In yesterday's post, I asked this question:

This is a companion discussion topic for the original blog entry at: http://www.codinghorror.com/blog/2008/12/finishing-the-game.html

In yesterday's post, I asked this question:

This is a companion discussion topic for the original blog entry at: http://www.codinghorror.com/blog/2008/12/finishing-the-game.html

I can remember having similar discussions about probability, and touching on these scenarios, with a friend of mine who is a maths and science wiz. The end result of those discussions is actually inline with the, supposedly incorrect, answer to the first scenario, regarding the probability of the sex of the unknown child. The underlying argument which my friend reinforced to me, and I have to say make sense, is that the history of an event is not known to the event and so cannot dictate the outcome of that event, unless the history has changed the circumstances/resources for any subsequent event. So when you flip a coin, regardless of how many times you have done so before, and/or the results of those flips, the new flip is independent - so the chances remain 50% Heads & 50% Tails. The only instance where this is not the case would be, say, picking red and blue balls out of a room containing a finite number of each. In which case, as you take each one out, you change the probability of the following balls being one color or the other.

Regardless, I guess this is, at the least, a reason for me to read more about probability.

Further on thisβ¦ I note that you state that the combinations of β*BGβ and β GBβ are different (hence providing three possible combinations out of the original four after ascertaining that the final combination must contain at least one βGβ). If that is the case, then you are stating that the order in which these children are placed (whether by age, or otherwise) is a factor here. In that case, should it not be four possible combinations out of an original six? Being βB1B2β, βB2B1β, βG1B1β, βB1G1β, βG1G2β, βG2G1β then reducing down to βG1B1β, βB1G1β, βG1G2β, βG2G1β or βBGβ, βGBβ, βGGβ, βGGβ and, once again, returning to the 50% chance of the other child being a βBβ.*

The question is badly worded: βtold you they had two children, and one of them is a girlβ

It should say βtold you they had two children, and at least one of them is a girlβ.

If just one of them were a girl it would mean the other is a not a girl with a probability of 1.

1 Like