23

Let's say I'm writing a simple luck game - each player presses Enter and the game assigns him a random number between 1-6. Just like a cube. At the end of the game, the player with the highest number wins.

Now, let's say I'm a cheater. I want to write the game so player #1 (which will be me) has a probability of 90% to get six, and 2% to get each of the rest numbers (1, 2, 3, 4, 5).

How can I generate a number random, and set the probability for each number?

Alon Gubkin
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4 Answers4

29
static Random random = new Random();

static int CheatToWin()
{
    if (random.NextDouble() < 0.9)
        return 6;

    return random.Next(1, 6);
}

Another customizable way to cheat:

static int IfYouAintCheatinYouAintTryin()
{
    List<Tuple<double, int>> iAlwaysWin = new List<Tuple<double, int>>();
    iAlwaysWin.Add(new Tuple<double, int>(0.02, 1));
    iAlwaysWin.Add(new Tuple<double, int>(0.04, 2));
    iAlwaysWin.Add(new Tuple<double, int>(0.06, 3));
    iAlwaysWin.Add(new Tuple<double, int>(0.08, 4));
    iAlwaysWin.Add(new Tuple<double, int>(0.10, 5));
    iAlwaysWin.Add(new Tuple<double, int>(1.00, 6));

    double realRoll = random.NextDouble(); // same random object as before
    foreach (var cheater in iAlwaysWin)
    {
        if (cheater.Item1 > realRoll)
            return cheater.Item2;
    }

    return 6;
}
Anthony Pegram
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5

You have a few options, but one way would be to pull a number between 1 and 100, and use your weights to assign that to a dice face number.

So

1,2 = 1
3,4 = 2
5,6 = 3
7,8 = 4
9,10 = 5
11-100 = 6

this would give you the ratios you need, and would also be fairly easy to tune later.

Matthew Vines
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    This is the simplest and most common approach. It's sometimes called a *probability distribution vector*. In practice, most implementations use a range dictionary, and perform a lookup in that based on a randomly generated value. – LBushkin Jun 10 '10 at 16:48
3

you can define array of distribution (pseudocode) :

//fair distribution

array = {0.1666, 0.1666, 0.1666, 0.1666, 0.1666, 0.1666 };

then roll the dice from 0 to 1, save to x then do

float sum = 0;
for (int i = 0; i < 6;i++)
{
   sum += array[i];
   if (sum > x) break;
}

i is the dice number.

now if you want to cheat change array to:

array = {0.1, 0.1, 0.1, 0.1, 0.1, 0.5 };

and you will have 50% to get 6 (instead of 16%)

Andrey
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0

And how do you like this castile?

Input data:

int range = { 0, 2, 4, 6, 8, 92}
int rand = Random.range(1, 100) 

The method itself:

public int Selectionbyrange()   
{
    List<Tuple<int, int, int>> ran_r_i  = new List<Tuple<int, int, int>>();
    // Creating 3 lists
    for (int i = 1; i < (int)range.LongLength;i++)                     
    {
        ran_r_i.Add(new Tuple<int, int, int>(range[i - 1], range[i], i));
        // Writing to the list of ranges
    }

    foreach (var ran_i in ran_r_i)
    {
    if (ran_i.Item1 < rand && rand <= ran_i.Item2)              
        return ran_i.Item3;
        // Range check
    }  

    return 1;    
}

As a result of random we get such data:

When rand > 0 and <= 2 we have 1

  • When rand > 2 and <= 4 we have 2

  • When rand > 4 and <= 6 we have 3

  • When rand > 6 and <= 8 we have 4

  • When rand > 8 and <= 92 we have 5

  • When rand > 92 and <= 100 we have 6

Jeremy Caney
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Sashka UA
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