/3 in the first instance and 1/2 in the second.
Jan 27, 2010

Good puzzle for someone who's never studied probability. Otherwise, this is a very fundamental example that anyone who studied probability should easily be able to see.
Jul 26, 2010

In my opinion, it's 1/2.
It's worded poorly. It asks what the probability of the other coin getting a heads is. In that case, it doesn't matter what the first one was. Thus, making the answer 1/2. If it's not, someone please answer why not.
Jul 26, 2010

I'm sorry. I misunderstood the question. : )
(user deleted)
Sep 08, 2010

I toiled over this. After reading the comments and solutions, I finally will agree to 1/3. But it makes me ask the question. If flipping one coin twice and taking and disregarding all tt results, is the probability of hh still 1/3?
Sep 24, 2010

On revisit, I'd say this question is poorly phrased when it says "the other coin"
Oct 28, 2011

Mykulvee, yes it's still 1/3. If the first flip is T then the second flip has to be H to count as a trial (to keep tru to the scenario). BUT YOU STILL HAVE TO TRY B/C IF IT COMES OUT T FOR THE SECOND FLIP THEN YOU DON'T COUNT IT AS A TRIAL!

A lot of people get caught up on that. If the first flip is H, then the second is allowed to be either H or T. There's your 3 possible outcomes. TH...HT,HH. Remember people that TT is not a possible outcome as per the scenerio description.
Oct 28, 2011

This teaser is worded perfectly. Write your program like this:
1: flip two coins
2: are they both tails?
Oct 28, 2011

This teaser is worded perfectly. Write your program like this:
1: flip two coins
2: are they both tails?
Oct 28, 2011

This teaser is worded perfectly. Write your program like this:
1: Flip 2 coins
2: Is at least one of them heads?
Oct 28, 2011

Sorry I don't know why it keeps adding the comment before I'm done. Anyway,

1: flip two coins
2: if both are tails then goto step 1 because our scenario gives information that does not allow for TT situations.
3: add 1 to counter "X"
remark: we know at this point that one of them is H and this is therefore a valid trial so we add 1 to our valid trial counter "X"
4: if both are H then add 1 to counter "Y"
remark: counter "Y" represents the number of HH outcomes
5: goto step 1

Now set this program to run 100 times. X should come out to be close to 75 (TT came up 25 times) and Y should come out to be close to 25. Now take Y/X.

So there is a 1/3 chance that both coins are H when you know that one of them is H (the program knows that one of them is H if they are not both T). Simple and unbiased.

Run this on your TI-85 or 83 or whatever you have before you comment. You'll see.
Jan 07, 2012

@Zem, please see my earlier comment. BTW, Monte Carlo is not needed to solve this.
Jul 23, 2013

Just an update: the answer is 1/2. The 1/3 answer would normally be correct if part of the question related to the order of the coins, but alas it does not. The error with the program and doing it yourself is you all are treating TH and HT as separate entities. TH and HT should be treated as the same because in the end, the only answer that matters is if both coins are heads, or if one coin is tails. With this in mind, TH and HT are to be treated as the same, so the two possible outcomes are TH (HT), or HH, making the anser 1/2 or 50%.
Jul 23, 2013

No, plosky, TH and HT are NOT the same. They are separate events and need to be counted as such. This has been explained to death already in the comments; go back and read them. The correct answer is 1/3.
Aug 22, 2013

You need to take into account that one of the two coins has a,
Chance to be Tails
Nov 06, 2013

TH and HT are treated the same because of conditional probability. Normally, the chance of two coins both landing heads would be 1/4 because Coin A has 50% chance to be heads, and Coin B has 50% chance to be heads (1/2 * 1/2 = 1/4).

However, since one of the coins is already heads, we can take it out of the equation.

There are two possible outcomes:
So, Coin A has 100% chance to be heads, and Coin B has 50% chance to be heads (1/2 * 1 = 1/2).


Coin B has 100% chance to be heads, and Coin A has 50% chance to be heads (1/2 * 1 = 1/2).

Order does not matter because of conditional probability, which makes one coin have 100% chance of being heads. It does not matter which coin it is, so the answer is 1/2, or 50%.
Nov 06, 2013

That's not how conditional probability works.

