More on Bertrand's box paradox

Question

I am confused about a conditional probability question from One Thousand Exercises in Probability by Geoffery Grimmett & David Stirzaker:

A man possesses five coins, two of which are double-headed, one is double-tailed, and two are normal. He shuts his eyes, picks a coin at random, and tosses it. What is the probability that the lower face of the coin is a head?

He opens his eyes and sees that the coin is showing heads; what is the probability that the lower face is a head?

He shuts his eyes again, and tosses the coin again. What is the probability that the lower face is a head?

I'm ok with the first two questions. For the third question, I think he asks for the probability that the lower face is a head on the second toss given the upper face is a head on the first toss. Let $D_H$ be the event that the coin is double-headed, $D_T$ the event that the coin is double-tailed, $N$ the event the coin is normal, $H_i$ the event the lower face is head on the $i$th toss, and finally $U_i$ the event the upper face is head on the $i$th toss.

Now we need to find $\mathbb{P}(H_2 \mid U_1)$. However, what we know from the second question is $$\mathbb{P}(D_H \mid U_1)= \frac{2}{3}, \quad \mathbb{P}(D_T \mid U_1) =0, \quad \mathbb{P}(N\mid U_1)=\frac{1}{3},$$

and I failed to deduce what $\mathbb{P}(H_2 \mid U_1) = \frac{\mathbb{P}(H_2 \; \cap\; U_1)}{\mathbb{P}(U_1)}$ is from what we know.

Answer 1

The underlying assumption is that given the choice of coin, the two tosses are independent. We will formalize this information.

So let $C\in \{hh,ht,tt\}$ be the choice of coin with probabilities $2/5,2/5$ and $1/5$, respectively. Let $U_i, L_i\in\{h,t\}$ be the upper and lower face in the $i$'th toss. Let $A=\{U_1=h\}$ be the event that we get an upper head in the first toss and $B=\{L_2=h\}$ a lower head in the second. We get by a standard application of Bayes that $P(C=hh|A) = 2/3$ and $P(CC=ht|A)=1/3$.

Now, the conditional independence means that $$P(A\cap B|C)=P(A|C)\; P(B|C)$$ which you may rewrite in the following way: $$ P(B\cap C|A) = P(B|C)\; P(C|A) $$ Then $$P(B|A) = \sum_C P(B\cap C|A) = \sum_C P(B|C)\; P(C|A) = 1\times \frac23 + \frac12 \times \frac13 + 0\times 0 = \frac56.$$

Answer 2

There are three types of coins $HH$, $HT$, and $TT$, with probabilities of $2/5$, $2/5$ and $1/5$ respectively. Let $L$, $U$ be the events that the lower/upper face is a head.

1. What is the probability that the lower face of the coin is a head?

$$P(L) = \frac{2}{5}\times1+\frac{2}{5}\times\frac{1}{2}+\frac{1}{5}\times0=\frac{3}{5}$$ (Or, simpler, there are six heads out of $10$ faces.)

2. He opens his eyes and sees that the coin is showing heads; what is the probability that the lower face is a head?

$$P(U)=P(L)=\frac{3}{5}$$ $$\therefore P(L|U)=\frac{P(L\cap U)}{P(U)}=\frac{2/5}{3/5}=\frac{2}{3}$$

3. He shuts his eyes again, and tosses the coin again. What is the probability that the lower face is a head?

$$P(HH|U_1)=\frac{P(HH\cap U_1)}{P(U_1)}=\frac{2/5}{3/5}=\frac{2}{3}$$ $$\therefore P(HT|U_1)=\frac{1}{3}$$ \begin{align}\therefore P(L_2|U_1)&=P(L_2|HH)P(HH|U_1)+P(L_2|HT)P(HT|U_1)+P(L_2|TT)P(TT|U_1)\\ &=\frac{2}{3}\times1+\frac{1}{3}\times\frac{1}{2}+0=\frac{5}{6}\end{align}

Answer 3

A less formal way of looking at it may help your intuition. At the beginning, the prior probability that the coin is double-headed was $\frac25$. After the second experiment, the posterior probability that the coin is double-headed is $\frac23$, and the posterior probability that the coin is normal is $\frac13$, so these are the prior probabilities for the third experiment.

We get the same answer if we assume there are $3$ coins, $2$ double-headed and $1$ normal. Now there are $6$ faces, and $5$ are heads, so the probability is $\frac56$.

Answer 4

Concerning first question.

There are $10$ equiprobable faces and $6$ of them are heads so the probability equals: $$\frac{6}{10}=\frac{3}{5}$$

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Concerning second question.

There are $6$ equiprobable heads and $4$ of them belong to a coin that is double headed so the probability equals: $$\frac{4}{6}=\frac{2}{3}$$

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Concerning third question.

The probability that he has a double headed coin in hand equals $\frac{2}{3}$ (as was shown above) and if that is the case then the probability that the lower face is a head equals $1$.

The probability that he has a normal coin in hand equals $1-\frac23=\frac{1}{3}$ and if that is the case then the probability that the lower face is a head equals $\frac{1}{2}$.

This leads to probability: $$\frac{2}{3}\cdot1+\frac{1}{3}\cdot\frac{1}{2}=\frac{5}{6}$$