Does having more options with equal probability increase success? Let us assume I'm playing a simple card game. The game has 3 cards, one with a value of 1, one with a value of 2, and one with a value of 3. The game begins with all cards face down (the numbers face down) in two piles. The left pile consists of one card and the right pile has two cards, one on top of the other. I can only draw one card at a time, and I can only draw off the top. To win, I must draw the cards so that I draw them in order (ie, 1, then 2, then 3).
I drew a probability tree and found that my chance of winning in each branch is .165 (33% for the first card and 50% for the second). But if I draw from the right pile first instead of the left, I have 2 ways of winning instead of 1. Does this mean I should draw from the right pile first, or is this a logical fallacy? 
 A: Imagine I have a deck of 3 shuffled cards with values $1,2,3$, and I shuffle them uniformly. I turn them face down in a pile to produce the sequence $\pi(1), \pi(2), \pi(3)$, and you draw from the top. You win if you turn over the cards in order -- so you win if $\pi(1) = 1$, $\pi(2) = 2$, $\pi(3) = 3$ (with probability $1/6$). You obviously have no choices to make, so you win if $\pi$ is the identity permutation.
In your specific situation, you have some choice of which cards to turn over in some order, but you are just applying another permutation $\tau$ to the cards by making these choices. You win if the shuffle ($\pi$) composed with your permutation $(\tau)$ is the identity permutation. Each permutation has a unique inverse, so whatever $\tau$ you pick, you still win with probability $1/6$, if $\pi = \tau^{-1}$. 
Note that this is true if we ignore the piles and you have complete freedom of which card to turn over at each step. There is also nothing special about $3$ here; with $n$ cards, you win this game with probability $1/n!$, and it doesn't matter what decisions you make.
A: It makes no difference.


*

*if your first choice is wrong, game over.

*if you make a hit on first choice from left pile, you have a $1/2$ chance of winning, and if you made a hit from the right pile, you have two choices but the same chance of $1/2$ !
