4 ms·
The best intuition I've heard for the two-child problem is as follows. Suppose you have two doors with two children behind them. The 1/2 chance of a boy versio
by tylerhou 10y ago
The best intuition I've heard for the two-child problem is as follows.
Suppose you have two doors with two children behind them. The 1/2 chance of a boy version works like this: Say you randomly open one of the two doors and there is a boy. That doesn't give you any information about the other boy, so you say that the probability of the other one being a boy is 1/2.
1/3: This time beforehand you know that there is at least one boy. Then you open a door and it happens to be a boy. That tells you something about the other door, which alters its probability to 1/3.
- pizza 10y agoFor the sequential case: P(2 boys | >= 1 boy) = P(>=1 boy | 2 boys) * P(2 boys) / P(>= 1 boy) where P(>= 1 boy of all 2 kids): 1 gg + 1 bg + 1 gb + 1 bb: = 3/4 P(2 boys): 1 gg + 1 bg + 1 gb + 1 bb: = 1/4 and trivially P(>= 1 boy | 2 boys) = 1 So P(2 boys | >=1 boy0 = 1 * (1/4) / (3/4) = 1/3
- notahacker 10y agoInteresting you mentioned doors, since my first thought on reading the article was of the Monty Hall Problem, specifically is this guy trying to parody the Monty Hall problem by coming up with purposely obtuse interpretations of purposely vague questions to complicate what would normally be considered a trivial conditional probability question? But the Monty Hall problem came later. This just feels like the probability equivalent of Xeno's Paradox, except revolving around arbitrarily shifting unstated assumptions about selection effects rather than arbitrarily reducing distances travelled.