Expected number of dimes

Question

My wallet has 2 nickels and 10 dimes. The coins are to be drawn out 1-by-1 and I'm to receive all the dimes that are drawn between the 2 nickels. What's the expectation?

My Approach:

Draw the sequence pattern for each different case and then get the probability

We know number of ways to arrange 10 dimes and 2 nickels = $\frac {12!} {10! 2!} $ = 66 ways

1st case - 10 dimes between 2 nickels - NDDDDDDDDDDN - only 1 way so probability = 1/66

2nd case - 9 dimes between 2 nickels - DNDDDDDDDDDN or NDDDDDDDDDND - 2 ways so probability = 2/66

Similarly 8 dimes between 2 nickels will have 3 ways, 7 dimes will have 4 cases, 6 dimes will have 5 cases etc until 1 dime has 10 cases and 0 dimes has 11 cases.

Using this approach, I get my expected value as 220/66 = 3.33 dimes

Not sure if this is he correct approach/answer tho

Answer 1

This can be solved more simply using indicator variables.

Let $X_i$ be an indicator random variable = $1$ if the $i_{th}$ dime is between the two nickels, and $0$ otherwise.

Consider dime $i$ in conjunction with the two nickels

There are three possibilities, $NDN\;NND\;DNN$ in only one of which does the dime lie between the nickels, so

$P(X_i=1) = \frac13$

The expectation of an indicator variable is just the probability of the event it indicates, so $\mathbb E[X_i] = \frac13$

and by linearity of expectation which operates even when the random variables are not independent,

$\mathbb E[X] = 10 \mathbb E[X_i] = 10\cdot\frac13 = \frac{10}3$

Answer 2

3 dimes 2 nickels ...

Total number of arrangements = $^5C_2 \times 2! \times 3!$

Probability of winning 0 dimes $ P(0) = \frac{4 \times 2! \times 3!}{^5C_2 \times 2! \times 3!} = \frac{4}{10}$

Probability of winning 1 dime $P(1) = \frac{3 \times 2! \times 3!}{^5C_2 \times 2! \times 3!} = \frac{3}{10}$

Probability of winning 2 dimes $P(2) = \frac{2 \times 2! \times 3!}{^5C_2 \times 2! \times 3!} = \frac{2}{10}$

Probability of winning 3 dimes $P(3) = \frac{1 \times 2! \times 3!}{^5C_2 \times 2! \times 3!} = \frac{1}{10}$

$X$ = average number of dimes won

$E(X) = \frac{4}{10} \times 0 + \frac{3}{10} \times 1 + \frac{2}{10} \times 2 + \frac{1}{10} \times 3 = 1$ dime.