Calculating mean after removing smallest items

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $(10\times 3.5−1.18−1.60)/8 \approx 4.03$