Probability of at least 1 even number using while loop

Find the probability of rolling at least 1 even number using 3 dice. You cannot use any vectors or matrices, or any vectorized operations. Instead, do it using loops. Notice that in each trail, to see if there is at least one even number, there is no need to always inspect the value of each of the dice. No need to inspect the remaining dice as soon as an even number is encountered. Thus a while-loop should be used to handle that part of the problem

I have no idea how to use the while loop in this problem

clear;clc;

nDice = 3; % Number of dice
nTrials = 1e6; % Number of trials

for i = 1:nTrials
    
end 

[A. Sawas]

Try breaking this problem into subtasks and solve them one by one.

1- Generating a random integer between 1 and 6

2- Verifying an integer is even or odd.

3- Seting a counter that will be only incremented when you get an even number.

[Guillaume]

I have no idea how to use the while loop in this problem

Well, you'll have to ask whomever set you that stupid assignment. The reasoning for using a while loop makes no sense. Nobody uses a while over just 3 elements (if that's what is meant).

You cannot use any vectors or matrices, or any vectorized operations. Instead, do it using loops.

Well, this does it without a vector, matrix, or vectorised operation:

(3*6*6 + 3*3*6 + 3*3*3) / 6^3   %(even digit as 1st die + (even digit as 2nd die with 1st die odd) + (even digit as 3rd die with 1st and 2nd odd)) / all possible rools

Using loops for this is completely pointless. It teaches you the wrong way of using matlab. If you couldn't work out the math, you could always do it with a vectorised method, which is the efficient way of using matlab:

rolls = cell(1, 3);
[rolls{:}] = ndgrid(1:6);
nnz(any(mod(cat(4, rolls{:}), 2) == 0, 4)) / 6^3 

[John D'Errico]

Done in one line? At least one even number? Something like this should work:

sum(sum(1 - mod(dec2base(0:215,6) - '0' + 1,2),2) > 0)/216
ans =
        0.875

Or, via simulation, here with 10 million sample rolls.

sum(sum(1 - mod(randi([1 6],[1e7,3]),2),2) > 0)/1e7
ans =
                 0.8750012

I hate HW assignments. Vectors not allowed. They teach you the wrong things, the wrong way to solve a problem, that you then need to unlearn to do anything useful.

[A. Sawas]

Guillaume and John D'Errico, you guys are awesome in finding short optimum solutions !

but I am not sure why you focused on finding the answer to the problem than the purpose of solving it using within the prescribed constraints. (using while loop, simulating trials, having there dices, not using vectors, etc.)

I think regardless whatever this problem and contraints seemed to be unlogical, one would learn something solving it.

[Guillaume]

I think regardless whatever this problem and contraints seemed to be unlogical, one would learn something solving it.

Yes, you learn the wrong way of using matlab. It teaches you to use loops where they're inefficient. It teaches you to think about this sort of problems in a way that is inefficient. Unfortunately, we can see the result in this forum, where we spend a lot of effort unteaching that mindset. People come here with code that have multiple loops saying that it is too slow.

This problem is better though of as:

• generate all possible dice rolls (there's only 216). That can be done efficiently in one line with ndgrid or dec2base. You could indeed use one (or three) loops to generate all these rolls. That'd be slower and more lines of code.
• find which of these dice rolls have even numbers. Again, the efficient way is a mod test on the whole lot. Here you're asked to use a while loop on a single roll. This in my opinion is completely absurd.

Answer 1

[Walter Roberson]

For each trial Initialize dice remaining to 3. While number of dice remaining is positive Decrement dice remaining. roll a die. If it was even add 6 to power of remaining dice to statistics and set dice remaining 0.

After all trials divide statistics by 216 times number of trials.

Answer 2

[Torsten]

Maybe of interest for checking the result of the Monte Carlo simulation:

P(rolling at least one even number with 3 dice) =
1- P(rolling only odd numbers with three dice) =
1 - 3^3 / 6^3