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I am trying to figure out probabilities for ten-sided dice for my RPG system just to make sure my stats are correct, but I thought I'd check my workings with better maths brains than mine.

The lowdown:

  • This system requires you to roll a pool of d10 of between 2 dice and 9 dice.
  • You look for matching pairs and matching triples. A pair is a success, and so is a triple.
  • (A triple is treated as a pair, with the third dice 'discarding' itself to give you a bonus effect of modifying its pair up or down by +1/-1, which is sometimes advantageous, as higher and lower pairs both have different advantages in system).
  • More pairs mean more successes. So we want as many as we can.
  • Naturally, with 2 dice rolled, the chance of a pair is 10%. There is an early biased curve of any success occuring, rising to 84.88% by the time you reach six dice rolled, and 99.64% at nine dice. This, I'm all good with.

Where I'm not sure I'm doing things right is working out the possibilites for a single triple, and then two pairs in a given pool.

What I've been doing for a triple is taking the chance of a pair, and then multiplying it by (1/10) + (1/10) for each extra dice besides the two dice already in the pair.

Hence, for a triple in six dice: $$0.8488 \times ((1/10)+(1/10)+(1/10)+(1/10))$$

And then, for the chance of two pairs in six dice, multiplying the chance of a pair in a pool of six dice by the chance of a pair in four dice.

So: $$(0.8488 \times 0.4960)$$

Which gives us this chart:

probability chart

Now, assuming all this is correct, it means that, once we reach six dice rolled, the chance of there being two pairs exceeds that of there being a triple.

Which surprises me. It sounds like an anomoly. I may just be being thick about this, and that really is truly the case, but I just want to make sure I'm not doing something terribly wrong here!

Thanks!

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  • $\begingroup$ Jax'ed the two formulae for you. $\endgroup$ – eharper256 Apr 17 '18 at 11:15
  • $\begingroup$ Two pairs are more common than a triple: For the same reason, two pairs are less worth than a triple in Poker. $\endgroup$ – Luke Apr 17 '18 at 11:22
  • $\begingroup$ Hmm, I didn't even think about scoring in Poker, but I guess that's a good point, someone must have had those chances in mind already. $\endgroup$ – eharper256 Apr 17 '18 at 11:29
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As you suspected, you didn't calculate the probability of a triple correctly.

Take the case of $3$ dice, for example. To get a triple, the first die can be anything, but then there is a $10$% chance the second die matches the first, and also a $10$% chance the third matches it as well. Hence, the probability of getting a triple with $3$ dice is $0.01$, rather than $0.028$

OK, so why can't you multiple the chance of getting a pair, and multiply by $0.1$ for the third die? It's because the probability of a pair is already taking all $3$ dice into account. It would make sense to multiply by $0.1$ the probability of the first two dice being a pair ... and indeed given that that probability is $0.1$, you end up with the correct $0.01$. Similarly, for more than $3$ dice, it only makes sense to multiply by $0.1$ the probability of the first $n-1$ dice to form a single pair, but the probability you are working with again takes all dice account already.

To get the formula for the probability of getting a triple, I would (like you do with the pair), calculate the probability of not getting a triple and subtract from $1$. OK, but how to calculate that? I am thinking that some generating function will be applicable here, but I am not good with those. So, let's do this the hard way.

Take $4$ dice. How not to get a triple? Well, they can all be different, or there can be one pair, or there can be two pairs. Now, you already calculated the chance of all different, which is $\frac{10\cdot9\cdot8\cdot7}{10^4}=0.504$. For exactly one pair: there are $10$ options for the number that occurs as a pair, and $9 \choose 2$ options for the other two numbers, there are $4 \choose 2$ ways for that pair to occur among the $4$ dice, and $2$ ways for the other two numbers, giving a total of $10\cdot{9 \choose 2} \cdot {4\choose 2}\cdot 2$ ways for exactly one pair to happen, thus giving a probability of $\frac{10\cdot{9 \choose 2} \cdot {4\choose 2}\cdot 2}{10^4}$ for that to happen. For two pairs: $10 \choose 2$ possibilities for the two numbers, and $4 \choose 2$ for those two pairs to be distributed among the $4$ dice, so ${10 \choose 2}\cdot {4 \choose 2}$ ways to get two pairs, giving a probability of $\frac{{10 \choose 2}\cdot {4 \choose 2}}{10^4}$

OK, that's not a closed formula .. but maybe you can do this process in Excel without too much trouble for more dice. Good luck!

