Probability Random Number Generator

Question

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?

+1 for "let's say I'm a cheater." – Zaid Masud Aug 08 '13 at 12:31 — Zaid Masud, Aug 08 '13 at 12:31

Anthony Pegram · Accepted Answer · 2010-06-10T16:53:41.500

25

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;
}

edited Jun 10 '10 at 16:53

answered Jun 10 '10 at 16:42

Anthony Pegram

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Oh, sure, a *customizable* way to cheat. I'll add one. – Anthony Pegram Jun 10 '10 at 16:52
1

@Andrey: And yet, since it's not pseudocode, it's not a pseudo-answer. :) – MusiGenesis Jun 10 '10 at 16:53
1

@Anthony: make sure you customize your method to accomodate any possible game of chance. Hang on a sec and I'll look up the rules of baccarat for you. – MusiGenesis Jun 10 '10 at 16:55
1

@MusiGenesis, cheating *must* be extensible. – Anthony Pegram Jun 10 '10 at 16:57
It returns six if its probability is 0 either. – Alon Gubkin Jun 10 '10 at 17:12

score 4 · Answer 2 · answered Jun 10 '10 at 16:42

4

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.

answered Jun 10 '10 at 16:42

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

score 3 · Answer 3 · answered Jun 10 '10 at 16:45

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%)

Probability Random Number Generator

3 Answers3

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