Confusion regarding the sample space of a experiment 
First, of all, I need to realize what sample space is and this is causing me some trouble. Perhaps I am misunderstading the problem? But this is my reasoning:
Since player 1 has two choices at each stage and there is only 4 games he can have provided he wins to player 2,3 and 4, then the size of the sample space is $2^4=16$ And for example, 
$$ P(X=0) = \frac{1}{16} $$
since there is only a way player 1 loses that is if he loses agains player 2 at the beginning. However, in my notes it says the sample space size is $5!$ and $P(X=0) =1/2$. How come this is true? I just don't see it. Am I overthinkin this problem?
 A: There are $5$ discreet numbers which can be randomly arranged in $5! = 120$ ways which is your sample space. 
I take it that $P(x=i)$ means $i =$ exactly $0,1,2,3,4$.
In the $120$ different sequences of $5$ numbers, we have $60$ sequences where the second number is larger than the first. ($1,2$ etc)
$P(0) = \frac{60}{120} = \frac{1}{2}$
Then we have $20$ sequences where the first number is larger than the second but smaller than the third. ($2,1,3$ etc)
$P(1) = \frac{20}{120} = \frac{1}{6}$
Then we have $10$ sequences where the first number is larger than the second and third but smaller than the fourth. ($3,1,2,4$ etc)
$P(2) = \frac{10}{120} = \frac{1}{12}$
$6$ sequences where the first number is larger than the second, third and fourth but smaller than the fifth. ($4,1,2,3,5$ etc)
$P(3) = \frac{6}{120} = \frac{1}{20}$
And finally, $24$ sequences (all start with $5$) where the first number is larger than the other $4$. ($5,1,2,3,4$ etc)
$P(4) = \frac{24}{120} = \frac{1}{5}$
Notice the sequences total $120$ and so the probabilities sum to $1$.
