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The question is based on the table below:

The table shows the results of a probability experiment. Forty-eight six-sided dice were rolled, and each die that showed $“2”$ on top (a deuce) was removed. The remaining dice were rolled again, and deuces were removed. This procedure was repeated until all the dice were gone. Each entry in the left column is a roll number. The corresponding entry in the right column is the number of dice that had not yet turned up deuces after this roll. These variables are not linearly related. What happens if you try to apply logarithms to straighten this data?

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I found the linear-regression line of the data points: $y=-2.078x + 32.149$. I know that if I apply logarithm to $0$, then it will not result in a value. However, I still went ahead to find in using logarithms to find a linear line between the points... $$\log y = \log(x^{-2.078}) + 10^{32.149}$$ $$y = 10^{32.149}x^{-2.078}$$ (If there are any mistakes in the two previous steps, please let me know.)

This does not create a linear line because it is a power function.

There is a follow-up question:

If forty-eight dice are rolled once, how many do you expect to remain after removing the deuces? How many dice do you expect to remain after two applications of the above procedure? After ten applications?

But I think that if I am able to answer the previous question with the logarithms, then I may be able to answer the continued question. Any help will be greatly appreciated.

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  • $\begingroup$ I suspect you are only expected to apply the logarithm to the number of dice remaining, not the number of rolls $\endgroup$ – Henry May 12 '18 at 23:06
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You would expect $\frac 16$ of the dice to come up $2$, so would expect that after $n$ rolls you would have $48(\frac 56)^n$ dice to remain. You should be taking the log of the number of dice remaining and have to ignore the point where they are all gone. You would expect the slope of the line to be about $\log(\frac 56)$

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  • $\begingroup$ Should I not make an equation that looks something like $\log y = -2.078x + 32.149$? $\endgroup$ – geo_freak May 12 '18 at 23:17
  • $\begingroup$ I think your $y$ is the number of rolls and your $x$ the number of dice remaining. You didn't specify, but the constant $32$ makes me think that. If so, you took the log of the wrong one. $\endgroup$ – Ross Millikan May 12 '18 at 23:25
  • $\begingroup$ Sorry about that, $y$ is the number of dice remaining and $x$ is the number of rolls. I used my calculator to find the LinReg of the data points in the two columns. $\endgroup$ – geo_freak May 12 '18 at 23:27
  • $\begingroup$ It can't be right.Even if you are using base $2$ logs, you would have $y \approx 2^{32}$ for $x=0$ $\endgroup$ – Ross Millikan May 12 '18 at 23:35
  • $\begingroup$ Key confusion here is that applying logarithm to the result of linear regression is not the same as applying logarithm first, followed by linear regression. $\endgroup$ – Jonny Lomond May 12 '18 at 23:45

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