# Coins problem-how to apply bayes

We choose arbitrarily one of the 12 apparent similar coins. However the coins, in reality, belong to two groups, the 9 are "fair" while the other ones have the probability to appear heads equal to $$\frac{6}{9}$$. We are making a throw and after having seen the result, we are given the following choices:

• To claim that we took a "fair" coin
• To claim that we took one of the "unfair" coins
• Not to announce any estimation for the coin.

If we make a wrong estimation we will need to pay (to someone who knows the type of the coin) 11 euros, and at the case of a right estimation 0 euro. If we do not answer we will need to pay 4,5 euros. I want to describe the procedure above with the decision theory and give the decision function of Bayes.

I have thought the following:

The probability of error is defined as:

$$P(errox \mid x)=P(\omega_1 \mid x) \text{ if we decide } \omega_2 \text{ and } P(\omega_2 \mid x) \text{ if we decide } \omega_1$$

The Bayes rule is optimum, that is, it minimizes the average probability error.

So we need to find the probability of errors.

If we claim that we took a "fair" coin, the probability that we had an "unfair cooin" is $$\frac{3}{12}$$.

If we claim that we took an "unfair" coin, the probability that we had a "fair coin" is $$\frac{9}{12}$$.

If we do not announce an estimation, how to we proceed?

Is the rest right? So the Bayes function will be distinguished by cases?

**EDIT:**Does it hold that the probability of error is $$\frac{1}{3}$$, since when we estimate that the coin was fair, then $$P(\text{"the coin was fair"} \mid \text{"estimation"})=\frac{1}{9}$$ and $$P(\text{"the coin was not fair"} \mid \text{"estimation"})=\frac{1}{3}$$ ?

• Is there any actual Bayesian statistics involved or is this just about conditional probabilities? Jan 21, 2021 at 8:59
• It's just about conditional probablities. I have thought the following: the probabilities of the throw are Fair coin - heads, fair coin - tails, biased coin - heads, biased coin - tails, and these are respectively $\frac{9}{12} \cdot \frac{1}{2}$, $\frac{9}{12} \cdot \frac{1}{2}$, $\frac{3}{12} \left( 1-\frac{6}{9}\right)$ and $\frac{3}{12} \cdot \frac{6}{9}$. Is this right? If so, how does it help? Jan 21, 2021 at 9:05
• How many times are you throwing each dice? Jan 21, 2021 at 9:06
• That is not given. Jan 21, 2021 at 9:07
• That is crucial information. If you only throw the dice once then it is useless, but if you can throw them 1 million times that helps a lot. Jan 21, 2021 at 9:08

If you throw each dice exactly once then there isn't much you can do, you don't really have a distribution, just one event. Considering this, the following strategies can be defined, which do not require you to even throw the dice.

$$A={\text{\{always claim the coin is fair\}}}$$
$$B={\text{\{always claim the coin is not fair\}}}$$
$$C={\text{\{say nothing\}}}$$

then the expected values are

$$E(A)=\frac{9}{12} \cdot 0 + \frac{3}{12} \cdot -11=-2.75$$

$$E(B)=\frac{9}{12} \cdot -11 + \frac{3}{12} \cdot 0=-8.25$$

$$E(C)=\frac{12}{12} \cdot -4.5=-4.5$$

So strategy A is quite good, and I don't think you can do better.

• Does it hold that the probability of error is $\frac{1}{3}$, since when we estimate that the coin was fair, then $P(\text{"the coin was fair"} \mid \text{"estimation"})=\frac{1}{9}$ and $P(\text{"the coin was not fair"} \mid \text{"estimation"})=\frac{1}{3}$ ? Jan 21, 2021 at 9:41
• @pingu My calculations do not include any estimation, I am only using known information, that there are 9 fair and 3 non-fair dice, and since each dice is equally likely to be chosen at random, then on average we will get it wrong in 3 out of 12 cases, if we always declare that the dice is fair. Jan 21, 2021 at 10:23