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Math Help - Expected value for dice

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    Expected value for dice

    If a dice is thrown until we get an even number, what is the expected no. of throws?

    Would this be the same as finding the expected value
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    Quote Originally Posted by acevipa View Post
    If a dice is thrown until we get an even number, what is the expected no. of throws?

    Would this be the same as finding the expected value
    Yes, same as expected value. I feel like I've encountered this infinite sum recently, but don't remember how to solve now, although numerically it's not hard to find the value is 2.

    There may be a way to solve without infinite sum as well.

    I intentionally omitted the sum, so that you can work it out on your own, or say where you got stuck..

    Edit: I was able to evaluate the sum by calling the sum L and then computing L + 2L which can be manipulated to 2 + 2L .... but it seems a bit messy.
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    Quote Originally Posted by undefined View Post
    Yes, same as expected value. I feel like I've encountered this infinite sum recently, but don't remember how to solve now, although numerically it's not hard to find the value is 2.

    There may be a way to solve without infinite sum as well.

    I intentionally omitted the sum, so that you can work it out on your own, or say where you got stuck..
    I'm actually a bit confused because so far I'm used to working out the expected value of the dice. But it's asking for the expected number of throws

    So, would you say, the probability of an even number is 0.5, therefore, the expected number of throws must be 2.
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    Quote Originally Posted by acevipa View Post
    I'm actually a bit confused because so far I'm used to working out the expected value of the dice. But it's asking for the expected number of throws
    So let X be the random variable: number of throws until even comes up.

    E(X) = 1*P(X=1) + 2*P(X=2) + 3*P(X=3) + ...

    You should be able to see that P(X=1) = 1/2. What about P(X=2)?
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    Quote Originally Posted by undefined View Post
    So let X be the random variable: number of throws until even comes up.

    E(X) = 1*P(X=1) + 2*P(X=2) + 3*P(X=3) + ...

    You should be able to see that P(X=1) = 1/2. What about P(X=2)?
    P(X=2)=0.5

    That just tells you the expected value of the dice. But what does it say about the expected no. of throws
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    Quote Originally Posted by acevipa View Post
    P(X=2)=0.5

    That just tells you the expected value of the dice. But what does it say about the expected no. of throws
    I don't know what you mean, but in fact P(X=2) = 1/4. This is because in order to end at 2 throws, we must throw odd then even, so (1/2)(1/2) = 1/4. Similarly, to end at three throws we must throw odd then odd then even, and P(X=3) = 1/8.

    We get the sum E(X) = 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + ...

    Above I mention a method of evaluating the sum, after which we will get E(X) = 2. Like I wrote above, the way I found seems kind of messy, but maybe that's a standard way, don't really know. If you want me to show you I can.
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    Quote Originally Posted by undefined View Post
    I don't know what you mean, but in fact P(X=2) = 1/4. This is because in order to end at 2 throws, we must throw odd then even, so (1/2)(1/2) = 1/4. Similarly, to end at three throws we must throw odd then odd then even, and P(X=3) = 1/8.

    We get the sum E(X) = 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + ...

    Above I mention a method of evaluating the sum, after which we will get E(X) = 2. Like I wrote above, the way I found seems kind of messy, but maybe that's a standard way, don't really know. If you want me to show you I can.
    Ok so essentially, P(Even)=0.5, P(Odd)=0.5

    Using a geometric distribution:

    P(X=k)=\left(\dfrac{1}{2}\right)^k which is the probability of getting an even number on the kth toss.

    Then we use a summation, to find the expected value:

    E[X]=\sum_{k=1}^{\infty}k\left(\dfrac{1}{2}\right)^k
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    Quote Originally Posted by acevipa View Post
    Ok so essentially, P(Even)=0.5, P(Odd)=0.5

    Using a geometric distribution:

    P(X=k)=\left(\dfrac{1}{2}\right)^k which is the probability of getting an even number on the kth toss.
    The equation is right, but the wording at the end is off; this is the probability of getting k-1 odds followed by a single even.

    Quote Originally Posted by acevipa View Post
    Then we use a summation, to find the expected value:

    E[X]=\sum_{k=1}^{\infty}k\left(\dfrac{1}{2}\right)^k
    Yes.
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