CODEWITHSUNDEEP
javaalgorithmprobabilitydice
HyperParameter
HyperParameter

Reputation: 284

Greater chance to roll 6 with dice in Java

I was wondering how can I program a dice which has a greater chance to roll a 6? I've tried everything but I just cant figure the algorithm. The probability of six is given by the user.

I just don't now how to program random which uses the given probability to roll a 6 or 1-5.

Upvotes: 0

Views: 852

Answers (3)

Phil Steitz
Phil Steitz

Reputation: 644

Divide what remains of the interval (0,1) when you take out a region the size of the probability of six into 5 equal pieces and assign these regions to 1 - 5:

/**
 * @param pSix probability that a six is returned
 * @param rnd Random instance
 * @return a random integer between 1 and 6 with 1-5 equiprobable and P(6) = pSix
 */
public int loadedDice(final double pSix, final Random rnd) {
    final double pOther = (1d - pSix) / 5d;
    final float val = rnd.nextFloat();
    if (val < pSix) {
        return 6;
    }
    return (int) Math.ceil((val - pSix) / pOther);
  }
}

Upvotes: 2

greedybuddha
greedybuddha

Reputation: 7507

I would suggest using a percentage based scheme. let the user pick the probability of a six, in this example lets say 30%. Then choose a random number between 0-1.0 (which is what java's Random#nextFloat does). If its below or equal to .3, then make it a six, otherwise make it a 1-5.

Random r = new Random();

float probability = r.nextFloat(); // get a value between 0 and 1
if (probability < probabilityOfSix){
    return 6;
} else {
    return r.nextInt(4) +1;
}

Upvotes: 3

jongusmoe
jongusmoe

Reputation: 676

Let's say a 6 has twice the probability to appear. Get a random number from 1 through 7, if your result is either a 6 or a 7, then you have a 6.

Same thing for three times the probability. Fetch a random number from 1-8: 6, 7, and 8 become a 6.

Upvotes: 2

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