Expected value if two coins are drawn simultaneously from a bag containing $2$ one unit and $3$ two units of money

Question

This is a question from the Resonance DLPD Algebra book (Page 250). To quote the book:

From a bag containing $2$ one rupee and $3$ two rupees coins, a person is allowed to draw $2$ coins simultaneously; find the value of his expectation.

My try:

There are three cases:

1. The person draws $2$ one rupee coins.
2. The person draws one $1$ rupee coin and one $2$ rupees coin.
3. The person draws two $2$ rupees coin.

The probability of case 1 is:

$$P_1=\frac25*\frac14 = \frac2{20}$$ The value of the two coins is $V_1=2$.

The probability of case 2 is:

$$P_2=\frac25*\frac34 = \frac6{20}$$ The value of the two coins is $V_2=3$.

The probability of case 3 is:

$$P_3=\frac35*\frac24 = \frac6{20}$$ The value of the two coins is $V_3=4$.

Hence his expected value should be:

$$V = P_1V_1+P_2V_2+P_3V_3 = 2.30$$

However, according to the answer given in my book, the correct value is $3.20$! But how? Where did I go wrong with my above approach?

Answer 1

For case $2$ you neglect the fact that you can get the two coins in either order. Thus the probability of that case is twice as high as what you compute and the expectation is now

$$\frac 2{20}\times 2+\frac {12}{20}\times 3+\frac 6{20}\times 4 =3.2$$ as desired.

Answer 2

For your case 2, the coins can be drawn in either order, so you need to multiply by 2.

Another approach:

The expected value of one coin is $$\frac{2 \times 1 + 3 \times 2}{5} = \frac{8}{5}$$

so the expected value of two coins is $2 \times (8/5) = 16/5$.