CODEWITHSUNDEEP
language-agnosticrandomprobability
Andreas Grech
Andreas Grech

Reputation: 108030

Simulating a roll with a biased dice

I did a search but didn't really get any proper hits. Maybe I used incorrect terms?

What I want to ask about is an algorithm for simulating a biased role rather than a standard supposedly-random roll.

It wouldn't be a problem if you can't give me exact answers (maybe the explanation is lengthy?) but I would appreciate &pointers to material I can read about it.

What I have in mind is to for example, shift the bias towards the 5, 6 area so that the numbers rolls would have a higher chances of getting a 5 or a 6; that's the sort of problem I'm trying to solve.

[Update]

Upon further thought and by inspecting some of the answers, I've realized that what I want to achieve is really the Roulette Wheel Selection operator that's used in genetic algorithms since having a larger sector means increasing the odds the ball will land there. Am I correct with this line of thought?

Upvotes: 1

Views: 8906

Answers (5)

Dr. belisarius
Dr. belisarius

Reputation: 61056

In general, if your probabilities are {p1,p2, ...,p6}, construct the following helper list: 

{a1, a2, ... a5} = { p1, p1+p2, p1+p2+p3, p1+p2+p3+p4, p1+p2+p3+p4+p5}

Now get a random number X in [0,1]

If 

        X <= a1  choose 1 as outcome  
   a1 < X <= a2  choose 2 as outcome 
   a2 < X <= a3  choose 3 as outcome  
   a3 < X <= a4  choose 4 as outcome  
   a4 < X <= a5  choose 5 as outcome  
   a5 < X        choose 6 as outcome

Or, more efficient pseudocode

   if     X > a5 then N=6
   elseif X > a4 then N=5
   elseif X > a3 then N=4
   elseif X > a2 then N=3
   elseif X > a1 then N=2
   else               N=1

Edit

This is equivalent to the roulette wheel selection you mention in your question update as shown in this picture:

enter image description here

Upvotes: 7

Adam Dymitruk
Adam Dymitruk

Reputation: 129744

Make a 2 dimensional array of possible values and their weights. Sum up all the weights. Randomly choose a value on the range of 0 to the sum of the weights. Now iterate through the array while keeping an accumulator of the weights seen so far. Once this value exceeds your random number, pick the value of the die represented here.

Hope this helps

Upvotes: 2

Jerry Coffin
Jerry Coffin

Reputation: 490623

One way that's usually fairly easy is to start with a random number in an expanded range, and break that range up into unequal pieces.

For example, with a perfectly even (six-sided) die, each number should come up 1/6th of the time. Let's assume you decide on round percentages -- all the other numbers should come up 16 percent of the time, but 2 should come up 17 percent of the time.

You could do that by generating numbers from 1 to 100. If the number is from 1 to 16, it comes out as a 1. If it's from 17 to 34, it comes out as a 2. If it's from 34 to 50, it comes out as a 3 (and the rest are blocks of 16 apiece).

Upvotes: 1

Scott C Wilson
Scott C Wilson

Reputation: 20036

Let's say the die is biased towards a 3.

Instead of picking a random entry from an array 1..6 with 6 entries, pick a random entry from an array 1..6, 3, 3. (8 entries).

Upvotes: 2

YXD
YXD

Reputation: 32521

Hmm. Say you want to have a 1/2 chance of getting a six, and a 1/10 chance of getting any other face. To simulate this, you could generate a random integer n in [1, 2, ... , 10] , and the outcome would map to six if n is in [6, 7, 8, 9, 10] and map to n otherwise.

Upvotes: 1

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