Oops, I changed it to a more unintuitive one right after you replied! In my original comment, I said “you flip two coins, and you only know that at least one of them landed on heads. What is the probability that both landed on heads?”
And… No! Conditional probability strikes again! When you flipped those coins, the four possible outcomes were TT, TH, HT, HH
When you found out that at least one coin landed on heads, all you did was rule out TT. Now the possibilities are HT, TH, and HH. There’s actually only a 1/3 chance that both are heads! If I had specified that one particular coin landed on heads, then it would be 50%
No. It’s still 50-50. Observing doesn’t change probabilities (except maybe in quantum lol). This isn’t like the Monty Hall where you make a choice.
The problem is that you stopped your probably tree too early. There is the chance that the first kid is a boy, the chance the second kid is a boy, AND the chance that the first kid answered the door. Here is the full tree, the gender of the first kid, the gender of the second and which child opened the door, last we see if your observation (boy at the door) excludes that scenario.
1 2 D E
B B 1 N
B G 1 N
G B 1 Y
G G 1 Y
B B 2 N
B G 2 Y
G B 2 N
G G 2 Y
You can see that of the scenarios that are not excluded there are two where the other child is a boy and two there the other child is a girl. 50-50. Observing doesn’t affect probabilities of events because your have to include the odds that you observe what you observed.
I was about to reply to you with a comment that started with “oh shit you’re right!” But as I wrote it I started rethinking and I’m not sure now.
Because I actually don’t think it matters whether we’re BB1 or BB2. They’re both only one generation of the four possible initial states. Which child opens the door is determined after the determination of which child is which gender. Basically, given that they have two boys, we’re guaranteed to see a boy, so you shouldn’t count it twice.
Still, I’m now no where near as confident in my answer as I was a moment ago. I might actually go and write the code to perform the experiment as I outlined in an earlier comment (let me know if you think that methodology is flawed/biased, and how you think it should be changed) to see what happens.
Yes! They responded to my comment before the edit, where I gave the coin example: “I flipped two coins, at least one of them was heads. What is the probability that both of them are heads?”
Before I read their reply, I edited it to the more confusing and infuriating two kids example. It’s annoying because it seems like it should be the same as saying “I have two children, and at least one is a boy. What is the probability that I have two boys?” In both the coin case and this one, the answer is 1/3, but when one child answers the door, it’s like sliding one quarter out from behind my hand. Now you know a particular (child, coin)'s (sex, face), and the answer is 50% again.
You assume that the probability of TH = HT = HH
When In fact, the probabilities are as follows:
P(HT)+P(TH) = 50% P(HH) = 50%
For all the probabilities being equal, you’d have to consider 4 cases:
HT, TH, HH (1) and HH (2).
The difference between HH (1) and HH (2) is which one you were told that was heads.
Then P(HH) = P(HH (1)) + P(HH (2)) = 2/4 = 50%
The issue is that you weren’t told a particular one was heads, only that at least one was heads. If I flipped a nickel and a dime, then the four possibilities are NtDt, NtDh, NhDt, and NhDh. If I say that at least one of them is heads but don’t tell you which one, then there are three possibilities: I flipped NhDt, NtDh, or NhDh. It’s only when I tell you that the nickel landed on heads that it collapses to NhDt and NhDh.
Sorry if the acronyms are hard to read, they’re much faster than typing something like “Heads (nickel) tails (dime)”
You’ve also ruled out TH by knowing one is heads. So the only possibilities are HT and HH. Is that not 50/50?
At least one is heads, but unless you know which one it is, you haven’t ruled out HT or TH
This is a ridiculous argument when taken to the extreme. Say you have three bags. Bag A contains 100 blue marbles. Bag B contains 99 blue marbles and 1 red marble. Bag C contains 100 red marbles. You reach into a random bag and draw a red marble. You’ve only eliminated bag A. Would you say it is a 50-50 whether you are left with a bag now containing 99 blue marbles or 99 red marbles? No, the fact that you drew a red marble tells you something about the composition of the bag you drew from. The odds that you drew out of bag B is 1/101, the total number of red marbles in bag B divided by the total number of red marbles across all bags. The odds that you are dealing with bag C is 100x that.
Now let’s say you have 4 bags. BB, BR, BR, and RR. You draw an R. There is a 50% chance you are dealing with bag 2 or 3 because together they contain 2 out of 4 R. There is also a 50% chance you are dealing with bag 4. So it is equally likely that you draw either color of marble if you take the remaining marble out of the bag you randomly selected despite there being twice as many BR bags as RR bags.
