Q.7.15 - A First Course in Probability

Question

Each coin in a bin has a value attached to it. Each time that a coin with value p is flipped, it lands on heads with a probability p. When a coin is randomly chosen from the bin, its value is uniformly distributed on ($0,1$). Suppose that after the coin is chosen but before it is flipped, you must predict whether it will land on heads or on tails. You will win $1$ if you are correct and will lose $1$ otherwise.  

(a)What is your expected gain if you are not told the value of the coin? 

(b) Suppose now that you are allowed to inspect the coin before it is flipped, with the result of your inspection being that you learn the value of the coin. As a function of p, the value of the coin, what prediction should you make? 

(c) Under the conditions of part(b), what is your expected gain?

Step-by-Step Solution

Answer

From the above information,

a) Expected gain if you are not told the value of the coin is $0$

b) $\text{We have function}p\text{. Predict heads if}p>\frac{1}{2}\text{.}$

c) The expected gain is $\frac{1}{2}$

Step 1 : Given Information (part a)

What is your expected gain if you are not told the value of the coin 

Step 2: Explanation (part a)

Your expected gain if you are not told the value of the coin is $0$

Step 3: Final Answer (part a)

The expected gain is zero.

Step 4: Given Information (part b)

As a function of p,the value of the coin, what prediction should you make

Step 5: Explanation (part b)

$\text{We have function}p\text{. Predict heads if}p>\frac{1}{2}\text{.}$

Step 6: Final Answer (part b)

As a function of p, the value of the coin, $\text{ Predict heads if}p>\frac{1}{2}\text{.}$

Step 7: Given Information (part c)

Under the conditions of part(b), what is your expected gain?

Step 8: Explanation (part c)

The expected gain is,

$E\left[\text{gain}\right]={\int }_0^{1}E[\mathrm{gain}\mid V=p]dp$

$={\int }_0^{\frac{1}{2}}\left[1\right(1-p)-1(p\left)\right]dp+{\int }_{\frac{1}{2}}^1\left[1\right(p)-1(1-p\left)\right]dp$

$={\int }_0^{\frac{1}{2}}(1-p-p)dp+{\int }_{\frac{1}{2}}^1(p-1+p)dp$

$={\int }_0^{\frac{1}{2}}(1-2p)dp+{\int }_{\frac{1}{2}}^1(2p-1)dp$

$={[p-\frac{2p^2}{2}]}_0^{\frac{1}{2}}{[\frac{2p^2}{2}-p]}_{\frac{1}{2}}^1$

$=[\frac{1}{2}-\frac{1}{4}]+[1-1-\frac{1}{4}+\frac{1}{2}]$

$=\left[\frac{2-1}{4}\right]+\left[\frac{-1+2}{4}\right]$

$=\frac{1}{4}+\frac{1}{4}$

=$\frac{2}{4}$

=$\frac{1}{2}$

Step 9: Final Answer (part c)

 The expected gain is  $\frac{1}{2}$