A counter intuitive probability question Here's the set-up: 
Suppose we have three people in a line, Andrew Smith, Ben Hardy and Ben Parsons, all of them are strangers to you. You are trying to guess the first name of the person at the front of the line, if you get it right, you get a prize. 
Then someone tells you that they happen to know one of the Bens, in particular, a Ben that is not at the front of the line. Should you let this person tell you what they know?
It appears not, as in the beginning we have a 2/3 chance of guessing correctly if we say that Ben is at the front. But if we condition on the last or the middle person being a Ben, then guessing that Ben is at the front now has only a 1/2 chance of being correct.
This seems counter-intuitive, given that we can infer from the that there must be a Ben in one of those positions, so it seems like we don't learn anything relevant.
Have I done the calculation incorrectly, or is this just another Monty-Hallesque quirk of probability that I need to wrap my head around?
 A: You can't properly do conditional probability unless you know exactly what would have happened under all circumstances.
For example: suppose that I know Ben Hardy, specifically, and also that I'm willing to help you out as much as possible. Then the following outcomes can happen:


*

*With probability $\frac13$, Ben Hardy is at the front, and I tell you where he is. This raises your probability that a Ben is at the front from $\frac23$ to $1$. 

*With probability $\frac23$, Ben Hardy is not at the front, and I tell you where he is. This lowers your probability that a Ben is at the front from $\frac23$ to $\frac12$, because now you know Ben Hardy is not at the front.



But now suppose that I know both Bens, and will pick the one not at the front to tell you about. If Andrew is at the front, I'll pick a random Ben. 
In this case, before the order is picked, you know that I will tell you about a Ben not in the front; moreover, both positions are equally likely. No matter who is at the front, you expect to hear the same thing: "a Ben is in the middle" with probability $\frac12$, and "a Ben is in the back" with probability $\frac12$.
In this case, the probability that the other Ben is in the front remains at $\frac23$, because I have not communicated any information to you. (This is the Monty-Hall-esque option.)
The only way your probability changes is if you expect to hear potential answers with different probabilities depending on who is in the front.
A: There are fewer unknown Bens left after you gain that information, so the probability that an unknown-name person will be Ben decreases. But the probability that you'll guess Andrew correctly increases.  
It's just the formula- favourable number of elementary events divided by total number of elementary events. The additional information does help you in decreasing the total number of possibilities but the number of events in favor that 'a particular person is Ben' also decreases. The overall value of a fraction decreases when you subtract the same value from both numerator and denominator.
A: There is a Ben not at the front of the line in every scenario, and the position of that Ben is irrelevant, so you would not gain any information. Out of 6 configurations, 4 have Ben at front. Probability that a Ben is in front is still 2/3. 
$$a,b_1,b_2 \quad a,b_2,b_1 \quad b_1,a,b_2 \quad b_1,b_2,a \quad b_2,a,b_1 \quad b_2,b_1,a$$
Each of the above has an equal probability. Thus
P(a,b_1,b_2)=1/12 
P(a,b_1,b_2)=1/12
P(a,b_2,b_1)=1/12
P(a,b_2,b_1)=1/12
P(b_1,a,b_2)=1/6
P(b_1,b_2,a)=1/6
P(b_2,a,b_1)=1/6
P(b_2,b_1,a)=1/6
where bold shows the Ben we are given. 
