Calculating mean after removing smallest items

I'm having a tough time wrapping my head around this question.

Lets say that I am doing an experiment where I roll 10 dice. Each time I roll the dice, I record the average value and repeat the process. If I find the average value of the results as I repeat the experiment, it should approach $3.5$ (because each value has has equal probability).

Here's my question: Lets say that I modify the experiment and after rolling the dice, I remove the two lowest dice and record the average of the remaining dice. How does the average value of the results change? What if I remove $N$ dice?

I'm sure that I'm just looking at it the wrong way. Any nudges in the right direction would be appreciated.

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As the number of dice rolls tends toward infinity, the removal of any finite number of dice will not affect the average. – Austin Mohr Sep 9 '12 at 23:24
@AustinMohr, the removal of the dice is not random though. The smallest values are removed each time, therefore breaking the assumption that each value will have equal probability (in other words, the probability of removing a 1 is substantially higher than the probability of removing a 6) – Stargazer712 Sep 9 '12 at 23:26
Think of it this way: If you roll sixty dice, you expect ten of each of the numbers 2 - 6 and eight 1's. This is a slightly higher average than usual. If you roll six million dice, you expect one million of each of the numbers 2 - 6 and 999,998 occurrences of 1. This is a very slightly higher average than usual. As the number of dice rolls tends toward infinity, the "very slightly higher" tends toward 0. – Austin Mohr Sep 9 '12 at 23:31
If you roll ten dice and deleted the smallest two numbers, and take the expected value of that, it's definitely larger than $3.5$. That larger number is what will be approached if the experiment is repeated ad infinitum. – Michael Hardy Sep 9 '12 at 23:31
@AustinMohr : You seem to be understanding the problem differently from the way I am. You're not removing the smallest two of six million; you're removing the smallest two every time you role ten dice. – Michael Hardy Sep 9 '12 at 23:33

Of the $6^{10}$ possible rolls, $6^{10}-5^{10}-10\cdot5^9$ have two $1$’s. Of the $10\cdot5^9$ rolls that have exactly one $1$, $10(5^9-4^9)$ have a $2$ as the lowest of the remaining nine dice; $10(4^9-3^9)$ have a $3$; $10(3^9-2^9)$ have a $4$; $10(2^9-1)$ have a $5$; and $10$ have a $6$.

Of the $5^{10}$ rolls that have no $1$’s, $5^{10}-4^{10}-10\cdot4^9$ have two $2$’s. Of the $10\cdot4^9$ rolls that have exactly one $2$, $10(4^9-3^9)$ have a $3$ as the lowest of the remaining nine dice; $10(3^9-2^9)$ have a $4$; $10(2^9-1)$ have a $5$; and $10$ have a $6$.

Of the $4^{10}$ rolls that have no $1$’s or $2$’s, $4^{10}-3^{10}-10\cdot3^9$ have two $3$’s. Of the $10\cdot3^9$ rolls that have exactly one $3$, $10(3^9-2^9)$ have a $4$ as the lowest of the remaining nine dice; $10(2^9-1)$ have a $5$; and $10$ have a $6$.

Of the $3^{10}$ rolls that have no $1$’s, $2$’s, or $3$’s, $3^{10}-2^{10}-10\cdot2^9$ have two $4$’s. Of the $10\cdot2^9$ rolls that have exactly one $4$, $10(2^9-1)$ have a $5$ as well, and $10$ have only $6$’s.

Of the $2^{10}$ rolls that have only $5$’s and $6$’s, $2^{10}-1-10$ have two $5$’s, and $10$ have a $5$ and nine $6$’s.

There is one roll whose smallest two dice are both $6$’s.

The sum of the numbers on the removed dice is therefore

\begin{align*} &2\left(6^{10}-5^{10}-10\cdot5^9\right)+10\sum_{k=2}^6\left((7-k)^9-(6-k)^9\right)(k+1)\\ &\qquad+4\left(5^{10}-4^{10}-10\cdot4^9\right)+10\sum_{k=3}^6\left((7-k)^9-(6-k)^9\right)(k+2)\\ &\qquad+6\left(4^{10}-3^{10}-10\cdot3^9\right)+10\sum_{k=4}^6\left((7-k)^9-(6-k)^9\right)(k+3)\\ &\qquad+8\left(3^{10}-2^{10}-10\cdot2^9\right)+10\sum_{k=5}^6\left((7-k)^9-(6-k)^9\right)(k+4)\\ &\qquad+10\left(2^{10}-1-10\right)+10\cdot11\\ &\qquad+12\\ &=2\sum_{k=1}^6k^{10}+10\cdot5^9+20\cdot4^9+30\cdot3^9+40\cdot2^9+50\\ &=168,066,052\;, \end{align*}

if I’ve made no computational errors.

The sum of all the dice in all $6^{10}$ possible rolls is $10\cdot3.5\cdot6^{10}=2,116,316,160$; subtracting the total of the two lowest numbers in each leaves $1,948,250,108$, for an average total per roll of $10$ dice of $32.22049477711$ and an average per die of $4.027561847139$. Henry’s simulation is right on the money in this case.

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Wow. Very thorough answer :) – Stargazer712 Sep 10 '12 at 1:32

Empirically (a sample of 1 million), the values are about

Nth_lowest  Expected    Mean_if_N_excluded
0           n/a         3.5
1           1.18        3.76
2           1.60        4.03
3           2.13        4.30
4           2.68        4.57
5           3.23        4.84
6           3.77        5.10
7           4.32        5.36
8           4.87        5.61
9           5.40        5.82
10          5.82        n/a


where for example with $N=2$ you have $(10\times 3.5−1.18−1.60)/8 \approx 4.03$

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There are $6^{10}$ possible outcomes when you roll ten dice.

The number of cases in which the minimum is more than $1$ is $5^{10}$, so the number of cases in which the minimum is $1$ is $6^{10}-5^{10}$.

The number of cases in which the minimum is more than $2$ is $4^{10}$, so the number of cases in which the minimum is $1$ or $2$ is $6^{10}-4^{10}$. In $6^{10}-5^{10}$ of those, the minimum is $1$, so in $(6^{10}-4^{10})-(6^{10}-5^{10})$, the minimum is $2$, i.e. in $5^{10}-4^{10}$.

Keep going. You will find the number of equally probable cases in which the minimum has each of the six possible values. Then you can find the expected value of the minimum.

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That gives the mean of the lowest as $\frac{71340451}{60466176} \approx 1.18$ but it is not so easy to find the mean of the second lowest to be about $1.60$ and thus the answer to the question to be about $(10\times 3.5 - 1.18 -1.60 )/8 \approx 4.03$. – Henry Sep 10 '12 at 0:35