[pdurrant]

Originally Posted by omk3

Here's my take:

Spoiler:

Question 1:
We have cylinders A, B, C. Let's say A is the one with the coin.
If I bet on A, the honest conman will randomly reveal either B or C, so the possible combinations are now AB, AC
If I bet on B, the honest conman will reveal C, so we only get AB.
If I bet on C, the honest conman will reveal B, so we only get AC.
So, four possible outcomes. Out of the four, in two I have the right cylinder from the start, in two I don't. It seems like a 1/2 chance.

Question 2:
Why not?

Question 3:
Let's say again that A is the one with the coin.
I may have chosen A,B,C. Let's call Ax, Bx, Cx the cylinder I've bet on.
We get the following possible combinations:
AxBC, ABxC, ABCx
Now one cylinder is randomly knocked over. We have the following possible combinations:
AxB, AxC, BC, ABx, AC, BxC, AB, ACx, BCx
As no coin is revealed, we know that BC, BxC, BCx are out. AC and AB are also out, because the cylinder I bet on is still standing.
So we are left with AxB, AxC, ABx, ACx. We have 2/4(1/2) chance to have bet on the right cylinder, so nothing changed.

This seems a little strange to me, so I may well be wrong....

Your answer to Q1 is wrong.

Hint:

Spoiler:

When you initially choose, you have a 1 in 3 chance of being right. The showman lifting a cylinder that doesn't have the coin under it doesn't change your odds, as he can lift such a cylinder every time. Splitting the first case (where you've chosen the right one initially) into two is erroneous.