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  • $\begingroup$ I see. So its actually even less than I suspected? Do you know would I express that as an excel formulae? At the moment I'm calculating the chance of it not being a pair, minusing that chance from 1, and then multiplying it by the (1/10)'s, which is clearly where I'm flopping. $\endgroup$ – eharper256 Apr 17 '18 at 11:26
  • $\begingroup$ @eharper256 Right, that's not correct, as I explained ... let me think about what the correct formula would be ... $\endgroup$ – Bram28 Apr 17 '18 at 11:27
  • $\begingroup$ So the brute force way to calculate exactly one pair is hence (10 x (9/2) x (4/2) x 2)/(10x10x10x10)? Or am I reading that entirely wrong? $\endgroup$ – eharper256 Apr 17 '18 at 11:58
  • $\begingroup$ Nah thats obviously wrong, that gives 0.018 which is clearly not 0.504. Derp; my B in GCSE maths isn't helping (laughs). $\endgroup$ – eharper256 Apr 17 '18 at 12:05
  • $\begingroup$ Nah, I'm facerolling my keyboard at the moment. I can't distill that into an excel formulae at all that makes sense. T_T $\endgroup$ – eharper256 Apr 17 '18 at 13:36
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Consider $n$ fair $10$-sided dice, where $n$ is a positive integer between $3$ and $20$. (By the pigeonhole principle, if there are more than $20$ dice, we are guaranteed to obtain at least one triplet.)

Now we are interesting in grouping the undesired outcomes (no triplets) by the number of distinct values $k$ showing when the dice are rolled, such that no value occurs more than twice. For $n = 3$, we can have $k \in \{2, 3\}$; for $n = 4$, we can have $k \in \{2, 3, 4\}$; for $n = 5$, we can have $k \in \{3, 4, 5\}$; and it is not too hard to see that for general $n$, we can have $$\lceil n/2 \rceil \le k \le \min(n, 10).$$

With $k$ distinct values on $n$ dice, there are $n-k$ values that appear exactly twice, and $2k-n$ that appear exactly once. There are $$\frac{10!}{(10-k)!}$$ orderings of the $k$ distinct values, each of which admits $$\frac{n!}{(2!)^{n-k} (n-k)!(2k-n)!}$$ ways to order the $n$ dice. Thus there are a total of $$S(n) = \sum_{k = \lceil n/2 \rceil}^{\min(n,10)} \frac{10!}{(10-k)!} \cdot \frac{n!}{2^{n-k} (n-k)!(2k-n)!} = \frac{10! \, n!}{2^n} \sum_{k=\lceil n/2 \rceil}^{\min(n,10)} \frac{2^k}{(10-k)!(n-k)!(2k-n)!}$$ outcomes without a triplet, and the desired probability is $$1 - \frac{S(n)}{10^n}.$$ This gives the table

$$ \begin{array}{|c|c|c|} \hline n & \text{probability} & \text{approx.}\\ \hline 3 & \frac{1}{100} & 0.01 \\ 4 & \frac{37}{1000} & 0.037 \\ 5 & \frac{107}{1250} & 0.0856 \\ 6 & \frac{197}{1250} & 0.1576 \\ 7 & \frac{6289}{25000} & 0.25156 \\ 8 & \frac{90673}{250000} & 0.362692 \\ 9 & \frac{604187}{1250000} & 0.48335 \\ 10 & \frac{3776179}{6250000} & 0.604189 \\ 11 & \frac{17896057}{25000000} & 0.715842 \\ 12 & \frac{202679881}{250000000} & 0.81072 \\ 13 & \frac{552756829}{625000000} & 0.884411 \\ 14 & \frac{2925783983}{3125000000} & 0.936251 \\ 15 & \frac{60553812757}{62500000000} & 0.968861 \\ 16 & \frac{123360298937}{125000000000} & 0.986882 \\ 17 & \frac{622134354217}{625000000000} & 0.995415 \\ 18 & \frac{624218460241}{625000000000} & 0.99875 \\ 19 & \frac{62485150744579}{62500000000000} & 0.999762 \\ 20 & \frac{624985150744579}{625000000000000} & 0.999976 \\ \ge 21 & 1 & 1 \\ \hline \end{array} $$ I have not given much thought to enumerating outcomes where we have at least two pairs, but it seems fairly simple. I encourage you to try it, now that you have seen how we reasoned in the triplet case. When I have time I might get around to amending this answer.

