# Probability of a two-headed coin given a few sample flips?

A probability problem was posed to me, and I am terrible at probability. It was first posed to me like this:

I have 5 coins. One of them is a two-headed coin, and the others are normal coins. I pick a coin, and then I flip that coin three times; each time is heads. What is the probability that the coin is two-headed?

My answer was, naturally, 1/5. I am still of the opinion that it's the correct answer to that particular phrasing (do you agree?). But of course I was told this is a wrong answer. After some thought, I came up with the following phrasing which is what was really meant:

You are given a coin and told that it has a 1/5 chance that it is double-headed, otherwise it is a normal coin. You are allowed three flips of the coin. Upon doing these, you receive three heads. What is the probability that your coin is double-headed?

Now I have two questions, first of all does that seem like a proper rewording to you? And secondly... What is the answer and the reasoning to reach that answer? I am terrible at probability, and while I can clearly see the problem and maybe spell out the first step or two, I am at a total loss as to how to arrive at an answer.

Edit: one more question, actually: is this an instance of the Monty Hall problem? I obviously see that it is very different, yet it feels somehow similar...

-
Why do you think the two phrasings are different? (Honest question.) Also, I disagree with your estimate of probability: here's why. –  Willie Wong Jan 27 '11 at 20:10
Well in the first case it seems to me that the probability is entirely dependent on the action of picking the coin, so I simplify it to just drawing one of five things from a bag at random. The second case seems different. Are they the same? –  Ricket Jan 27 '11 at 20:14
What if the problem had said, "I pick a coin, flip it, and get a tail. What is the probability that the coin is two-headed?" I suspect you would answer "zero" without hesitation. –  mjqxxxx Jan 27 '11 at 21:15
@mjqxxxx Good point but in my mind it still doesn't make logical the opposite case. Obviously flipping a coin and getting tails eliminates any chance of being a double-headed coin, but flipping a coin and getting heads doesn't seem to say anything about the other side of the coin. Flipping a coin many times and getting heads each time, however, seems to go against the 50% chance of a normal coin landing on heads, that's when you might wonder if the coin is two-headed, but I still seem to reject that it actually affects the probability. This has really piqued my interest in probability though. –  Ricket Jan 28 '11 at 3:22

I think the rewording is fine, however, I believe your anwser is not correct.

Indeed, without throwing coins at all, there is a $1/5$ chance you get the double headed coin, and a $4/5$ chance you get a normal coin.

Now, the probability of getting three heads on three throws of a regular coin is $1/8$, as you can surely check. And as there is a $4/5$ chance you do get a regular coin, so you have a $1/8 \cdot 4/5 = 1/10$ chance you get three heads flipping a regular coin.

So to sum up the outcomes:

• $1/5$ chance of getting a double headed coin

• $1/10$ chance of getting a normal coin and three heads

• $7/10$ chance of getting a normal coin and anything but three heads.

However, note that you are asking: What is the probability that your coin is double-headed assuming you GET three heads? So the only options you are considering are the first two (as you know the third didn't occur).

Now there is a $1/5 + 1/10 = 3/10$ chance you get three heads. So there is a $\frac{1/5}{3/10} = 2/3$ chance of getting a double headed coin.

-

With the Bayesian interpretation of the question (your second phrasing), and leaving it as an exercise to figure out my notation: $$p(HHH|DH)=1$$ $$p(DH)=\frac{1}{5}$$ $$p(DH|HHH)=\frac{p(DH \wedge HHH)}{p(HHH|DH)p(DH)+p(HHH|N)p(N)}$$ $$=\frac{1/5}{1/5+(1/8)(4/5)}=\frac{2}{3}$$

-

Both your choice of the coin and the outcomes of the subsequent flips are random variables (though they are not independent). You want to calculate the probability of the coin choice given specific observed outcomes; i.e., you want a conditional probability. Here, you want $P(\text{two-head} \vert HHH) \equiv P(\text{two-head} \wedge HHH) / P(HHH)$. The numerator and the denominator are $$P(\text{two-head} \wedge HHH) = P(\text{two-head})P(HHH \vert \text{two-head}) = \left(\frac{1}{5}\right)(1) = \frac{1}{5}$$ and $$\begin{eqnarray} P(HHH) &=& P(\text{two-head} \wedge HHH) + P(\text{normal} \wedge HHH) \\ &=& P(\text{two-head})P(HHH \vert \text{two-head}) + P(\text{normal})P(HHH \vert \text{normal}) \\ &=& \left(\frac{1}{5}\right)(1) + \left(\frac{4}{5}\right)\left(\frac{1}{8}\right) \\ &=& \frac{1}{5} + \frac{1}{10} = \frac{3}{10}. \end{eqnarray}$$ The desired result is $$P(\text{two-head} \vert HHH) = \frac{P(\text{two-head}\wedge HHH)}{P(HHH)} = \frac{1/5}{3/10} = \frac{2}{3}.$$

-

Urn models!

Tossing (Head,Head,Head) with one coin is like drawing a black ball out of an urn with 1 black and 7 white balls. The Head/Head coin is then like an urn with 8 black balls, let's imagine them as marked with a white dot.

Now, we put 5 urns (4 representing a normal coin and 1 representing the Head/Head coin) into a box. Because all urns contain the same number of balls, we can smash them without changing the probabilities: We can draw each of the 40 balls with same probability, and it will be with 1/5 probability from the urn representing the Head/Head coin.

If we draw any ball out of the box, that ball will be marked with the probability of 8/40, because 8 of 40 balls are marked. It will be black with 12/40. Under the precondition that we draw a black ball, it will be marked with the probability of 8/12.

It is confusing, that we had the probability 1/5 before. So, what happened? How could we ignore the white balls?

• One way to ignore the white balls is this:

We state, that we already know the case of tossing something else than (Head,Head,Head): 0 of 28 white balls are marked, so that means that we would know with the probability 28/40 that the coin has the probability 0/28 of being that Head/Head coin.

That way we ask: "We know that none of the white balls are marked, but how many black balls are marked?"

• The other way lets us rephrase the question:

You are given a coin and told that it has a 1/5 chance that it is double-headed, otherwise it is a normal coin. You are allowed three flips of the coin. Upon doing these, you receive either three heads or you give it back to restart this game. What is the probability that your coin is double-headed?

That way we can ignore or remove the white balls out of the box.

-

Here's another way of thinking to get the correct result given above.

Say you take each of the coins (knowing whether it's H/H or H/T) in turn, and do 1000 runs of three flips.

For the four H/T coins the probability of a HHH result is $(\frac{1}{2})^3$ or 0.125, so you would expect to get 125 HHH runs from each of the four fair coins, or a total of 500 HHH runs.

For the one H/H coin, you would get, of course, 1000 HHH runs.

You'e now determined that $\frac{1000}{1500}$ or $\frac{2}{3}$ of the HHH runs will come from the H/H coin. Someone coming along and picking one particular HHH run from the whole experiment will find that $\frac{2}{3}$ of the time he was looking at a result from the H/H run.

-