If X is "both coins are heads", and Y is "at least one coin is a head", then the problem becomes:

P(X given Y) = P(X) / P(Y)

P(X) = 1/4 [HH ht th tt]
P(Y) = 3/4 [HH HT TH tt]

So, P(X given Y) = 1/3.
Nov 07, 2013

Sorry, that should be P(X given Y) = P(X and Y) / P(Y). P(X and Y) is still equal to 1/4.
Dec 18, 2013

I swear you don't know what conditional probability is. Notice the word "conditional" before probability.

Let's use your example:
X is both coins are heads
Y is one coin is heads

X = HH th ht tt
Y = HH TH/HT tt

So, X/Y = 1/2

TH and HT are the exact same in this conditional probability problem because, for the thousandth time, TH and HT are the exact same thing.
Dec 21, 2013

If you flip 2 coins 100 times then: 25 times you'll get 2 heads, 50 times you'll get 1 head, and 25 times you'll get 0 heads (or 2 tails).
Of the 75 coin flips where at least 1 coin lands heads, both coins land heads in 25 flips (or 1/3).
Spikethru4's probability analysis is correct given the condition "at least one coin is a head",

However the condition isn't "at least one coin is a head" but "I tell you that at least one coin is a head". If the coins land heads and tails I could tell you ""at least one coin is a tail" instead.

Unless it is a requirement that I must tell you "heads" if the coins land mixed (presumably I don't tell you anything if they both land tails), and as no such requirement is specified it can't be assumed, then the answer is 1/2.
Jan 06, 2014

Plosky, you clearly need to revisit conditional probability. Here's some reading to get you started:

And I'd like you to show some source (other than your own stubbornness) to back up your claims that TH and HT are the same event. Conditional or otherwise, they are DIFFERENT events, albeit with the same result. If you are going to count them as the same, then you need to account for the fact that HT (in any order) is twice as likely as HH.

Palmer, this question has nothing to do with multiple trials. ONE pair of coins is flipped; there are four possible and equally likely outcomes. We are given extra information that eliminates one (and only one) of the outcomes. So, our desired result is one out of the three remaining outcomes.

Oh, and just so you can have some confidence that I do know what I'm talking about, I obtained my degree in Mathematics from Oxford University over twenty years ago. Doesn't mean I remember everything I was taught, but I can certainly do basic probability.
Jan 08, 2014

One trial or multiple trials, it makes no difference to the conditional probability calculation.
The information you have is " I TELL you that at least one of the coins came up heads".
Of the 4 possible outcomes from a pair of flipped coins, if:
a) HH Prob("I tell you at least 1 heads) = 1
b) HT Prob("I tell you at least 1 heads) = 1/2
c) TH Prob("I tell you at least 1 heads) = 1/2
d) TT Prob("I tell you at least 1 heads) = 0

So the answer is 1/2. The answer can only be 1/3 if I am required to tell you "at least one heads" if I flip HT or TH, since this isn't stated in the problem it can't be assumed.

Oh, and just so you can have some confidence that I do know what I'm talking about, I obtained my degree in Mathematics & Statistics from London University
Jan 09, 2014

I'm sorry Palmer, I don't understand any of your last post (apart from the bit about your degree). You seem to be adding in extra assumptions that are not part of the question. Nowhere is it defined what we would or would not have been told if two tails had come up; it's not relevant.

All we know is that two coins were flipped and that at least one is a head (presumably the flipper is telling the truth). As I said earlier, the question becomes, "What is the probability both coins are heads, GIVEN THAT at least one is a head?" and the conditional probability formula gives us an answer of 1/3.
Jan 09, 2014

Also, I am not assuming the flipper is required to tell me anything about the coins. I'm just working with what I have been told.
Jan 09, 2014

The condition isn't "GIVEN that at least one lands heads" (in which case the answer is 1/3) but "I TELL you that at least one lands heads". They are different conditions.
If I flip HT or TH I could tell you "at least one landed tails". You're assuming I can only tell you heads, I'm making no such assumption.

The problem with " just working with what I have been told." is that if you apply that reasoning to the Monty Hall Problem (I pick Door 1, Monty opens Door3) or the 3 Prisoner Problem (the warden tells Tom that Dick will be pardoned) you get an (incorrect) answer of 1/2 to both problems.

