I am stuck on this question:

A box has three coins. One has two heads, one has two tails, and the other is a fair coin with one head and one tail. A coin is chosen at random, and comes up head.

a) What is the probability that the coin chosen is the two headed coin

b) What is the probability that if it is thrown another time it will come up heads

c) What is the probability that the coin chosen is the two headed coin, supposing that the coin is thrown a second time and comes up heads again

I have solved part a, and the answer is $\frac 23$.

However I am stuck on parts b and c. This is my thought process:

part b: P(Heads on the second throw) = P($H_2$|$H_1$) $\cdot$ P(fair coin) + P($H_2$|$H_1$) $\cdot$ P(coin with two heads) = $\frac 12$ $\cdot$ $\frac 13$ + 1 $\cdot$ $\frac 13$ = $\frac 12$, and I'm not sure how to even start with part c.

The answer to part b should be $\frac 56$, but can anyone explain why? Thanks.


3 Answers 3


Your answer to part (a) is incorrect. If you pull out a random coin and flip, you have six scenarios:

  1. Head 1 of 2-headed coin
  2. Head 2 of 2-headed coin
  3. Head of fair coin
  4. Tail of fair coin
  5. Tail 1 of 2-tailed coin
  6. Tail 2 of 2-tailed coin

3 of those are "heads", and 2 of those 3 correspond to the 2 headed coin. Thus, the answer to part (a) is $\frac 23$

Part (b): Probability of getting double headed * getting head from that double headed + Probability of getting fair * getting head from that fair coin $$\frac 23 \cdot 1+\frac 13 \cdot \frac 12=\frac 56$$

Part (c): Now you have $3\cdot 2\cdot 2=12$ possiblities; check all of them and see which of the ones involving two heads involve the double headed coin.

You can also look at Bayes's Theorem; it covers problems like this one.

  • $\begingroup$ ok, i realized that i forgot that Head 1 and Head 2 of 2-headed coin should be counted as 2 possibilities. Thanks! $\endgroup$
    – peco
    Sep 30, 2018 at 2:28

Part $a$ does not give $p=\frac{1}{2}$. Suppose the two heads/tails on the rigged coins are marked 1,2.

Break it down into 6 possibilities:

  • Fair Coin - Came up heads
  • Fair Coin - Came up tails
  • Two Heads - Came up heads 1
  • Two Heads - Came up heads 2
  • Two Tails - Came up tails 1
  • Two Tails - Came up tails 2

You know that it came up heads so there are $3$ possibilities left. $2$ have the unfair coin and $1$ with the fair coin.

As an extreme example to get the reasoning, replace the coins with dice with 100 sides. One dice has all 100's, one is fair and another has all 1's. A DM behind the screen rolls one of them up to him/her and tells you they got 100. Do you bet they used the rigged dice?

You already know that the first flip gave heads so in part b, it is not $\frac{1}{3}$ each.

$$ P(H_2 \mid Fair,H_1) P(Fair \mid H_1) = \frac{1}{2} \frac{1}{3}\\ P(H_2 \mid Two Heads,H_1) P(Two Heads \mid H_1) = 1 \frac{2}{3}\\ P(H_2 \mid Two Tails,H_1) P(Two Tails \mid H_1) = 0 \times 0\\ P(H_2 \mid H_1) = \sum above = \frac{1}{6} + \frac{4}{6} = \frac{5}{6}\\ $$


First let's define clearly the events:

  • $C_{Heads}$ = "Heads-only coin is drawn from the box"
  • $C_{Tails}$ = "Tails-only coin is drawn from the box"
  • $C_{Fair}$ = "Fair coin is drawn from the box"

  • $H_i$ = "Coin gives Heads when tossed the $i$-th time"

  • $T_i$ = "Coin gives Tails when tossed the $i$-th time"

a) We are looking for $P(C_{Heads}|H_1)$. Just use Bayes theorem:

$P(C_{Heads}|H_1) = \frac{P(H_1|C_{Heads})* P(C_{Heads})}{ P(H_1) }= \frac{1*1/3}{1/2} = 2/3$

with $P(H_1)=1/2$ because 3 heads and 3 tails in total are available in the box and $P(C_{Heads}) = 1/3$ because it's 1 coin out of 3 we can pick from the box.

b) We are looking for $P(H_2|H_1)$. Let's use the rule of total probability: $P(H_2|H_1) = P(H_2|H_1,C_{Heads}) * P(C_{Heads}|H_1) + P(H_2|H_1,C_{Tails}) * P(C_{Tails}|H_1) + P(H_2|H_1,C_{Fair}) * P(C_{Fair}|H_1) = 1*2/3 + 0 + 1/2 * 1/3 = 5/6$

c) We are looking for $P(C_{Heads}|H_1,H_2)$. Again let's use Bayes theorem:

$P(C_{Heads}|H_2,H_1)=\frac{P(H_2,H_1|C_{Heads})*P(C_{Heads} )}{P(H_2,H_1)}= \frac{P(H_2,H_1|C_{Heads})*P(C_{Heads} )}{P(H_2|H_1)*P(H_1)} = \frac{1*1/3}{5/6*1/2}= 4/5$

So if you get Heads twice, there is 80% probability that you picked the Heads-only coin.


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