4 cards are shuffled and placed face down. Hidden faces display 4 elements: earth, wind, fire, water. You turn over cards until win or lose. Question: $4$ cards are shuffled and placed face down in front of you. Their hidden faces display 4 elements: water, earth, wind, fire. You turn over cards until win or lose. You win if you turn over water and earth. You lose if you turn over fire. What is the probability that you win?
I understand that wind is effectively absent from the sample space. Does not affect your chances of winning or losing. I also know that $\frac13$ (because we removed wind), you can pick fire where you lose the game.
 A: Since there are 4 cards, there are 4! = 24 possible orderings of the cards.
You lose in the following cases:  
1) Fire is the first card. There are 3! = 6 ways for this to occur. 
2) Wind is the first card and Fire is the second. There are 2! = 2 ways for this to occur.  
Hence, the probablity of losing = $\frac{6+2}{24}=\frac{1}{3}$
So your answer is correct.
A: There are two intepretations here. For both you can throw out the Wind card and only focus on the three other cards. Restating the question:
You shuffle four cards. There is a bijection between your four cards and Water, Earth, Wind, and Fire; i.e., unique assignments.


*

*You win if you flip over $W$ and $E$ (both) before an $F$.
There are only $3\cdot 2 \cdot 1 = 6$ permutations: $\{WEF, EWF, WFE, EFW, FWE, FEW\}$. Only $WEF, EWF$ result in wins so $P(win) = \frac{1}{3}$. This is the same as the probability of getting $F$ as the last card.

*You win if you flip over either $W$ or $E$ (inclusive) before an $F$.
Looking at the permutations in 2a, you lose if $F$ is the first card. Since only $FWE, FEW$ result in a loss, $P(win) = \frac{2}{3}$.
A: Probablity of losing is not 1/3. It should be 2/3.
As Below,
He wins if,


*

*First three cards are Water/Earth/Wind = 3!

*First two cards are Water/Earth and third is Fire = 2


so the total = 6 + 2 = 6
Probability of winning  = 8/24 = 1/3 , so loosing 2/3.
