Problem 5.13 Easy Free 2 variations

Which Coin Was Drawn?

A box contains $1000$ coins: $999$ fair coins and $1$ coin with heads on both sides. You pick a coin uniformly at random and flip it $10$ times, observing heads every time. Find the probability that the coin you picked is the double-headed one.

Work it out, then compare with the answer.

SolutionShow the worked solution

Let $D$ be the event that the double-headed coin was picked, and $H_{10}$ the event of ten heads. The prior odds of $D$ are

\frac{\mathbb{P}(D)}{\mathbb{P}(D^c)}=\frac{1/1000}{999/1000}=\frac{1}{999}.

The likelihoods are $\mathbb{P}(H_{10}\mid D)=1$ and $\mathbb{P}(H_{10}\mid D^c)=2^{-10}$ , because the flips of a fair coin are independent. Their ratio is therefore $2^{10}=1024$ .

By Bayes’ theorem, posterior odds equal prior odds times the likelihood ratio:

\frac{\mathbb{P}(D\mid H_{10})}{\mathbb{P}(D^c\mid H_{10})}=\frac{1}{999}\cdot 1024=\frac{1024}{999}.

Converting odds $r$ into a probability $r/(1+r)$ gives

\mathbb{P}(D\mid H_{10})=\frac{1024}{1024+999}=\frac{1024}{2023}\approx 0.506.

\boxed{\dfrac{1024}{2023}}

Key idea. Each head doubles the odds in favour of the double-headed coin. Ten heads multiply the odds by about $1000$ , which roughly cancels the prior odds of $1:999$ , so the posterior is close to $\tfrac12$ .

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 5.13a

A bag contains $19$ fair coins and one double-headed coin. You draw a coin uniformly at random and flip it repeatedly, observing heads every time. Find the smallest number of consecutive heads needed for your posterior probability that the coin is double-headed to exceed $95\%$ .

Work it out, then compare with the answer.

Solution

The prior odds of the double-headed coin are $1/19$ . Each head multiplies the odds by $2$ , so after $n$ heads the posterior odds are $2^n/19$ .

A posterior probability above $0.95$ means odds above $0.95/0.05=19$ . The condition is therefore $2^n/19>19$ , that is, $2^n>361$ .

Check the boundary values:

$n=8$ : $2^8=256<361$ , and the posterior is $256/275\approx 0.931$ .
$n=9$ : $2^9=512>361$ , and the posterior is $512/531\approx 0.964$ .

\boxed{9}

Variation 5.13b

Two coins have heads probabilities $p_1$ and $p_2$ with $0<p_1<1$ , $0<p_2<1$ , $p_1+p_2=1$ , and $p_2>p_1$ . One coin is chosen uniformly at random and flipped twice, giving two heads. Given that this makes it $4$ times as likely that Coin $2$ was chosen rather than Coin $1$ , find $p_2$ .

Work it out, then compare with the answer.

Solution

The two coins are equally likely a priori. The posterior odds of Coin $2$ against Coin $1$ therefore equal the likelihood ratio

\frac{p_2^2}{p_1^2}.

Setting this equal to $4$ and using $p_1,p_2>0$ gives $p_2=2p_1$ . Combining with $p_1+p_2=1$ gives $3p_1=1$ , so $p_1=\tfrac13$ and $p_2=\tfrac23$ . This satisfies $p_2>p_1$ .

\boxed{p_2=\tfrac{2}{3}}

Spotted an error? Email us