Me:
The testdata for genetics is wrong i think

The probability that Johnny's siblings will get the dieseas is 0.75

There are 8 cases we want to consider, 4 cases for the chromosome of Johnny's mother and 2 cases for which chromosome Johnny gets. Out of these, there are 4 cases that Johnny gets the dieseas, which is (Xx,left), (xX,right), (XX,left), (XX,right). the probability of the mother's chromosome is 25% Xx, 25% xX, 50% XX. So the probability for Johnny's siblings to get is 75%? I'm quite sure 66.667% is wrong but it passes the testdata

Jury:
Chromsome Y is 100% sick because the entire tree has Y as sick. So now the question is what they get from their mother. Their mother can only have Xx, xX or XX. xx is not possible because it is given that Johnny has the disease. So you have 1/3 probability for each. The probability of X being sick is 1/3 * 1/2 + 1/3 * 1/2 + 1/3 * 1 = 2/3 = 0.666. Good job getting accepted!

Me:
From the statement, I am assuming that below is a valid C++ simulation for what is happening in the world. The mother is chosen from each possible chromosome combination first, then johnny's chromosome is chosen second.

int success=0;
int total=0;
for (int x=0;x<1000000;x++){
	int arr[2]={rand()%2,rand()%2};
	int johnny=rand()%2;
	int sibling=rand()%2;
	
	if (arr[johnny]==1){
		success+=arr[sibling];
		total++;
	}
}
cout<<(double)success/total<<endl;

We see that the output is around 0.75. Unless I have intepreted what it means for the chromosome combination of everyone to be uniform?

Jury:
The way to think about it is as follows: You have a world with 25% xx women, 25% xX women, 25%Xx women and 25% XX women. Now in Johnny s case, we know that his mom is not from the first 25%, so his mom must be one of the other cases. Which all have equal probability to be (because they were all 25 in the first place). Since probabilities have to sum up to 1, the probability is 1/3.

Jury:
The simulation program is wrong because it does not compute the probabilities GIVEN THAT Johny is sick. The problem asks you for the conditional probability: P[brother is sick | Johny is sick].

https://en.wikipedia.org/wiki/Conditional_probability.

The problem statement says "We consider that all possible chromosome combinations have a uniform distribution". The "xx" is not a possible combination because we know that Johny is sick. So the possible combinations for the mother given that Johny is sick are Xx, xX, XX, with a probability of 1/3 for each one.

Me:
I know what conditional probability is. I don't understand what "The simulation program is wrong because it does not compute the probabilities GIVEN THAT Johny is sick" means.

I firstly randomly select a random universe where the chromosomes of the mother is randomly chosen over all possible chromosome pairs, and Johnny and his sibling are randomly chosen a chromosome from their mother. Then the if statement only checks those worlds where "Johnny is sick" or he has received X from his mother.

"So the possible combinations for the mother given that Johny is sick are Xx, xX, XX, with a probability of 1/3 for each one." How can you assume that they have equal probability? In a universe where Johnny's mother has XX, Johnny is more likely to get dieseas than if he was in a universe with xX. That is why we weight them accordingly in the weights 25%,25%,50%.

So I don't understand what exactly is wrong with my simulation program

Jury (after contest):
Yes, I agree with you. In a proper mathematical setup, the probability space is defined at the beginning, and after the observation that Johnny is ill, the mother's distribution becomes 1/2, 1/4, 1/4 and the sibling's probability is 3/4 as your experiment showed.

However, our intentions were that this problem should be one of the easiest. Hence, we tried to explain in the problem statement that "We consider that all possible chromosome combinations have a uniform distribution" with the intention to make the solution easier. That is, the contestant should just realize that xx is not valid and that we told them that we have a uniform distribution over {xX, Xx, XX}.

I agree that the problem statement was ambiguous, and together with the fact that "our definition" was different than the conventional definition in the probability theory, the statement became much more ambiguous. We are very sorry for this, but we are glad that you got an AC in the end. Again, we are very sorry for the penalty and the frustration. We hope that, apart from this problem, you enjoyed the contest and we are keen to see your performance tomorrow.

int success=0;
int total=0;
for (int x=0;x<1000000;x++){
	//sick dna is represented with 1, not sick with 0
	int arr[2]={rand()%2,rand()%2}; //dna of the mother
	int johnny=rand()%2; //which dna of the mother johnny gets
	int sibling=rand()%2; //which dna of the mother sibling gets
	
	if (arr[johnny]==1){ //in this universe, johnny is sick, which fufils our assumption
		success+=arr[sibling];
		total++;
	}
}
cout<<(double)success/total<<endl;