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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- Aug 2nd 2010, 08:33 PMacevipaExpected 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 - Aug 2nd 2010, 09:27 PMundefined
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. - Aug 2nd 2010, 09:35 PMacevipa
- Aug 2nd 2010, 09:38 PMundefined
- Aug 2nd 2010, 09:49 PMacevipa
- Aug 2nd 2010, 09:54 PMundefined
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. - Aug 2nd 2010, 10:02 PMacevipa
Ok so essentially, P(Even)=0.5, P(Odd)=0.5

Using a geometric distribution:

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

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

$\displaystyle E[X]=\sum_{k=1}^{\infty}k\left(\dfrac{1}{2}\right)^k$ - Aug 2nd 2010, 10:09 PMundefined