Expectation of amount of money won in a game

[songoku]

Homework Statement

A man throws $n$ dice and will receive $ $\frac{1}{2}x$ where $x$ is number of sixes occurred ($x=0,1,2,...,n$). Prove that the expectation of the amount receives is $ $\frac{1}{12}n$

Relevant Equations

Linear Combination of Random Variable: E(aX) = a.E(X)

Binomial Distribution: X ~ B (n, p)

Expectation of Binomial Distribution = np

This is what I did:

Let Y = number of sixes occurred when $n$ dice are thrown

Y ~ B (n, 1/6)

E(Y) = $\frac{1}{6}n$Let Z = amount of money received → Z = $\frac{1}{2}Y$

E(Z) = E(1/2 Y) = 1/2 E(Y) = $\frac{1}{12}n$I got the answer but I am not sure about my working because I didn't take the amount of money into account (whether the man receives $0.5 or $1 or $1.5 , etc)

Is my working correct? Thanks

 

[Delta2]

It seems to me your working is correct. You took into account the amount of money received in an implicit way by using the random variable Z.

 

[songoku]

It seems to me your working is correct. You took into account the amount of money received in an implicit way by using the random variable Z.

Actually I am more confident if I use probability distribution table to calculate the expectation, but I just can't solve it

This is what I tried:
For Z = 0 → P(Z = 0) = $\binom n 0 \left ( \frac{1}{6} \right) ^0 \left ( \frac{5}{6} \right) ^n$

For Z = 0.5 → P(Z = 0.5) = $\binom n 1 \left ( \frac{1}{6} \right) ^1 \left ( \frac{5}{6} \right) ^{n-1}$

For Z = 1 → P(Z = 1) = $\binom n 2 \left ( \frac{1}{6} \right) ^2 \left ( \frac{5}{6} \right) ^{n-2}$
.
.
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For Z = $\frac n 2$ → P(Z = n) = $\binom n n \left ( \frac{1}{6} \right) ^n \left ( \frac{5}{6} \right) ^0$Expectation
$$=0 \times \binom n 0 \left ( \frac{1}{6} \right) ^0 \left ( \frac{5}{6} \right) ^n + 0.5 \times \binom n 1 \left ( \frac{1}{6} \right) ^1 \left ( \frac{5}{6} \right) ^{n-1} + 1 \times \binom n 2 \left ( \frac{1}{6} \right) ^2 \left ( \frac{5}{6} \right) ^{n-2} + ... + \frac{n}{2} \times \binom n n \left ( \frac{1}{6} \right) ^n \left ( \frac{5}{6} \right) ^0$$
$$=\sum_{r=1}^n \frac{r}{2}\binom n r \left ( \frac{1}{6} \right) ^r \left ( \frac{5}{6} \right) ^{n-r}$$

Is this correct? If yes, how to prove that the sum will be $\frac{1}{12}n$?

Thanks

 

[Delta2]

This should be in your textbook notes on how to calculate the expectation value of a variable that follows the binomial distribution. That sum is the expectation value of binomial with the factor 1/2 in front of it.

Now that I think of it I should be able to calculate that sum but I can't at the moment, I am stuck as you :D.

 

[PeroK]

Or, simply: the expected return on each die is $0.50/6. He throws $n$ dice, so the expected return is $0.50n/6 = $n/12.

 

[PeroK]

PS a way of seeing that this is valid is to imagine throwing the $n$ dice one after the other; rather than all at the same time. Then, taking the expected gain after $n$ consecutive throws.

The expected gain for $n$ consecutive throws must be $n/12; and, it must be the same as throwing all the dice at the same time.

 

[Delta2]

The expected gain for n consecutive throws must be $n/12; and, it must be the same as throwing all the dice at the same time.

That was something that was troubling me, if the two experiments are equivalent from a probabilistic point of view. Why are they if I may ask?

 

[PeroK]

That was something that was troubling me, if the two experiments are equivalent from a probabilistic point of view. Why are they if I may ask?

In both cases there are $n$ independent dice.

 

[songoku]

Thank you very much for all the help and explanation Delta2 and PeroK