Problem 16 Solution

Yes, the answer is 2/3. Numerous people have tried to explain why they think the answer is 1/2, arguing that since both coins have a head then seeing a head doesn't rule out anything and thus it could be either coin with equal probability. While it is true that seeing a head does not rule out anything it still provides valuable information that can not be ignorred.

To illustrate why the 1/2 answer is wrong let propose a similar problem. Suppose a soup company canned 2,000,000 cans of tomato soup. Suppose further that they put the wrong label on half of the cans. The company then sent 999,999 correct cans and 1 incorrect can to one store and 999,999 incorrect cans and 1 correct can to another store. To determine which store has the correct cans you go to one store at random and open one can. You see it does indeed have tomato soup inside. What is the probability you went to the store with 999,999 correct cans? It is not reasonable to assume the probability is 50/50 that you are in either store. The odds that you would pick the one correct can in the store with 999,999 incorrect cans are extremely remote. It is true that before opening a can the odds were 50/50 but once you do open a can that information gives you a clue that can not be ignorred. In fact the probability that you are in the store with the correctly labled cans is 99.9999%.

Now let me introduce Bayes' theorem on conditional probability to explain why the answer is 2/3:

Let the events A1, ...,Ak for a partition of the space S such that Pr(Aj > 0 for j=1,...,k and let B be any event such that Pr(B) > 0. Then, for i=1,...,k,

Pr(Ai | B) = Pr(Ai) * Pr(B|Ai) / [ Σ (for j=1 to k) Pr(Aj)*Pr(B|Aj) ] .

For the problem in question we can reword this to: Pr(two headed coin|head chosen) = Pr(choosing two headed coin) * Pr(choosing heads given that two headed coin was chosen) / [Pr(choosing two headed coin) * Pr(choosing heads given that two headed coin was chosen) + Pr(choosing one headed coin) * Pr(choosing heads given that one headed coin was chosen)] =

1/2 * 1 / [ (1/2 * 1) + (1/2 * 1/2) ] =

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

If this all went over your head think if it another way. There were 3 heads to begin with. 2 of them have another head on the other side. You could have chosen any head with equal probability and since 2 out of 3 have a head on the reverse side the answer is 2/3. Of course this answer would probably not merit very much in partial credit in probability class.

Some other people have questioned my exact wording of this problem, mainly objecting over the tense of the verb I use for when the coin is chosen. What appears now was taken almost exactly from a similar problem in Probability and Statistics (second edition) by Morris H. Degroot on page 63, problem number 5. I changed cards to coins and eliminated what would be the coin with two tails. The book is a commonly used college text on the subject and should be above reproach. Here is how they stated a similar problem:

A box contains three cards. One card is red on both sides, one card is green on both sides, and one card is red on one side and green on the other. One card is selected from the box at random, and the color on one side is observed. If this side is green, what is the probability taht the other side of the card is also green?

Essentially the same problem, but worded using sunken boats, secret compartments, and bars of gold, is asked on page 603 of Statistics for Business and Economics by Edwin Mansfield.

A reader sent in the following explanation that he thought might help:

You have probably received similar e-mails to the one I am now sending, but
just in case, ...

Perhaps the best way to demonstrate that 2/3 is the correct answer to this
problem is to approach it in a simple intuitive manner, such as:

Denote the double headed coin A and the regular coin B.  Further, denote
the sides of the double headed coin A1 (a head) and A2 (another head), and
the sides of the regular coin B1 (a head) and B2 (a tail).

We then have the following possible outcomes:

coin selected           side shown              other side
         A                 A1 (H)                   A2 (H)
         A                 A2 (H)                   A1 (H)
         B                 B1 (H)                   B2 (T)
         B                 B2 (T)                   B1 (H)

We exclude the last possibility from consideration since it is not a head.
Three possibilities remain, all equally likely, two of which have a head on
the other side.  The required probability is thus 2/3.


Here is my favotite e-mail I received on this problem. I x'd out some private information.

Dear Michael,

I am a biology teacher at Xxxxxxx High School in Xxxxxxx, MA. I recently was searching for an extra credit question to put on a test for tenth graders. I used the three coins in a bag problem (#16). Unfortunately, none of my students got the correct answer. They all answered 1/2. I have done my best to explain the problem and some of the students now understand why the correct answer is 2/3. The debate has spread throughout the whole school. Most teachers and students beleive the answer to be 1/2. So I started an organization called Team 2/3 - The Probability Masters. Students take an oath to become official members of the team. The mission of the team is to recruit as many new members as possible. Team members get a membership card that has your original question on the back which helps them recruit new team members. Many people think I have gone off the deep end, but I find it amazing that so many people including most teachers and the principal still do not see the truth. If fact, the day after I started the Team 2/3, the principal got on the morning announcements and stated the the answer is 1/2 and the group that thinks otherwise should rethink the problem.

One student created a computer program that simulates 10,000 coin picks in 1 second. The data clearly show the 2/3 answer. But that same student insists that the probability is 2/3 only if many trials are conducted. He doesn't understand that the probability is the same in the first and last trial. People are stuck on the idea that you must have either the H-H or H-T coin and say it is 1 out of two. I have tried to simplify your solution without the T-T coin to help people understand.

If two coins: H-H, H-T

Conditional Prob = 1/2 divided by 3/4 = 2/3

Team 2/3 will recruit for two more days and then reveal the truth. We will try to get the principal to announce that Team 2/3 was correct all along. All team members will hold their heads high if that happens. I only have one math teacher on my side at this time. One of my student team members had a hat made for me that says "Team 2/3 Captain". I wore it on Friday and will wear it this week. We have almost everyone in the school discussing your coin problem. Many of the students are giving it a great deal of thought.

We are very confident that the answer is 2/3, however most people still disagree.

It would be most helpful if you would place a call to the school and talk to Mr. Xxxxxxx Xxxxxxxxx, the principal. This would be greatly appreciated by the team, your team, Team 2/3.

Xxxxxxx High School 1-(xxx)-xxx-xxxx


Troy X. Xxxxxxx
Biology Teacher P.S. Please feel free to call me at home: 1-(xxx)-xxx-xxxx

I did call the principal and tried to explain why the answer was 2/3. He seemed pretty entertained that somebody would call all the way from Baltimore about this. After going over the problem he seemed to partially understand but added that "he didn't want to be confused with facts." Later the Biology teacher wrote me back, here is his second e-mail:

I really appreciate you talking to the principal. He did talk to me afterward and he seemed very unsure of the correct answer. Team 2/3 has 42 oath-taking official members including an English teacher and the guidance secretary. We allowed people to join unofficially by signing the unofficial membership list. That way they still get a membership card without taking the oath; it's our "chicken list". The Latin teacher, the US history techer, and the alternative education teacher are on that list. Our membership card states-"TEAM 2/3 the Probability Masters" and there is a picture of a man holding two coins. On the back of the card is your original problem. Another team formed (not as organized as our team) called Team 1/2. They also have membership cards. Just this past Thursday, a good week after your call, I reminded the principal that we needed closure to the debate. I just assumed he believe the answer was 2/3. And I told him it would be nice to recognize the courage of team 2/3 and congratulate them. He got on the announcements and stated "After consulting with several mathematical experts" (and other people heard about the controversy and contacted him with explanations for 2/3) "the answer to the coin probability problem is 'probably' 2/3, so congratulations to Team 2/3, you are probably correct. While we would have liked a stronger statement(without the probablies), we have accepted the principal's announcement as victory and feel that we have been successful in our mission.

Thanks for your help,


Michael Shackleford, A.S.A.