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  • $\begingroup$ The astute reader may note that the expression $10!/((10-k)!(n-k)!(2k-n)!)$ is actually a multinomial (trinomial) coefficient, which may be written $$\binom{10}{10-k, n-k}.$$ $\endgroup$ – heropup Apr 18 '18 at 6:23
  • $\begingroup$ Thanks for providing all of that, the table especially is appreciated. It still goes way over my head on how to obtain S(n) though, and how to enuciate all of this as an excel formulae. But at least I have the results to use now. $\endgroup$ – eharper256 Apr 18 '18 at 11:08
  • $\begingroup$ I completed an 80 sample 6-dice rolling test and ended up with 12.35% trebles, so allowing for deviation, the 15.76% is quite close. (in this test, I also got 45.68% one pair, and 16.05% two pairs, lending credence to the idea of two pairs being somewhat more frequent, though of course one sample is hardly sufficient as proof). $\endgroup$ – eharper256 Apr 18 '18 at 11:11
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One approach is to use an exponential generating function.

Let's say we want to compute the probability that there are no triples when rolling $n$ ten-sided dice. There are $10^n$ possible outcomes, all of which we assume are equally likely. We would like to count the number of outcomes in which there are no triples; call this number $a_n$. We define the exponential generating function, $f(x)$, by $$f(x) = \sum_{n=0}^{\infty} \frac{a_n}{n!} x^n$$ In order for there to be no triples, each number from one to ten must appear zero, one, or two times in the $n$ dice. So $$f(x) = \left[ 1 + x + \frac{1}{2!} x^2 \right]^{10} $$ Expanding the right-hand side with a computer algebra program, we have $$f(x) =1+10 x+50 x^2+165 x^3+\frac{1605 x^4}{4}+762 x^5+1170 x^6+1485 x^7+\frac{12645 x^8}{8}+\frac{5695 x^9}{4}+\frac{4363 x^{10}}{4}+\frac{5695 x^{11}}{8}+\frac{12645 x^{12}}{32}+\frac{1485 x^{13}}{8}+\frac{585 x^{14}}{8}+\frac{381 x^{15}}{16}+\frac{1605 x^{16}}{256}+\frac{165 x^{17}}{128}+\frac{25 x^{18}}{128}+\frac{5 x^{19}}{256}+\frac{x^{20}}{1024}$$ So, for example, $$a_6 = 6! \times 1,170 = 842,400$$ which is the number of possible outcomes in rolling six ten-sided dice with no triples. Therefore the probability of getting no triple in six dice is $$\frac{a_6}{10^6}=0.8424$$ and the probability of getting at least one triple is $$1 - 0.8424 = 0.1576$$ The probabilities of getting at least one triple for other numbers of ten-sided dice are easy to calculate by the same method, using the expanded form of $f(x)$ above. Notice that $a_n=0$ for $n>20$, which captures the fact that there is always at least one triple when rolling more than $20$ dice, as we know from the Pigeonhole Principle.

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  • $\begingroup$ Thanks, that's helpful as well. Any idea how I would express this as an excel formulae as well? $\endgroup$ – eharper256 Apr 21 '18 at 8:37
  • $\begingroup$ @eharper256 You could type the coefficients of the expanded form of $f(x)$ into a table in Excel which could then be referenced in a VLOOKUP formula, for example, provided you used two columns, one containing $n$ and the other containing the coefficient of $x^n$. $\endgroup$ – awkward Apr 21 '18 at 11:54

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