# different interpretations?

• Apr 30th 2008, 07:11 AM
heathrowjohnny
different interpretations?
Suppose you have a 2-headed coin, a 2-tailed coin, and a fair coin. You choose a coin randomly. What is the probability that you get heads? What is $P \left(\text{2-headed coin}| \text{heads} \right)$

This is similar to the Monty-Hall problem right? Namely, you have $P(H) = \begin{cases} 1 \ \ \text{if 2-headed coin} \\ 0 \ \ \text{if 2-tailed coin} \\ \frac{1}{2} \ \ \text{if fair coin} \end{cases}$.

The answers are $\frac{2}{3}$ and $\frac{1}{3}$ respectively?

But is this only in the classical interpretation? Can we use another interpretation and get another answer?
• Apr 30th 2008, 07:14 AM
topsquark
Quote:

Originally Posted by heathrowjohnny
Suppose you have a 2-headed coin, a 2-tailed coin, and a fair coin. You choose a coin randomly. What is the probability that you get heads? What is $P \left(\text{2-headed coin}| \text{heads} \right)$

This is similar to the Monty-Hall problem right? Namely, you have $P(H) = \begin{cases} 1 \ \ \text{if 2-headed coin} \\ 0 \ \ \text{if 2-tailed coin} \\ \frac{1}{2} \ \ \text{if fair coin} \end{cases}$.

The answers are $\frac{2}{3}$ and $\frac{1}{3}$ respectively?

But is this only in the classical interpretation? Can we use another interpretation and get another answer?

How is this similar to Monty Hall? This is just adding up probabilities. By randomly picking a single coin and flipping it there are 3 ways to get a head and 6 possibilities in all. So the probability of getting heads as a result is just 1/2.

-Dan
• Apr 30th 2008, 07:31 AM
heathrowjohnny
but thats the frequentist approach. I am asking if you could get the above answers using a different approach.
• Apr 30th 2008, 08:16 AM
heathrowjohnny
my fault, I made an error in calculation: but you can also think of it as $\frac{1}{3}(1) + \frac{1}{3}(0) + \frac{1}{3}(\frac{1}{2}) = \frac{1}{2}$.
• Apr 30th 2008, 11:23 AM
topsquark
Quote:

Originally Posted by heathrowjohnny
my fault, I made an error in calculation: but you can also think of it as $\frac{1}{3}(1) + \frac{1}{3}(0) + \frac{1}{3}(\frac{1}{2}) = \frac{1}{2}$.

That's fine. There may be more than one way to calculate it, but no matter how you look at it you will get the same answer.

-Dan
• Apr 30th 2008, 12:48 PM
Plato
However, the question is not the probability of flipping heads.
The question is for the probability that the double headed coin was chosen given that heads is showing.
The answer to that is $\frac{2}{3}$
• Apr 30th 2008, 02:42 PM
heathrowjohnny
That is the second part of the question, yes.