I disagree. And this, in fact, is exactly what happens with this question. I quote myself above:
|people seem to either insist on 1/2, or insist on altering the question until the answer is 1/2.|
Okay, I have picked a number from 0 to 3, with an equal chance of each.
Now, I'm telling you it isn't 3.
What is the chance that it's zero? I say that the chance it is zero is 1 in 3.
I am now going to write the numbers out in binary.
It isn't 3 (i.e. it isn't 11). What is that chance that it's zero (00)?
|If it helps... rephrase your original problem to say "look at coin 1" instead of "look at one of the coins". The result of coin 1 will rule out two possibilities, not just one.|
That doesn't help. That's the exact opposite of what I want. The answer is 1 in 3. I want the version where you do NOT know which coin I looked at, because that is a different question.
Let HH, TH, HT, and TT be the coin configurations, and TT' be you saying "it's not double tails".
P(HH | TT') = P(TT' | HH)P(HH)/(P(TT' | HH)P(HH) + P(TT' | TT)P(TT) + P(TT' | TH)P(TH) + P(TT' | HT)P(HT)) =
= P(TT' | HH)/(P(TT' | HH) + P(TT' | TT) + P(TT' | TH) + P(TT' | HT)) =
= 1/(1 + 0 + 1/2 + 1/2) =
I believe it's
the answer to a different question. Here's my version, starting with the simple statement of Bayes theorem:
P(A|B) = P(B|A) P(A) / P(B)
Let's say A is HH (both heads), and B is TT' (not tails-tails) i.e.
P(HH|TT') = P(TT' | HH) P(HH) / P(TT')
P(TT' | HH) is the probability of not tail-tail, given that it's head-heads. Well, that's one. If it IS heads-heads, it's definitely NOT tails-tails, so this is a certainty. P(TT' | HH) = 1
P(HH) = 1/4 (this is just the simple probability of heads-heads)
P(TT') = 3/4 (this is just the simple probability of anything other than tails-tails)
So, P(HH|TT') = P(TT' | HH) P(HH) / P(TT')
= 1 x (1/4) / (3/4)