# How to calculate probability of dice rolls [closed]

I'm playing board game of Zombicide. One of the actions in there is attack, where you roll a number of 6 sided dice, and you get a hit for every dice that showed more than "m".

For example - roll 4 dice, and you get hit for every dice that showed 3 or more.

Calculating odds for single dice is simple - 1 dice, hit from 3, gives me 66% chance of hitting. But I can't figure out correct formula to calculate probability of "m hits using n dice, and x threshold value".

I was able to "calculate" (well, more guesstimate) that when using 2 dice, hits from 4, I get:

• 75% chance of 1 (or more) hits
• 50% chance of 2 hits.

But how to make a formula for this?

Anyone can help?

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## closed as unclear what you're asking by Lost1, Yiorgos S. Smyrlis, Sami Ben Romdhane, TMM, Claude LeiboviciJan 18 '14 at 15:51

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what does 'hit from 3' mean? how is that 66%? –  Lost1 Jan 18 '14 at 12:19
@Lost1: possibly the probability of throwing a 3, 4, 5 or 6 from one die as in $\frac46$. It makes you wonder about "more than" –  Henry Jan 18 '14 at 12:20
If you call rolling greater than "m" on a single die a "success", and if you roll $n$ times, then the number of successes in the $n$ rolls has a Binomial distribution (Google). –  David Mitra Jan 18 '14 at 12:20
This link looks like a good introduction. –  David Mitra Jan 18 '14 at 12:27
Hmm .. not sure what is not clear, let me show by example. Let's assume we're throwing 1 dice, and we have "3" as thres hold. This means that if the dice will roll to 3, 4, 5 or 6, There will be one hit. This means that I have 66% of chance of getting 1 hit. With two dices, each has 66% chance of generating hit, so I can have 0, 1 or 2 hits, with probabilities (if I calculate correctly) - 11%, 88% and 44% (it doesn't sum to 100, because you have "1" hit even if dice gave you 2, 3 or more. –  user122207 Jan 18 '14 at 17:58

Note your hitting probability is $p=\frac{6-x+1}{6}$

If you want to hit exactly $m$ times out of $n$, you need binomial distribution $B(n,p)$

from this, hitting $m$

$P(\text{hitting exactly m times})=\frac{n!}{m!(n-m)!}p^m(1-p)^{n-m}$

Probability of hitting at least m times out of n is

$P(\text{hitting at least m times})=\sum\limits_{i=m}^n \frac{n!}{i!(n-i)!}p^i(1-p)^{n-i}$

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While it might look cool, it gives me literally nothing, as I just don't understand the formulas. Sorry - was hoping something simpler can be provided. Well, I can always write a program that will "roll the dice" a million times, and get percentages off it. –  user122207 Jan 18 '14 at 18:02

This is simple.

We have $n$ dice, get a hit if its $x$ or more and you want to know the probability of getting exactly $m$ hits.

Lets start with a few specific examples: $x = 3$, $n = 4$

The probability from one dice of getting a hit is $\frac{6-x+1}{6} = \frac{4}{6}$

and we need this to occur m times.

The probability of no hit is $\frac{6-x-1}{6} = \frac{2}{6}$

and we need this to occur n-m times.

So for $m = 1$ this can occur on any of our four dice

\begin{align} \text{Probability} & = \frac{4}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} + \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} + \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} + \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{4}{6}\\ & = 4 \cdot \frac{4}{6} \cdot \left( \frac{2}{6} \right)^3 \end{align}

Because we need it to occur at least one and that can be on any of the 4 dice.

For $m = 2$ we need this to happen twice on any of the 4 dice

\begin{align} \text{Probability} & = \frac{4}{6} \cdot \frac{4}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} + \frac{4}{6} \cdot \frac{2}{6} \cdot \frac{4}{6} \cdot \frac{2}{6} + \frac{4}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{4}{6} \\ & + \ \frac{2}{6} \cdot \frac{4}{6} \cdot \frac{4}{6} \cdot \frac{2}{6} + \frac{2}{6} \cdot \frac{4}{6} \cdot \frac{2}{6} \cdot \frac{4}{6} + \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{4}{6} \cdot \frac{4}{6} \\ & = 6 \cdot \left(\frac{4}{6} \right)^2 \cdot \left(\frac{2}{6} \right)^2 \end{align}

The first number is $6$ because there are six ways to pick two dice from four and we know this by noting that the first dice we pick can be any of 4 and the second can be any of the other three this gives us $4 \times 3 = 12$ ways we can pick the dice if we care about the order so we divide this by the number of ways we can pick two dice from two to get six.

You may want to try a few more examples if you haven't spotted the pattern yet but you should see: $$\text{Probability} = \frac{\prod^{m}_{n - m + 1}}{m!} \cdot \left(\frac{n-m+1}{n}\right)^m \cdot \left(\frac{n-m-1}{n}\right)^{n-m}$$

Note $\frac{\prod^{m}_{n - m + 1}}{m!}$ can also be written as $\frac{n!}{(n-m!)m!}$

If instead you want at least this many hits you would have to add up exactly m hits, exactly m+1 hits etc

$$\text{Probability} = \sum_{i=m}^{n} \frac{\prod^{i}_{n - i + 1}}{m!} \cdot \left(\frac{n-i+1}{n}\right)^i \cdot \left(\frac{n-i-1}{n}\right)^{n-i}$$

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i was half way writing it but i got bored but why is it a product at the end? –  Lost1 Jan 18 '14 at 13:35
our answer somewhat disagree –  Lost1 Jan 18 '14 at 13:35
The product is because if you want to select say 3 from six there are $\frac{6 \cdot 5 \cdot 4}{1 \cdot 2 \cdot 3}$ you could also write this as $\frac{6!}{3!\cdot 3!}$. I prefer the product notation because if we wanted to select 2 from 1million $\frac{1000000 \times 999999}{1 \times 2}$ avoids calculating $999998!$ –  Warren Hill Jan 18 '14 at 13:51
@Lost1, Your correct I was double counting some hits. I've now corrected my post. The answer is now the same as yours. I'll leave my post as it explains where the numbers come from but as you got the correct answer first I've upvoted it. –  Warren Hill Jan 18 '14 at 14:44
you should check out binomial distribution. –  Lost1 Jan 18 '14 at 15:37