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I would like to simulate some real probabilities using dice. I know how to calculate the probability of getting a certain dice roll, but has anyone shown it in reserve? By that I mean here is a way to use dice that meets a criteria at 99%? 98%... What kind of dice roll would mimic all probabilities roughly? Wish to get as accurate as can to tenth of a decimal. Would there be large gaps?

I was thinking these parameters. To simulate getting a certain percentage the person may be told to roll up to five dice. To meet a cercentage of 1/6 it may be roll one die and get a 6. For others it might be roll 5 dice and get at least all 5 and 6. Or for others it may allow two rerolls.

Anyone have a suggestion how to make this chart or do I just try different ideas on dice combinations and make my own chart.

Also the game may come with more than five dice. Maybe some dice are not 1 2 3 4 5 6 but like 1 1 1 2 2 3

This is for simulating real statistics of something in a board game.

Robert Cousins
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    What do you actually want to mimic? – Mesmerized student Jun 18 '17 at 23:16
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    Think about what you actually want to do with these dice and then write what you want in an articulate manner. If you really want to mimic, you may want to look at programming, several languages will have a random number generator and will allow you to "mimic" dies. – mdave16 Jun 18 '17 at 23:25
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    What does "Would tellers be large gaps?" mean? What does "has anyone shown it in reserve? " mean? Please reread your question and make it clear. – Ross Millikan Jun 19 '17 at 00:12

2 Answers2

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To make things easier to write, I'll use the word "win" to mean "meet the criterion" and "lose" for "not meet the criterion."

There are a lot of different ways to designate "winning" and "losing" results on dice. For example, some people might roll a single die $4$ times (or roll $4$ different-colored dice, always reading the red die first, then the orange die, and so forth). This generates $6^4 = 1296$ possible results, all equally likely, so there is always some set of outcomes whose probability is within $0.1\%$ of any probability you want between $0$ and $100\%.$

If you don't like that technique, by rolling two dice and adding the numbers you get a result from $2$ to $12.$ The probabilities of each sum are such that you can precisely model any multiple of $1/36$ probability by assigning some subset of the possible sums to the "winning" category. For example, to model a probability of $13/36,$ roll two dice and win if the sum is $3,$ $6,$ or $7.$

To model any multiple of $1/1296$ probability, you roll once with some set of "winning" sums, roll again if the sum is $12$ (with a possibly different set of "winning" sums on the second roll); any other roll loses.

If you are willing to put numbers greater than $6$ on your dice then you can (for example) label one die $1,2,3,4,5,6$ and another die $1,7,13,19,25,31,$ which lets you roll any sum from $2$ through $37$ (inclusive) with equal probability. This makes a table easier to write: for example, for a probability of $30\%$ you would specify the number sequence $27,9,$ meaning that you win if you roll greater than $27$ on the first roll and lose if you roll less, but if you roll $27$ exactly then you roll again and win if you roll at least $9.$

If you allow a second re-roll then you can get much closer than $0.1\%$ to any probability. In fact, there are schemes for modeling any rational probability exactly, as long as you are willing to keep re-rolling until some condition is achieved (which will happen eventually with probability $1.$) See Eight-sided dice from six-sided dice and Creating unusual probabilities with a single dice, using the minimal number of expected rolls for examples.

There are, in fact, many possible ways you could organize the rolling of dice to model any particular probability (either approximately or exactly); there is not one canonical way to do it. It is a matter of personal preference to decide what kind of scheme you will use.

David K
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The general way to produce a given probability is to use a uniform distribution of the unit interval and the event $X\le P$ (or $X<P$).

So the question is how to approximate a uniform distribution. It's simple with a $d10$, just write $0.$ and then throw and write down the result minus one and repeat. For example to approximate the distribution with (roughly) and throw a $6$, a $8$ and a $3$ you would have produced the random number $0.572$.

Now if you're using a $d6$ you'll do the same thing, but you will have to work in base $6$ instead.

Now if you want to simulate $99\%$ probability you have to know what $0.99$ is in base $6$. This is done by repeatedly multiplying with $6$ and take and drop the integer part:

$$0.99\times 6 = 5.94$$ $$0.94\times 6 = 5.64$$ $$0.64\times 6 = 3.84$$ $$0.84\times 6 = 5.04$$ $$0.04\times 6 = 0.24$$

etc, so in base $6$ it would be $0.55350\cdots$. You trow dice and once you get above or below the "magic" sequence of $6,6,4,6,1,\cdots$ you stop and if you comes high you failed and if you stop below you succeed.

For the probabilities in each step you have:

$$\begin{matrix} \text{exited with higher value} & \text{still in iteration} & \text{exited with lower value} \\ \hline 0 (0\%) & 1/6 & 5/6 (83.33\%) \\ 0 (0\%) & 1/36 & 35/36 (97.22\%) \\ 2/216 (0.92\%) & 1/216 & 213/216 (98.61\%) \\ 12/1296 (0.92\%) & 1/1296 & 1283/1296 (98.92\%) \\ 77/7776 (0.99\%) & 1/7776 & 7698/7776 (99.00\%) \\ \end{matrix}$$

If you don't mind the case that you will throw the magic sequence indefinitely (with probability $0$) you can achieve the exact desired probability.

skyking
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