Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

There is a game I play that utilizes a particular mechanic that I am trying to assess mathematically. Me being a layman, I am having some problems figuring out one of the aspects for myself.

The basic mechanic works like this: you roll a number of 6 sided dice. Any die that rolls a 5 or a 6 nets you a success. You count your total number of successes, and the more of them there are the better.

Now, I know how to calculate the chance that I will succeed when I am trying to roll a fixed number of successes. But, sometimes the game has you roll against someone else's roll. For example:

Player A rolls X number of dice with 6 sides. Player B rolls Y number of dice with 6 sides. The person who rolls more results of 5 or 6 succeeds.

How do I calculate the probability that player A will succeed?

Thank you very much in advance!

share|improve this question
Finding the exact probability seems really ugly. As far as getting a good approximation (statisticians, please correct me if I'm wrong, as I'm no expert): If X and Y are large enough, the differences between the numbers of successes of players A and B approximately follow a normal distribution with mean $\frac{X-Y}{3}$ and variance $\frac{2X+2Y}{9}$. If we want A to win, then we want to look at the area to the right of 0 under a normal curve with the aforementioned mean and variance. –  Nick D. Dec 18 '13 at 3:21
(cont.) That is, the probability is approximately the area under the standard normal curve to the right of $\frac{Y-X}{\sqrt{2X+2Y}}$. –  Nick D. Dec 18 '13 at 3:32
@NickD. In practice, $X$ and $Y$ are probably not large enough to approximate the standard normal well. –  Austin Mohr Dec 18 '13 at 3:50
Probably true. Unless you find people who $really$ love rolling dice. –  Nick D. Dec 18 '13 at 3:51

1 Answer 1

On any die, you either "win" by rolling a $5$ or a $6$, or you don't.

The probability of rolling at least $n$ winning dice out of $d$ rolled is

$$P(n,d) = \sum_{k=n}^{d} {d \choose k} \left(\frac{1}{3}\right)^k \left(\frac{2}{3}\right)^{d-k}.$$

Further, if $n > d$, $P(n,d) = 0.$

So if you have $x$ dice and your opponent has $y$ dice, then the probability of you winning is

$$P_{win}(x,y) = \sum_{k=0}^{x} {x \choose k} \left(\frac{1}{3}\right)^k \left(\frac{2}{3}\right)^{x-k} [1-P(k,y)].$$

share|improve this answer
Ok. That is HUGELY helpful. Only problem is, I don't understand some of the notation (due to my status as a layman). Could you please explain to me, what does the X over K in the second bracket mean? –  Dave Dec 18 '13 at 4:50
Also, would k=o throughout the entire equation? I understand what it stands for in the ∑ part of the equation, but I am uncertain about its usage throughout the rest of the formula... –  Dave Dec 18 '13 at 4:57
The $x k$ in the column represents a binomial coefficient. It's also read "$x$ choose $k$." If you had $75$ bingo balls and pulled $15$ of them, the number of different combinations of balls is $75$ choose $15$, or $75!/(60! \cdot 15!)$, where the $!$ is the factorial operator. In this context, you'd want to count how many ways you can get, say, three winning dice out of seven. It's $7$ choose $3$. –  John Dec 18 '13 at 6:06
Regarding your second comment: When you roll your $x$ dice, you can have anywhere from $0$ to $x$ winning dice. So what the equation does is add up the individual probabilities of your opponent rolling less than $k$ winners out of his $y$ dice, given that you rolled exactly $k$ winners, for all possible values of $k$ that you can roll ($0$ to $x$). –  John Dec 18 '13 at 6:11
Thanks! Just to make sure, that equation calculates the chance that you will roll MORE "wins" than player Y, not equal to or more, correct? –  Dave Dec 18 '13 at 23:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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