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I am playing at making my own table-top gaming system/rules and I wanted to have a better handle on how likely different dice combinations will give a higher result than one another. I know that a six sided die roll averages to 3.5, and an eight sided die roll averages 4.5, but I still don't quite have a grasp on just how likely it is an 8 sided die comes up with a higher result than a 6 sided one.

I would also like to know how adding integers to die results effects their comparative advantage as well, like how often would the sum of 3 six-sided dice with a 1 added to the final result give a higher outcome than just 3 six-sided dice?

Thanks in advanced for any advice, I'm just not really sure where to start with this, I focused mostly on algebra/calc/trig in school and never really did any probability/stat.

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  • $\begingroup$ you might start be reading my answer to a similar question here. The answer I gave there discusses how to compute the average result of arithmetic combinations of results of dice as well as how to calculate the variance (similarly the standard deviation). While calculating the variance of two different scenarios (e.g. throwing three d6 vs three d4) won't tell you exactly the probability directly, it will give you intuition via en.wikipedia.org/wiki/Chebyshev%27s_inequality $\endgroup$ – JMoravitz Apr 16 '15 at 16:07
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For your first question, six-sided die vs. eight-sided, make a $6$ by $8$ table, with values $1, 2, 3, 4, 5, 6$ in one direction and $1, 2, 3, 4, 5, 6, 7, 8$ in the other direction.

The resulting $48$ small squares in the table determine $48$ possible outcomes of the two dice. You can then see for how many squares does six-sided beat eight-sided; how many ties; and how many times eight-sided beats six-sided. Assuming the dice are fair, you can then divide by $48$ to get the desired probabilities.

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  • $\begingroup$ This is a very useful approach for comparing a single die result vs another single die result. Unfortunately, when comparing sum of 3 dice, to a sum of a different 3 dice, it becomes much harder to visualize as a convenient 2 dimensional grid (as it is more natural to consider as a six-dimensional grid, one direction for each of the six dice). $\endgroup$ – JMoravitz Apr 16 '15 at 16:09
  • $\begingroup$ Thanks very much, that is an excellent way to approach the problem. I do a fair bit of coding so I could totally mock up multi-dimensional arrays and check every possibility that way. $\endgroup$ – axel Apr 16 '15 at 16:14
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There are online utilities that can do the calculations for you. This one for example looks good to me: you can combine dice rolls (subtract 1d6 from 1d8 for example) to get a probability curve for all the possible results.

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  • $\begingroup$ I have taken a look at anydice before, it as a useful utility but I am not sure how I could derive the probability of 2 competing sets of dice using it, $\endgroup$ – axel Apr 16 '15 at 16:19
  • $\begingroup$ @axel for example: set the formula to "output 1d8 - 1d6" and Data to atleast shows that 20.83% chance of a 4 or more, 56.25% chance of 1 or more (i.e. d8 rolled higher than the d6). $\endgroup$ – Avon Apr 16 '15 at 16:21
  • $\begingroup$ oh! OH! ok, I get it now, I was spending the last few minutes trying to wrap my head around how that helps me but I get it now. and I imagine the % of 1 or more minus 0 or more gets me the chance they come up equivalent? $\endgroup$ – axel Apr 16 '15 at 16:31
  • $\begingroup$ Even simpler: the % of 0 total is the chance they are equal. $\endgroup$ – Avon Apr 16 '15 at 16:33
  • $\begingroup$ (And by that I mean under the 'Normal' Data output instead of the 'Atleast'.) $\endgroup$ – Avon Apr 16 '15 at 16:42
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Because you're making a gaming system, I guess you might be comfortable with some simulations that require little programming. Here is a program with an approximate answer to your question about 3 rolls of a six-sided die with and without a bonus point. (The third decimal place may be off by 1 or 2.)

The probability is about 0.546 that three rolls with a bonus point beats three rolls without. (About 0.36 of being greater than or equal.)

m = 10^6;  x1 = x2 = numeric(m)
for(i in 1:m) {
  x1[i] = sum(sample(1:6, 3, rep=T))
  x2[i] = sum(sample(1:6, 3, rep=T)) + 1 }
mean(x2 > x1);  mean(x2 >= x1)
## 0.546143
## 0.636343
mean(x2 - x1)
## 1.000961  # Reality check: theoretical value 1

This kind of simulation might make it very easy for you to investigate various options quickly and informally. Of course, you'd have to do the math for exact probabilities to many places.

[The small program above is written in R, excellent statistical software available free at r-project.org.]

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