The standard solution to all of these types of problems (which are all variants of the much older Bertrand's Box Paradox) is to consider not only what you've been told, but also what else you could have been told . In the MHP, Monty could have opened Door2 (if you picked the car). the warden could have told Tom that Harry will be pardoned (if Tom was to be executed)
Jan 09, 2014

OK, thanks for clarifying your argument; I understand where you're coming from now.

You say you're not making assumptions about the condition, but I think you are. How do we know the probability of being told there is at least one head is 1/2? How do we know that the flipper *must* tell us something? We know that he *does*, which changes the telling from a random event to a certainty and means that what he tells us becomes the condition.

I'm pretty sure the MHP is based on the premise that Monty *must* offer to open a door, so it does not compare directly to this problem. However, I'm going to go back and re-read it.
Jan 09, 2014

As I see it, this question differs from MHP in a couple of ways:

Firstly, Monty opening a door and revealing a goat is defined in the rules of MHP, and he only has a choice of which door to open if you chose the door with the car. The information that at least one coin is heads is volunteered by the flipper and we have no way of knowing how likely he was to give that information as opposed to any other kind, including none at all.

Secondly, MHP asks "Is it in your interests to switch doors?", not "What is the probability you have chosen the car?". Different question, different answer.

If you remain convinced that the answer is 1/2, I would be interested to see your explanation of the events, their probabilities, and your derivation of the answer.
Mar 17, 2014

I really don't understand what's so hard about understanding this. The teaser says "at least one coin is heads". One coin is definetely heads. There is no changing this.

So, if we already know that one of the coins is heads, all we have left to do is figure out the other coin, right? Yes. The other coin can either be heads or tails. There is 50% chance it will be heads, and 50% chance it will be tails. So, overall it's 50%.

Now, you might say, "But what about the order? Doesn't that matter?" No, it does not. There is nothing in the teaser that states the order matters. The coin that is definitely heads could either be the first one, or the second one, but all the teaser wants to know is if both coins are heads.

So, one coin is heads for sure, and the other coin is not known. 50%.
Apr 02, 2014

Plosky, I understand it perfectly. You're the one having difficulty.

Two coins are tossed. Possible outcomes are:

2 heads - 25%
2 tails - 25%
1 head and 1 tail (in any order) - 50%

Note that "1 head and 1 tail" is TWICE AS LIKELY as "2 heads".

If "at least one coin is heads" then the possible outcomes are:

2 heads
1 head and 1 tail

These constitute 75% of the original outcomes; therefore, the probability of each is:

2 heads = 25%/75% = 1/3
1 head and one tail = 50%/75% = 2/3

Note, again, that "1 head and 1 tail" is still twice as likely as "2 heads". And therein lies the flaw in your argument, because you don't recognise that HT and TH are separate outcomes that MUST be counted twice.

Also, you cannot say "one coin is heads, so what is the other?", because you don't know WHICH coin is heads. So yes, the order DOES matter.

Let me make the point clearer.

I toss two coins, and put one in each hand (after looking at them). Now the outcomes are:
(left hand)-(right hand)

If I tell you that the coin in my left hand is a head, the probability that the other is also a head IS 1/2 because you can discard two outcomes (T-H and T-T).

If I tell you that at least one coin is a head, you can only discard one outcome (T-T) and the probability of a second head is 1/3.

Can you see that it doesn't matter where the two coins are physically located? There are always four outcomes and the condition "at least one is a head" only dismisses one of them.
Jun 21, 2014

Except there are NOT four possible outcomes. There are only two.

Outcome 1: One coin is heads (the one we already know to be heads), and the other (that we don't know) is tails.

Outcome 2: One coin is heads (the one we already know to be heads), and the other (that we don't know) is heads.

Two possible outcomes. These are literally the only possible outcomes knowing that one coin is already heads.
Jul 15, 2014

Plosky, your 2 outcomes are the same!

If you toss 2coins there are 3 possible outcomes, 2 heads, 2 tails, or 1 of each, and there's a 50% chance it'll be 1 of each. If you flip 2 coins 400 times, then 200 times you'll get 1 head and 1 tail.

If you pick 1 of those 400 flips such that at least 1 is heads (that is if both coins are tails then you pick again) the probability that both coins are heads is 1/3.
Try it.
Jul 15, 2014

Ignore the 1st line of my last post, I misread your comment.
Sep 04, 2014

Plosky, again, you miss the important point.


