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I'm trying to figure out the odds of a random generator generating the same list of numbers $455$ times in a row. The random numbers are in a range from $0-1.999999$ ($6$ decimal points)

I figure that the odds of getting a match on any number in the sequence are $1$ out of $10^7$. Raising that value to $455$ should yield my desired result. Correct?

However Excel won't generate that answer for me. Does anyone have a tool that can calculate this in scientific notation?

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  • $\begingroup$ Wow that was fast! But I think I made an error. Shouldn't it be (.5 * 10^6)^455 ? The first digit can only be 1 or 0. $\endgroup$ Aug 15 '13 at 17:26
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You write that your random numbers are from the range from $0$ to $1.999999$ with six decimal places. Hnece there are $2000000=2\times 10^6$ possible numbers. Then the final result is much bigger of course: $$\approx 1.0748601772107342002865544942320363407 \times 10^{-2867}$$

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  • $\begingroup$ 2 x 10^6 looks correct, but how did you generate the approximation? $\endgroup$ Aug 15 '13 at 17:31
  • $\begingroup$ A calculator. (Know anything of logarithms?) $\endgroup$
    – anon
    Aug 15 '13 at 17:33
  • $\begingroup$ Sadly, I've forgotten most of that stuff :( $\endgroup$ Aug 15 '13 at 17:34
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    $\begingroup$ @Whisk: With good-enough scientific calculator (base change formula yields $\log_{10}2=\ln 2/\ln 10$): $$\begin{array}{ll} (2\times 10^6)^{-455} & =10^{-455(6+\ln2/\ln10)} \\ & =10^{-2866.9686\cdots} \\ & =10^{0.0313\cdots}\times10^{-2867} \\ & =1.07486\cdots\times10^{-2867}.\end{array}$$ $\endgroup$
    – anon
    Aug 15 '13 at 17:40
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You don't need a computer to put $ \left( \dfrac {1}{10^7} \right)^{455} $. This is just $ \dfrac {1}{10^{7 \cdot 455}} = \boxed {\dfrac {1}{10^{3185}} = 10^{-3185}} $.

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  • $\begingroup$ Probably because you put too much faith in OP's original understanding of the problem (and didn't check it yourself), and subsequently failed to correct the answer in light of new information. Specifically, the $10^{-7}$ figure is incorrect. The other two answers using that figure were deleting by their owners. $\endgroup$
    – anon
    Aug 17 '13 at 10:26
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Actually you need to calculate $$(1/2 * 10^{-6} )^{455}$$ Which is by no means easy :
$$1.07486017721073420028655449423203634073126755189808458... × 10^{-2867}$$ (Using Wolfram alpha)

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