A man has 5 coins, two of which are double-headed,
one is double-tailed, and two are normal.
He picks a coin at random and tosses it.
What's the probability that the lower face is a tail?
P(H) = P(H|2H)P(2H) + P(H|2T)P(2T) + P(H|N)P(N) = $\dfrac {3}{5}$
He sees that the coin is showing head.
What's the probability that the lower face is a head?
P(${H_l}|{H_h}$) = $\dfrac {P({H_l} \cap {H_h})}{P({H_h})}$ = $ \dfrac {3}{4} $
I got those questions. Now, the man tosses the coin again.
What is the probability that the lower face is a head?
My instructor gave the following:
Let ${H_h}{H_l}$ be the event that he sees a head up on the first toss and a head down on the second toss.
P(${H_h}{H_l}|{H_h}$) = $\dfrac {P({H_h}{H_l} \cap {H_h})}{P({H_h})}$ = $\dfrac {P({H_h}{H_l})}{P({H_h})}$ = $\dfrac {5}{6}$
I understand his calculation, but his interpretation is different from mine.
Since he isn't holding the two-tailed coin, P(2H) = P(N) = $\dfrac {1}{2}$
$P({H_l})$ = $P({H_l}|2H)P(2H) + P({H_l}|N)P(N)$ = $\frac {1}{2} + \frac {1}{2} (\frac{1}{2})$ = $\frac {3}{4}$
Can you comment if my answer is entirely wrong or if I simply interpreted the question differently? Thanks!
Your answer is wrong, I’m afraid: you’re twice as likely to be holding a double-headed coin as a normal coin. You know that you’re not holding the double-tailed coin, so when you flipped it the first time, there were $8$ possible outcomes, all equally likely, corresponding to the $8$ sides of the $4$ coins that have at least one head. Six of those outcomes are heads, and of those $6$, $4$ come from the double-headed coins and only $2$ from the normal coins. Thus, the odds are $4:2$ that your coin is double-headed. In terms of probability that means that with probability $\frac23$ you have a double-headed coin and are certain to get a head on the next toss, and with probability $\frac13$ you have a normal coin and a probability of $\frac12$ of getting a head on the next toss. Your overall probability of getting a head on the second toss is therefore
$$\frac23\cdot1+\frac13\cdot\frac12=\frac56\;.$$
It takes a bit of finagling to devise a situation in which your calculation gives the right answer. Suppose that after you toss the first coin, an observer says, ‘Okay, that eliminates the double-tailed coin as a possibility. Here, let me eliminate it from the set.’ He then throws it away and lets you pick one of the remaining $4$ coins at random. In this scenario it’s true that you get a normal coin with probability $\frac12$ and a double-headed coin with probability $\frac12$, so your probability of getting a head on your second toss (with a possibly different coin) is
$$\frac12\cdot1+\frac12\cdot\frac12=\frac34\;.$$