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I am interested in calculating odds of winning UK Lottery. The format is that 6 numbers from 1-59 are drawn. I am interested only (at this stage) in the odds of winning the jackpot (matching six balls). As an aside, I'm interested in the odds for a total balls count of 49, and 59, to see the change in chance of winning.

The mathematical formula for calculating the odds is (where 49 is total balls, 6 is number drawn:

\$\text{Odds of winning} = \dfrac{49!}{(6!*(49-6)!}\$

The main method of my code is to collect input from the user on parameters of the draw.

I have a class called DrawInfo to store information about the draw. I have a simple method to return the Factorial of a number.

I have a method to calculate the odds of winning the jackpot. This is currently all in the one class, as a small, simple app. I do appreciate that DrawInfo could live in its own class.

class Program
        static void Main(string[] args)
            Console.WriteLine("Enter the total number of balls in the draw: ");
            int totalBalls = int.Parse(Console.ReadLine());

            Console.WriteLine("enter the number of balls drawn: ");
            int ballsDrawn = int.Parse(Console.ReadLine());

            DrawInfo di = new DrawInfo(totalBalls, ballsDrawn);

            int totalWinOdds = FindJackpotWinningOdds(di);
            Console.WriteLine(String.Format("the odds are 1/{0:n0}", totalWinOdds));

        static int FindJackpotWinningOdds(DrawInfo di)
            BigInteger totalBallsFactorialSum = Factorial(di.TotalBalls);
            BigInteger ballsDrawnFactorialSum = Factorial(di.BallsDrawn);

            BigInteger JackpotWinningOdds = 0;
            JackpotWinningOdds = totalBallsFactorialSum / ((ballsDrawnFactorialSum * Factorial((di.TotalBalls - di.BallsDrawn))));
            return (int)JackpotWinningOdds;

        static BigInteger Factorial(BigInteger i)
            if (i <= 1)
                return 1;
            return i * Factorial(i - 1);

    public class DrawInfo
        public int TotalBalls { get; set; }
        public int BallsDrawn { get; set; }

        public DrawInfo(int totalBalls, int ballsDrawn)
            this.TotalBalls = totalBalls;
            this.BallsDrawn = ballsDrawn;
share|improve this question
up vote 6 down vote accepted

Quick remarks:

  • Don't abbreviate needlessly: di.
  • Why assign totalBallsFactorialSum and ballsDrawnFactorialSum, when you are only using them once? You're not even consistent: in the case of Factorial((di.TotalBalls - di.BallsDrawn)) you don't assign the result to a variable.
  • Don't overdo it with the brackets: there's no point for the inner ones in Factorial((di.TotalBalls - di.BallsDrawn)).
  • JackpotWinningOdds doesn't folow the capitalization conventions.
  • Why is it called FindJackpotWinningOdds? Wouldn't CalculateJackpotOdds be better?
  • The this in this.TotalBalls = totalBalls; and this.BallsDrawn = ballsDrawn; is superfluous.
  • TotalBalls and BallsDrawn should be private set;.
  • Why even assign the result to JackpotWinningOdds? This whole method can be reduced to a one-liner, though perhaps it would be best to split it over multiple lines to increase legibility:

        return (int)(
           / (
              * Factorial(di.TotalBalls - di.BallsDrawn)

    This method could even just be a method on DrawInfo -- together with BigInteger Factorial(BigInteger i), of course, and Factorial() could then even be a private method.

share|improve this answer
Thanks. I was having problems calculating the 'JackpotWinningOdds' so was trying to break it down into individual variables to see where I was going wrong. Ideally I'd what you wrote above, as long as it's clear. – kafka Feb 1 at 15:36
assigning a meaningful name to a variable even if that variable is subsequently only used once is a very good way of documenting its semantic intent. – Alnitak Feb 1 at 18:23

A quick note to start:

Console.WriteLine(String.Format("the odds are 1/{0:n0}", totalWinOdds));

Is exactly the same as:

Console.WriteLine("the odds are 1/{0:n0}", totalWinOdds);

Now on to some maths fun... There's a multiplicative version which means you don't have to compute such massive numbers so don't need to use BigIntegers:

private int GetBinomialCoefficient(int totalNumberOfBalls, int numberOfBallsDrawn)
    // range checking of arguments omitted.
    var total = 1;
    for (var i = 1; i <= numberOfBallsDrawn; i++)
        total *= (totalNumberOfBalls + 1 - i) / i;
    return total;

GetBinomialCoefficient(59, 6) == 45057474


I have changed the name of the method away from GetJackpotOdds based on 200_success's excellent answer.

share|improve this answer
Thanks. I like the paths part of this answer, as I was wondering if there was a better way than having to work with massive numbers. – kafka Feb 1 at 16:06

You should fix your terminology.

The number of possible combinations is:

$$\left(\begin{array}{c}n\\r\end{array}\right) = \frac{n!}{r!\ (n-r)!}$$

The probability that any one ticket has the winning combination is:

$$\frac{1}{\left(\begin{array}{c}n\\r\end{array}\right)} = \frac{r!\ (n-r)!}{n!}$$

However, odds are a different convention for expressing probabilities. For example "1:1 odds" means a 50-50 chance; "1:3 odds" means a 25% chance. Therefore, the odds of winning the jackpot are

$$1\ :\ \frac{n!}{r!\ (n-r)!} - 1$$

In other words, you have an off-by-one error.

Calculating the result by actually computing \$49!\$ is nuts.

$$ \require{cancel} \begin{align} \frac{49!}{6!\ (49-6)!} - 1 &= \frac{49 \cdot 48 \cdot 47 \cdot \ldots \cdot 1}{(6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)(43 \cdot 42 \cdot \ldots \cdot 1)} - 1 \\ &= \frac{49 \cdot 48 \cdot 47 \cdot 46 \cdot 45 \cdot 44 \cdot \cancel{43} \cdot \cancel{42} \cdot \ldots \cdot \cancel{1}}{(6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)(\cancel{43} \cdot \cancel{42} \cdot \ldots \cdot \cancel{1})} - 1 \end{align} $$

That's 10 multiplications and one division. The numerator is a bit over 10 billion, which fits comfortably inside a C# long type.

share|improve this answer
Thanks. For clarify I'd rather just the probability of the winning (1 in 13,983,816 say) – kafka Feb 2 at 9:26

The chance the first ball matches an of your number is:


Assuming you hit this you have five numbers to match and 48 balls in the machine that can be picked.

So the second ball has a probability of:


Continue for all six balls picked.

You then have a probability of winning of:

$$= \frac{6}{49} \cdot \frac{5}{48} \cdot \frac{4}{47} \cdot \frac{3}{46} \cdot \frac{2}{45} \cdot \frac{1}{44}$$

$$= \frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{49 \cdot 48 \cdot 47 \cdot 46 \cdot 45 \cdot 44}$$

$$= \frac{720}{10,068,347,520}$$

$$= 1: 13,983,816$$

$$= 1 \space \text{in} \space 14 \space \text{million}$$

share|improve this answer
Thanks the maths here is what I have in my head, rather than what I found from Googling (and subsequently based my code on). – kafka Feb 2 at 9:27

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