AlliedModders - [Solved] Suggestions on randoming

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Suggestions on randoming

sup folks,

need suggestions for best random experience.
i.e. I have 5 items with chances of: 97 90 54 35 2

How do I pick one of those items based on chance?

Re: Suggestions on randoming

5 items with chances > pick one based on chance?

What chance?

Re: Suggestions on randoming

Get a random number starting from a range of the minimum and the maximum number of an array, and then find the closest number(the random number result) on that array.

Code:

new min = 0, max = 100, result
result = random_num(min, max)
find_the_closest_number_on_array(data[], len, result)

Re: Suggestions on randoming

Quote:

Originally Posted by Relaxing (Post 2603358)

Get a random number starting from a range of the minimum and the maximum number of an array, and then find the closest number(the random number result) on that array.

Code:

new min = 0, max = 100, result
result = random_num(min, max)
find_the_closest_number_on_array(data[], len, result)

Is there such function and how do I actually use it?

Re: Suggestions on randoming

Here's an algorithm I found on web: https://www.geeksforgeeks.org/find-c...-number-array/

Re: Suggestions on randoming

If you are picking a single item randomly between a few, the sum of all chances must be 100%. Otherwise its just wrong, like in your example :D

Re: Suggestions on randoming

If you're looking for a chance between two options (yes and no) then you can use this:

Code:

#define chance(%1) ( %1 > random(100) )

chance(90) will return true 90% of the time (on average).

But, as maqi mentioned, if you are trying to give one of the 5 items, the probabilities need to add up to 100 and then you would use a different algorithm to determine which item they get.

EDIT:

Here is a function that will select from a list of items depending on their probability. The probabilities must add up to 100. It returns the index of the item that was chosen randomly.

Code:

stock random_item(itemChances[], count=sizeof itemChances)

{

        static rand, i, sum

        rand = random(100)

        i = sum = 0

        for( i = 0; i < count; i++ )

        {

                sum += itemChances[i]

                if( sum > rand )

                        break

        }

        return i

}

Examples:

Code:

new chanceItem = random_item([10, 70, 20], 3)

10% of the time, chanceItem will be 0
70% of the time, chanceItem will be 1
20% of the time, chanceItem will be 2

Re: Suggestions on randoming

Quote:

Originally Posted by fysiks (Post 2603426)

If you're looking for a chance between two options (yes and no) then you can use this:

Code:

#define chance(%1) ( %1 > random(100) )

Code:

stock random_item(itemChances[], count=sizeof itemChances)

{

        static rand, i, sum

        rand = random(100)

        i = sum = 0

        for( i = 0; i < count; i++ )

        {

                sum += itemChances[i]

                if( sum > rand )

                        break

        }

        return i

}

Examples:

Code:

new chanceItem = random_item([10, 70, 20], 3)

10% of the time, chanceItem will be 0
70% of the time, chanceItem will be 1
20% of the time, chanceItem will be 2

I can't make chances add up, it would be too complicated.

I'm thinking about this:
4 different rarities - 1 (1-10%), 2 (11-30%), 3 (31-60%), 4 (61-100%).

random(100) to get % and check rarity, then just pick random item who has that rarity?
Not sure how fair it would be

Re: Suggestions on randoming

Quote:

Originally Posted by Airkish (Post 2603431)

I can't make chances add up, it would be too complicated.

I'm thinking about this:
4 different rarities - 1 (1-10%), 2 (11-30%), 3 (31-60%), 4 (61-100%).

It's super easy:

Code:

item = random_item([10, 20, 30, 40], 4) + 1

Quote:

Originally Posted by Airkish (Post 2603431)

random(100) to get % and check rarity, then just pick random item who has that rarity?
Not sure how fair it would be

That doesn't make any sense. Use my function as shown above.

Re: Suggestions on randoming

You could map your probabilities so they all add up to a total of 1, whatever their values are. I'll be expressing probabilities in the range [0, 1] but you can always multiply it by 100 to display the chance in percents.

Untested, but the same algorithm did work in JavaScript for me:

Code:

GetRandomItem(const Float:itemChances[], count = sizeof itemChances) {

        new Array:mappedChances = ArrayCreate(1, 0);

        new index = 0;

        new Float:sum = 0.0;

        

#if AMXX_VERSION_NUM > 182

        ArrayResize(mappedChances, count);

#endif



        // Put all elements into mappedChances and accumulate

        for(index = 0; index < count; index++) {

                ArrayPushCell(mappedChances, itemChances[index]);

                sum += itemChances[index];

        }

        // Map so the sum of all chances is 1

        for(index = 0; index < count; index++) {

                ArraySetCell(mappedChances, index, ArrayGetCell(mappedChances, index) / sum);

        }

        sum = 0.0;

        new const Float:rand = random_float(0.0, 1.0);

        for(index = 0; index < count; index++) {

                sum += ArrayGetCell(mappedChances, index);

                if(rand <= sum) {

                        break;

                }

        }

        

        ArrayDestroy(mappedChances);

        return index;

}

Code:

new const Float:itemChances[] = { 0.5, 0.1, 0.05, 0.15, 0.2, 0.4, 0.04, 0.06, 0.5 };

new const result = GetRandomItem(itemChances, sizeof itemChances);

As you can see, the chances in the array add up to 2.0 (or 200% percent), but in that case if you divide them by 2 they will add up to 1.0.
It's pretty much what Fysiks did but with the mapping step so input values don't have to add up to 100%.