Calculating the expected value of a dice roll

[Addez123]

Homework Statement

In a boardgame you move forward the amount of steps that the dice roll. Except if the dice show 1 you move 6 steps. Calculate the Expected Value.

Relevant Equations

E(x) = sum(g(k) * p(k))

I write p(k) as:
$$p(k) = 1/6, k = 2,3,4,5$$
$$p(k) = 2/6, k = 6$$

Is that wrong?
Because then the expected value becomes
$$1/6 * 4 + 2/6 * 6 = 8/3$$

While my book says 11/3

 

[PeroK]

Homework Statement:: In a boardgame you move forward the amount of steps that the dice roll. Except if the dice show 1 you move 6 steps. Calculate the Expected Value.
Relevant Equations:: E(x) = sum(g(k) * p(k))

I write p(k) as:
$$p(k) = 1/6, k = 2,3,4,5$$
$$p(k) = 2/6, k = 6$$

Is that wrong?
Because then the expected value becomes
$$1/6 * 4 + 2/6 * 6 = 8/3$$

While my book says 11/3

Well, $8/3 < 3$, which is less than you'd expect if $1$ counted as $1$ and not $6$.

Your problem seems to be that you didn't count throws of $2,3$ or $5$.

 

[PeroK]

PS The book's answer looks wrong. Assuming I've understood the rules.

 

[etotheipi]

I don't get the book answer.

Edit: Beaten by @PeroK

 

[haruspex]

$$p(k) = 1/6, k = 2,3,4,5$$
$$p(k) = 2/6, k = 6$$
Is that wrong?
Because then the expected value becomes
$$1/6 * 4 + 2/6 * 6 = 8/3$$

No, it doesn’t become that.
In your last step, you multiplied the 1/6 by the number of cases with that probability, not by the number of steps taken in those cases.

 

[Addez123]

Ah yea that was why!
Should've calculated 1/6*2 + 1/6 * 3 etc.