5 ms·
There are problems that I find even more counterintuitive than Monty Hall, such as the sister's paradox. Your neighbours has two children, one of which is
by plopilop 8y ago
There are problems that I find even more counterintuitive than Monty Hall, such as the sister's paradox.
Your neighbours has two children, one of which is a girl.
What is the probability of the other child to be a girl?
A. One may say 1/2, but actually, using basic Bayes theorem: P(2 girls | at least 1 girl) = P(2 girls AND 1 at least girl) / P(at least 1 girl) = (1/4) / (3/4) = 1/3.
I think this problem is actually equivalent to the Monty Hall problem. But it can get much weirder.
Your neighbour has 2 children, one is a girl.
You know the other child is born on a Sunday,
what is the probability that this other child is a girl?
A. Similarly, counting all possible cases, one does not get 1/2 or 1/3, but rather 13/27.
- cjslep 8y agoYou are horribly butchering Bayes' Theorem. Both your examples are incorrect because your two events ('2 girls' and 'at least 1 girl') are not oberved independently of one another, which is a necessary condition to using Bayes' Theorem with a valid result. Great tool but wrong job.
- plopilop 8y agoWhat? Bayes' theorem does not require anything about the events, even more so it is useful when the events are not independent. If events are independent, Bayes' theorem has no usefulness. https://en.wikipedia.org/wiki/Bayes%27_theorem#Statement_of_theorem https://en.wikipedia.org/wiki/Bayes%27_theorem#Statement_of_... Edit: however now that I look into it, I should rather have used "conditional probabilities" name rather than Bayes'. I always mix both of them, my bad.
- kqr 8y agoIn the first case, I guess what's happening is that when we pick two people randomly with uniform distribution, we expect, with equal probability, any combination of boys and girls: MM MF FF FM. When we observe a girl, we can exclude the case of two boys, so the remaining cases to consider are MF FF FM, where obviously only 1/3 of the cases is two girls. So I guess what makes this counterintuitive is that we think of the example as independent variables actually affecting each other, but what happens is really that we're drawing from a collection without replacement, and obviously the mix of the collection is going to change as we do so. This should apply also on a humankind, world-wide scale: if you have met three women in a row, the fourth person you meet is more likely to be a man. Gambler's fallacy! Or just drawing without replacement. A roulette table is drawing with replacement. :)
- KozmoNau7 8y agoBut MF and FM in this case are the same?
- cecilpl2 8y agoNo because children are distinguishable. If you flip two coins, you get HH, HT, TH, and TT with equal probability. Try it.
- KozmoNau7 8y agoBut the sequence is irrelevant. HT is equal to TH, because in this situation we only care whether the two coins are equal to each other or different from each other.
- Dylan16807 8y agoIt's the percent chance that matters, so it doesn't matter if you can distinguish them. The cases still exist with equal chance.
- deleted 8y ago[deleted]
- KozmoNau7 8y agoThe first child is not an influencing factor on the second child. It's either a boy or a girl, 50/50.
- KozmoNau7 8y agoTo elaborate, the situation as stated in the question is currently Hx, so HH and HT are the only possible results.
- Dylan16807 8y agoYou can calculate 50/50 but it's not by ignoring the difference between HT and TH.
- jeremejevs 8y agoLooks like this is the Boy or Girl paradox [0], which has ambiguity problems. [0] https://en.wikipedia.org/wiki/Boy_or_Girl_paradox https://en.wikipedia.org/wiki/Boy_or_Girl_paradox
- grkvlt 8y agoI love how Wikipedia helpfully informs us that 'Biologically, according to studies of the human sex ratio, it is more likely for any one person to be born into the male sex than the female sex.' as though it is somehow relevant or has anything to do with the actual paradox being discussed... NOTE I've since edited the Wiki page, and deleted the sentence - since it isn't relevant and there was no citation for the claim!
- DavidSJ 8y agoYour neighbours has two children, one of which is a girl. What is the probability of the other child to be a girl? This is oft-mentioned but with this wording it's ambiguous. Here are three possible interpretations: 1. Your neighbor has two children. We observed both and found that there was exactly one girl. What is the probability that both are girls? (Answer: 0) 2. Your neighbor has two children. We observed one and found that she was a girl. What is the probability that the other (unobserved) child is a girl? (Answer: 1/2) 3. Your neighbor has two children. We observed both and found that there was at least one girl. What is the probability that both are girls? (Answer: 1/3) Interpretation #1 is kind of silly but it serves to illustrate the ambiguity of the plain English wording. Interpretation #2 is how most people interpret the problem, and it's bolstered by the use of the word "other" in your prompt. For there to be an "other" child, there must have been a specific girl observed originally, rather than a general observation that there was at least one girl. Interpretation #3 is how you intended the problem.
- plopilop 8y agoyup, you're right.
- olavk 8y ago> Your neighbours has two children, one of which is a girl. What is the probability of the other child to be a girl? The way you have phrased the question, the probability is 1/2.
- Gh0stRAT 8y agoActually, it's ever so slightly higher than 1/2 because there is a non-zero chance of them being identical twins. If we assume an identical twin incidence rate of 0.4%, then the actual probability would be 63/125 that the second child is also a girl.