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A die is thrown and the number is noted. Then the die is thrown again repeatedly until a number at least as high as the number obtained on the first throw is thrown. Find the mean number of times the die is thrown, including the first throw.

mean_number_of_throws_calculation

The answer is $3.45$, but I am getting $2.53$.

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  • $\begingroup$ You seem to be trying to calculate something, but it is not entirely clear what your numbers below are trying to calculate. It almost looks like the number you calculated is the expected amount of points you get if you roll the die two times and get $7-k$ points if your first roll was $k$ and your second roll was greater than or equal to $k$. This calculation has almost nothing to do with the problem asked however where you roll the die repeatedly until getting a number at least as large as the first you rolled. $\endgroup$ – JMoravitz Nov 7 '17 at 22:18
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    $\begingroup$ Lets look at a single part of the larger problem. If your first roll is a $4$, how many rolls on average will it take to roll a number at least as large as a $4$ again? Think using a hypergeometric distribution. $\endgroup$ – JMoravitz Nov 7 '17 at 22:19
  • $\begingroup$ So am I going about this completely wrong? This question was in the conditional probability section of my book so I was trying to base it off that. $\endgroup$ – Bey Nov 7 '17 at 22:27
  • $\begingroup$ @JMoravitz do you mean the geometric distribution? The hypergeometric distribution describes the distribution of successes in a number of draws without replacement. $\endgroup$ – Theoretical Economist Nov 7 '17 at 22:27
  • $\begingroup$ @TheoreticalEconomist ah, yes indeed I did mean the geometric distribution. @ Lil B, yes it is tangentially related to conditional probability, but more specifically it has to do with conditional expected value, $E[X\mid Y]$, and being able to break down the calculations for that. $\endgroup$ – JMoravitz Nov 7 '17 at 22:29
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If your first roll was a $4$ then for each roll thereafter there is a $\frac{3}{6}$ chance to roll a number at least a large as a $4$ again. Looking at this specific case, the expected number of rolls until doing so would be $\frac{6}{3}=2$.

In general, if you have chance $p$ for success, it will take on average $\frac{1}{p}$ many independent attempts to get your first success.

Noting that having rolled a four as your first roll only accounts for $\frac{1}{6}$ of the time and calculating the rest of the related probabilities and finally accounting for the initial roll we get the final answer.

$1+\frac{1}{6}(\frac{6}{1}+\frac{6}{2}+\frac{6}{3}+\frac{6}{4}+\frac{6}{5}+\frac{6}{6})=3.45$ The $1$ comes from the initial roll, the $\frac{1}{6}$ comes from the chance to be in each respective case, and each $\frac{6}{k}$ comes from the expected number of rolls until rolling a $7-k$ or greater.

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    $\begingroup$ The phrase "each $\frac{6}{k}$ comes from the expected number of rolls until rolling a $k$ or greater" is somewhat misleading since $\frac{6}{k}$ is the expected number of rolls until rolling a $7 - k$ or greater. $\endgroup$ – N. F. Taussig Nov 8 '17 at 10:20

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