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!

  • $\begingroup$ 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. $\endgroup$
    – Nick D.
    Commented Dec 18, 2013 at 3:21
  • $\begingroup$ (cont.) That is, the probability is approximately the area under the standard normal curve to the right of $\frac{Y-X}{\sqrt{2X+2Y}}$. $\endgroup$
    – Nick D.
    Commented Dec 18, 2013 at 3:32
  • $\begingroup$ @NickD. In practice, $X$ and $Y$ are probably not large enough to approximate the standard normal well. $\endgroup$ Commented Dec 18, 2013 at 3:50
  • 1
    $\begingroup$ Probably true. Unless you find people who $really$ love rolling dice. $\endgroup$
    – Nick D.
    Commented Dec 18, 2013 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

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

  • $\begingroup$ 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? $\endgroup$
    – Dave
    Commented Dec 18, 2013 at 4:50
  • $\begingroup$ 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... $\endgroup$
    – Dave
    Commented Dec 18, 2013 at 4:57
  • $\begingroup$ 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 $3$. $\endgroup$
    – John
    Commented Dec 18, 2013 at 6:06
  • $\begingroup$ 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 $x$). $\endgroup$
    – John
    Commented Dec 18, 2013 at 6:11
  • $\begingroup$ 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? $\endgroup$
    – Dave
    Commented Dec 18, 2013 at 23:46

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