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.
00
01
10
11
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.
How's this?
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) =
= 1/2 |
I believe it's
incorrect 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)
= 1/3