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You have a lottery number with ten slots. On each slot is an equally probably number from 0 to 1. You are paid the maximum number of the ten slots. What is the expected payout of the lottery number.

My strategy was to use the integral definition for expected value. The probability $p(x)$ that a single value $x$ is the highest of all the ten slots means that all the other values must fall within the range from $0$ to $x$, which happens with probability of $x^9$. Since there are ten different spots that this happens is $10x^9$. Then using the definition of expected value, we get $\int_0^1 x10x^9 \,dx$ = $\frac{10}{11}$.

Does this look correct?

GTOgod
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1 Answers1

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Based on your description of the problem, it is expected that the expectation value will be 5. Since it is equal probability to choose a number between 0 and 1 for each box, the expectation value in each box is 1/2. Thus summing over 10 boxes would make the value 5. More rigorously, you can represent the expectation value for each box as an integral then sum over the boxes $$E(X) = \sum_{n=1}^{10}\int_0^1x\,dx=\sum_{n=1}^{10}\frac{1}{2}=5$$

MasterYoda
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  • The problem I'm having is that x is uniformly distributed on the interval and can hence be any value. So I do need to integrate somehow to account for the fact that x can range in [0,1] I would think? – GTOgod Nov 28 '17 at 21:44
  • Rereading, I am fairly certain the finiteness of the slots is kept in check by the probability function itself in the fact that i used x^9 to model the highest probability – GTOgod Nov 28 '17 at 22:01
  • Oh okay. I think I might understand your problem now. I rewrote my solution. – MasterYoda Nov 28 '17 at 23:33
  • I think you misread the question. I am not looking for the sum of the values in the digits, if it were, then the answer would be 5. I am looking for the expected payout, which is the highest digit out of the ten. As such the answer must be between 0 and 1. – GTOgod Nov 28 '17 at 23:37