I have a probability question that reads:


A box has three coins. One has two heads, another two tails and the last is a fair coin. A coin is chosen at random, and comes up head. What is the probability that the coin chosen is a two headed coin.

My attempt:

P(two heads coin| given head) = P(two heads coin * given head)/P(given head)
= 1/3/2/3 = 1/2

Not sure whether this is correct?


Hint: No, it isn't. Let $H \equiv$ obtaining heads, $A \equiv$ picking a two-headed coin, $B \equiv$ picking a two-tailed coin, and $C \equiv$ picking a fair coin. Then observe that the probability of obtaining a head is: $$ \begin{align*} P(H) &= P(A)\cdot P(H \mid A) + P(B)\cdot P(H \mid B) + P(C)\cdot P(H \mid C) \\ &= \dfrac{1}{3} \cdot \dfrac{2}{2} + \dfrac{1}{3}\cdot \dfrac{0}{2} + \dfrac{1}{3}\cdot \dfrac{1}{2} \end{align*} $$

  • $\begingroup$ so my numerator is correct and my denominator is wrong. am i right? $\endgroup$ – lakesh Aug 9 '13 at 10:34
  • $\begingroup$ No, the numerator is wrong as well. Using my notation, you have calculated $P(A)$ instead of $P(A \text{ and } H)$. $\endgroup$ – Adriano Aug 9 '13 at 10:35
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    $\begingroup$ I don't think you're answering the question being asked. You correctly find $P(H)=1/2$, but the question was about $P(A\mid H)$ which your hint seems to say nothing about at all. $\endgroup$ – Henning Makholm Aug 9 '13 at 12:17
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    $\begingroup$ @HenningMakholm I was hoping that the OP knew the formula: $$ P(A \mid H) = \dfrac{P(A) \cdot P(H \mid A)}{P(H)} $$ so I helped him out with the harder part, which was calculating the denominator. $\endgroup$ – Adriano Aug 9 '13 at 16:32

For such a small number of options its easy to count them

The possible outcomes are:

heads or heads using the double head coin
tails or tails using the double tail coin
heads or tails using the fair coin

All these outcomes are equally likely. How many of these are heads and of those how many use the double headed coin?

$$Answer = \frac{2}{3}$$


The answer is 2/3.

P(A∣H)=P(A) * P(H∣A) / P(H) --> Bayes Theorem

P(A) = 1/3, because 1 out of 3 coins are 2-headed

P(H|A) = 1, because given you have the 2-headed coin, then it is for sure (100%) you will get a head

P(H) = 1/2, because there are 3 heads and 3 tails in total, so it's 50% chance to get heads

Therefore, P(A|H) = (1/3) * 1 / (1/2) = 2/3


The symmetry of this problem provides another way to look at it: you choose a coin at random and look at one of its faces at random. That face shows something (maybe heads, maybe tails).

First question:

What is the probability that the other face of the coin is the same as the face you can see?

The answer to that question is clearly $\frac23$, since two of the three coins have the same thing on both faces and the third coin has something different on the other face no matter which face you look at.

Second question:

In which case is the probability greater that the other face is the same as the face you can see: (A) you see heads; (B) you see tails; or are the probabilities the same?

By symmetry (because the conditions of the experiment are exactly the same even if we relabel every "heads" as "tails" and every "tails" as "heads"), neither probability is greater; they are the same.

So it doesn't matter which face you see when you first look at the coin; there is a $\frac23$ probability the other face is the same, even if the first face is heads (as it is in the original question).


2/3 is the answer, definitely not 1/2. Oddly, these 3 coins are very similar to the Monty Hall game show problem with 3 curtains. You want the curtain with the prize, not the 2 with goats. You initially choose curtain A. Monty opens curtain B, and there's a goat. Monty turns to you and asks, do you want to switch your choice? Mathematicians went crazy saying how odds were just 1/2 whether you switched or not. Even Paul Erdos, most prolific mathematician in recent history, got it wrong. That super IQ woman got it right, Marilyn, in her Parade magazine column. So here's the analogy. Initially odds are 1/3 that it's a double headed coin. You want the double headed coin (for some bizarre reason, maybe to use to cheat someone). You choose coin A. Monty shows you coin B is 2 tails. Now do you want to switch? Does it matter? Are the odds just 50 50? Computer simulations show you switch your choice, as you have a 2/3 chance of getting it right with a switch. The probabilities are now 2/3 and 1/3 for where the 2 headed coin is, after seeing one head. One head makes it twice as likely the coin is 2 headed, not a fair coin.


1/2 is the only answer!! 50 50 chance, that's it period!!! The double tails is out for sure, so discard it. What is left besides the normal coin and 2 headed coin? Nothing! So, it's between two coins only. You can't rip the heads or tails from the coins and shake them up in the box. That would give you 3 heads and only 1 tail. That would be a 3 out of 4 chance! BUT THAT IS OUT! These are coins with no magical way of knowing if the coin chosen has a tail on the hidden side. So it's a 50 50 or 1/2 chance. Both the normal and 2 tail coins are out because it HAS to be a 2 headed coin chosen! So if you randomly draw out a coin with heads. You have a 50/50 or 1/2 chance. THAT'S IT, 1/2. No formula needed. I'm not great at math myself. This is common comprehension! Thanks Del

  • $\begingroup$ Indeed, the probability that the coin is two-headed, given that you know it has at least one head on it, is $\frac12.$ But that is not what the question asked. In the question as asked, you see only one side of the coin. It is possible then to draw the fair coin and not know that it has any heads face on it. $\endgroup$ – David K May 24 '15 at 19:10

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