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Calculating the expected value of a dice roll

  • Thread starter Addez123
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  • #1
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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
 

Answers and Replies

  • #2
PeroK
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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##.
 
  • #3
PeroK
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PS The book's answer looks wrong. Assuming I've understood the rules.
 
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  • #4
etotheipi
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I don't get the book answer.

Edit: Beaten by @PeroK
 
  • #5
haruspex
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$$p(k) = 1/6, k = 2,3,4,5$$
$$p(k) = 2/6, k = 6$$
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.
 
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  • #6
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Ah yea that was why!
Should've calculated 1/6*2 + 1/6 * 3 etc.
 

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