2
$\begingroup$

These questions come up regarding a RPG in which you roll multiple dice, but only keep the highest value. So, suppose I were to roll 5 dice, each of which has 12 sides (numbered from 1 to 12). If I only keep the highest roll, what would the average end up being?

To throw in an additional level of complication, what would happen if the dice "explode" or "penetrate"? Explode/Penetrate=If you roll a die and roll the highest value (if you roll an 8 on an eight sided die), re-roll the die and add the results together. Repeat until you do not roll the maximum value.

$\endgroup$
  • $\begingroup$ Average of how many rolls? $\endgroup$ – barak manos Nov 7 '14 at 16:17
  • $\begingroup$ If it wouldn't make things extremely complicated, assume a true average over an infinite number of rolls. $\endgroup$ – user190683 Nov 7 '14 at 16:20
  • $\begingroup$ It's certainly not a good answer, but if you want brute force, it's only $12^5$ combinations and a simple $max$ function on each for the first case, and that's easily done in any computer program :). $\endgroup$ – Alan Nov 7 '14 at 16:23
  • $\begingroup$ This seems related: math.stackexchange.com/questions/355820/… $\endgroup$ – Jason Knapp Nov 7 '14 at 16:29
2
$\begingroup$

For your first question (expected value for max value of 5 rolls of a 12-sided die):

Let $X$ be the maximum value.

Then $P(X=k)=\left(\frac{k}{12}\right)^5-\left(\frac{k-1}{12}\right)^5$ for $k=1,2,...,12$.

The reasoning is that for the maximum to be $k$, you need all the rolls to be less than or equal to $k$, but not all less than $k$.

So $$E[X]=\sum_{k=1}^{12}k \left( \left(\frac{k}{12}\right)^5-\left(\frac{k-1}{12}\right)^5\right)=\frac{2604108}{12^5}\approx10.4653$$

$\endgroup$
1
$\begingroup$

Here is the answer for the first part of your question:

There are one event, where the maximal number is $1$ (all dices showing $1$).

There are $2^5$ events where all dices showing either $1$ or $2$. Subtract the one event, where all dices showing $1$, you get $2^5-1$ events, where maximal number is $2$.

There are $3^5$ events, where all dices showing $1$, $2$ or $3$. Subtract the $2^5$, where the maximal number is $1$ or $2$ and you get $3^5-2^5$ events with maximal number $3$.

Continue that idea and you get the average:

$\frac{1}{12^5}\sum\limits_{i=1}^{12}(i^5-(i-1)^5)\cdot i $

(Note that $12^5$ is the whole number of possible events)

$\endgroup$
1
$\begingroup$

Let $P_{n}$ denote the probability to get a value of $n$:

  • $P_{ 1}=(\frac{ 1}{12})^5 =\frac{ 1}{248832}$
  • $P_{ 2}=(\frac{ 2}{12})^5-(\frac{ 1}{12})^5=\frac{ 31}{248832}$
  • $P_{ 3}=(\frac{ 3}{12})^5-(\frac{ 2}{12})^5=\frac{ 211}{248832}$
  • $P_{ 4}=(\frac{ 4}{12})^5-(\frac{ 3}{12})^5=\frac{ 781}{248832}$
  • $P_{ 5}=(\frac{ 5}{12})^5-(\frac{ 4}{12})^5=\frac{ 2101}{248832}$
  • $P_{ 6}=(\frac{ 6}{12})^5-(\frac{ 5}{12})^5=\frac{ 4651}{248832}$
  • $P_{ 7}=(\frac{ 7}{12})^5-(\frac{ 6}{12})^5=\frac{ 9031}{248832}$
  • $P_{ 8}=(\frac{ 8}{12})^5-(\frac{ 7}{12})^5=\frac{15961}{248832}$
  • $P_{ 9}=(\frac{ 9}{12})^5-(\frac{ 8}{12})^5=\frac{26281}{248832}$
  • $P_{10}=(\frac{10}{12})^5-(\frac{ 9}{12})^5=\frac{40951}{248832}$
  • $P_{11}=(\frac{11}{12})^5-(\frac{10}{12})^5=\frac{61051}{248832}$
  • $P_{12}=(\frac{12}{12})^5-(\frac{11}{12})^5=\frac{87781}{248832}$

Now let's calculate the expectancy (mean average):

$\sum\limits_{n=1}^{12}n\cdot{P_n}=\frac{1\cdot1+2\cdot31+3\cdot211+4\cdot781+5\cdot2101+6\cdot4651+7\cdot9031+8\cdot15961+9\cdot26281+10\cdot40951+11\cdot61051+12\cdot87781}{248832}\approx10.47$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.