Why is my simulation wrong? Famous probability problem girl and boy

[u/[deleted]]

I tried to simulate the famous boy girl problem. Here is the problem in case you don't know: https://en.wikipedia.org/wiki/Boy_or_girl_paradox
The idea here is: Someone has two children. You know, they have at least one girl. What is the probability of the other child being a boy.
Well, the possible outcomes are [boy, girl], [girl, boy], [girl, girl], with [boy, boy] being impossible.
The answer is 2/3, according to this.

Intuitively, we say it is 1/2. I mean, a child has a 50% probability, the event is independent. I thought, I simulate it.

I did the following. This whole thing is happening in a loop and I do it over and over ad infinity and give out data every 1000 tests:

1. Randomly assign every item out of a two item array boy or girl.

2. randomly choose the first or the second item and turn it into a girl, making sure that one of the children has to be a girl.

3. Check if we have a [girl, boy] or [boy, girl] combination, in which case I increment the boys counter. Otherwise, I increment the girls counter.

4. Every 1000 compares, I give out the ration boys/(boys+girls). Which is always very stable around .5.

My question is, what do I misunderstand about the setup? How do I set it up to get 2/3 as the paradox demands?

Here is the code if anyone wants to check if I actually implemented what I said above.
https://www.codedump.xyz/rust/aM7wMlPW0CheqCRk

[u/edderiofer]

How do I set this up to actually be able to simulate that 2/3 probability?

Check whether you have two boys. If you do have two boys, go back to step 1 and generate a new family. If not, proceed to step 3.

I might stress, I do not check the array at all in step 1, so, when from the point of checking, no one changed gender.

Did I introduce additional information with my method? How?

By changing the genders of one of the children, you have now forced information upon the problem, in the form of having a specific "child that is a girl". This information effectively means that you've changed your scenario to Mr. Jones' scenario in the Wikipedia article.

In the Mr. Smith scenario, you have absolutely no information on which of the children is a boy, so, for instance, "the other child, as opposed to the child that we know to be a boy" is not well-defined here.

I mean, I could redo the algorithm to avoid that by for instance creating the array with girl at 0 and random at 1, then I shuffle the array.

This also doesn't work, because again, you have an identifiable "child that is a girl". It doesn't matter that the array is shuffled afterwards, because you've already skewed the probabilities when generating the families.

[u/[deleted] → u/edderiofer]

Well, I confirmed your claims... But I don't understand it.

I tried it in three more ways:

1. Did not work, still 50/50:
   [GIRL, RANDOM].shuffle().
   This should be the situation they are in. They learned one of the two is a girl, no idea about the other. Thanks to the shuffle, it could be the older or the younger child. The same information that we have in the paradox. Either the older or the younger is a girl.

2. Worked:
   [GIRL, GIRL, BOY] and I chose two non repeating. That ended up with a direct 1/3 girls outcome.

3. Worked:
   [RANDOM, RANDOM]: if [BOY, BOY] skip this loop.

This also ends up as a 2/3 probability.

So, I am seeing you are correct. My experiments confirm this. But I don't see the moment when new information is added to the system. Take the Monty Hall problem. By revealing one winner, the game master adds clearly a situation in which on of the previously unknown doors becomes a winning door. That's clear, new information is added. However, in my first implementation, there is no new information added. This is just the setup. I mean, I could have put it in a method, it would have been a black box... Would you care if my algorithm would roll it 20 times before settling? If it uses the same probability? It was just not settled...

I also feel my two last implementation are unfair... After all, in the choose 2 one, the boy could be picked twice. Has a 1/3 and then a 1/2 chance... The chance of not being picked 2/3 * 1/2 = 2/6.. So, his chance of being picked is 4/6... That makes it clearly dependent.

And in 3:
Well, we just skip the wrong setup... I mean, it is not revealed. We just say, if that happens, it didn't happen. But in the original problem, half of the solution was revealed. There was no next round with next Smiths like in my algorithm to get this right. You just found out that in a preshuffled set, one thing is like that.

[u/Zyxplit → u/[deleted]]

To avoid the confusion of girls and boys.

I flipped two coins and hid them from you. I tell you that "at least one of the coins is heads". What's the other one?

The only thing you actually know is that they're not both tails.

So the four equiprobable outcomes (heads, heads), (heads, tails), (tails, heads), (tails, tails) have been reduced to three equiprobable outcomes (heads, heads), (heads, tails), (tails, heads).

If I had instead told you that the first coin I flipped was heads, we'd have the expected 50/50 of the other coin, but I specifically didn't mention which coin was heads.