Probability of exactly one even number

Problem: Given n number of dice to throw, find the probability that in a single throw of the dice there is exact one even number. You cannot use any vectors or matrices, or any vectorized operations. Instead, use a double for-loop (one over the different trials, and another one for the different number of dice used).

I am confused on how to make sure that there is only one even number per trial.

[Torsten]

You might want to compare your Monte-Carlo simulation results with the "correct" value for the probability.

It is given by P = n * 0.5^n.

Answer 1

[Image Analyst]

Very close but you need to check if just one even number was thrown OUTSIDE the inner loop over j. Also, a tip: to determine if something is even you can use rem(value, 2) rather than all that dividing stuff you're doing. Try this:

clear;
clc;
nDice = 3; % Number of dice
nTrials = 1e6; % Number of trials
singleCount = 0; % Count of how many trials have 1 even number
for i = 1:nTrials
	% For this trial, initialize the number of even numbered dice thrown.
	evenNumberCount = 0;
	% Make nDice throws detecting if the throw was an even number.
	for j = 1 : nDice
		thisDicesValue = randi(6);
		evenNumberCount = evenNumberCount + (rem(thisDicesValue, 2) == 0);
	end
	% See if only ONE of those throws was an even number during this trial.
	if evenNumberCount == 1
		% Only one even number was thrown
		singleCount = singleCount + 1;
	end
end
prob = singleCount / nTrials
fprintf('Found %d experiments where exactly 1 die in %d throws of %d dice had one even number.\n',...
	singleCount, nTrials, nDice);

and adapt as needed.

Answer 2

[Jos (10584)]

I suggest you use two counters: one counting the numbers of even values in a single throw, and one counting the number of throws that contain a single even number. The first one you need to reset to 0 before every throw. The later one you need to update by 1 if the first one is 1 after the throw-loop

Additional remarks:

• take a look at the function REM: rem(X, 2) == 0 is true for an even value of X

• this question is best answered using arrays and can even be handled by a one-liner

Prob = sum(sum(rem(randi(6, nTrials, nDice), 2) == 0, 2) == 1)) / nTrials