Hacker News new | ask | show | jobs
by bad_user 5830 days ago
> Let's toss two coins until at least one shows a head. By your reasoning the odds of them both being heads is 1/2. It's not. Try it.

You haven't read the article then ... the problem as stated in the article is that you know one coin is going to be a head, so what's the probability of the other one also being a head?

Of course ... the events aren't connected ... the second coin toss doesn't depend in any way on the first coin.

That's why I think there's something wrong about the article ... saying that the probability is 33% fails both intuition and elementary probabilistic.
1 comments
RiderOfGiraffes 5830 days ago
Here's the problem as Gary originally stated it, and as the article quotes it:

    > I have two children, one of whom is a son born
    > on a Tuesday.  What is the probability that I
    > have two boys?
You say:

the problem as stated in the article is that you know one coin is going to be a head,

No. The point of the article is that you don't know how or why you are given this information.

Suppose I toss two coins until I get one that's a head, then I tell you that I have two coins, and one is a head. I've complied with the problem as described. The probability that they are both heads is 1/3.

I was there when this problem was posed. I was in the room when the questions were asked, and Gary clarified. I had lunch with Gary afterwards, and he said it was deliberate that it was ambiguous.

It seems to me that you're missing the point. Perhaps you should explain clearly exactly how you think the situation arises where the information given is as described.

link
bad_user 5830 days ago
> The probability that they are both heads is 1/3.

The probability they are both heads is 1/2.

You're trying to formulate this as ....

  P(head AND head | one is a head) =
  P(head AND head) / P(one is a head) =
  1/4 * 4/3 = 4/12 = 1/3
But this is wrong ... you know that you have a head ... which makes ...

  P(one is a head) = 1 && P(head_a AND head_b) = P(head_a) = P(head_b)
1/3 would be the probability only on the first try (instead of stopping when you've got a head).
link