Outcomes are:
1. Coin 1 is Heads, coin 2 is Tails;
2. Coin 2 is Heads, coin 1 is Tails;
3. Both coins are Heads.
Dec 22, 2014

This is a variation of a very old problem given by Martin Gardner back in the 1950s called the Boy or Girl paradox. The strictly correct answer is that the question is underdetermined. Without some additional information about how we came to learn that one of the tosses is a head, in the event that one was and one wasn't, the answer could be 1/3 or 1/2 or anything in between. If the questioner has decided in advance that should one toss be H and one T then she will certainly inform us that at least one is a H, then the asnwer is 1/3. If the questioner has decided to choose at random which one to let us know, then the answer is 1/2.
Jan 10, 2015

Jesus you all are so dumb and cannot understand the independent mathematics.

The two coin flips, eve though they are at the same time, are independent of each other. That means that they IN NO WAY AFFECT EACH OTHER.

So, if they do not affect each other, then each coin flip is it's own independent event, and assuming we're not taking percentage of the coin landing on its side, then the coin can only land on heads or tails.

One coin flip we already know the outcome of, so it is ignored. (the coin that we are told is heads) The second coin flip can only land on heads or tails, and the first coin flip does not influence this coin flip at all, so it is 50%.

Stating that TH, HT and HH make it 33% does not make sense because the coin flips are independent of each other.

THE COIN FLIPS DO NOT AFFECT EACH OTHER. THERE ORDER DOES NOT MATTER. Honestly, take a probability class before posting here, please.
Jan 16, 2015

Congratulations, plosky, you've finally got something right.

Yes, the two coins are indeed independent events.

However, the relevance of this little nugget of information to the problem is, sadly, minimal.

I'm not quite sure what gives you the right to tell people to attend a probability class when you clearly have so little grasp of the subject yourself. Incidentally, though, since last I wrote here, I have, on teaching prac, attended several probability classes and, surprisingly, none of those classes even entertained the possibility that two heads, one head and one tail, and two tails are equally likely.

Your arguments here always start from an incorrect base. Your perception of probability is not the correct mathematical definition. Two independent coin tosses always provide FOUR equally likely outcomes. That is the way it is. No amount of you shooting your mouth off is going to change that. If you can find some independent research that challenges that then I would be very interested to read it.


Now, Dezaxa, on to you. We do not need any additional information about how we came to learn that at least one coin is heads. We are told that we are told this. The motives of the teller are not relevant. You might as well ask how was it decided that two coins would be tossed. You work with the facts, which are: (a) Two coins are tossed; (b) One, possibly two, came up heads. There's no need to invent other random events about which we know nothing.
Mar 01, 2015

In my opinion there is one more Event that must be count in the whole scenario and it`s the strength that force the coins to flip over. So if we flip the coins at the same time we will have 2 small events in 1 bigger ( the 1/3 answer of the question) ,but if we flip them one after another we will have 1 small event in 1 bigger and another 1 small event in 1 bigger,which makes them 2 totally independent events (the 1/2 answer of the question).
Mar 18, 2015

Dezaxa is quite correct in his analysis. If the questioner picks a coin at randon if the coins land Heads/Tails then the answer is 1/2, if he reports Heads in all cases then the answer is 1/3.

It is similar to the logic in the Ignorant Monty Hall problem - Monty still reveals a goat, but in this case there is no advantage in switching doors.

Both solutions are easily verifiable by manual or computer simulation, and proven mathematically by Bayes Theory
Feb 04, 2016

Palmer, what about if the questioner selected the number of coins to be tossed at random from a pot containing n coins?
Feb 11, 2016

spikethru4, it's unfortunate you dismissed Dezaxa's comment as irrelevant. Even though you are right in saying that it doesn't matter for this specific question, it does point at the heart of this controversy.

Smart researchers, as spikethru4 mentions, have been having the exact same debate we are having here, more than 70 years ago, and it would be a shame to reinvent the wheel.

" In response to reader criticism of the question posed in 1959, Gardner agreed that a precise formulation of the question is critical to getting different answers for question 1 and 2. Specifically, Gardner argued that a "failure to specify the randomizing procedure" could lead readers to interpret the question in two distinct ways:"

It's unlikely anyone can add any new information to this discussion, not already mentioned in the above wikipedia page.

Case closed?
Feb 11, 2016

Correction: 'as Dezaxa mentions'

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