In 5vs5 game what is the probability to get ACE (5 kills)?

Question

Let's assume the common cyber sport game where 10 players play 5vs5. Let's say that one of them, Player 1, wins the gunfight with prob p=0.5. What is the probability that he will get an Ace (all 5 kills) in this game? My initial thought was that it would be 1/32, because he needs to win 5 times, but I have a feeling that we have to? count the chances that teammates can kill enemies too and steal your kills.

Answer 1

Let the chance of Player 1 wiping out the rest of the enemy be $F(m,n)$, where $m$ men including Player 1 are alive on team $A$, and $n$ are alive on team $B$.
We want $F(5,5)$.
There is one chance in $m$ that the next fight involves Player 1, so he has to win; and $m-1$ chances in $m$ that someone else is, in which case the player on team $A$ has to die.

$$F(m,n)=\frac1{2m}\left(F(m,n-1)+(m-1)F(m-1,n)\right)$$

The boundary cases are $F(m,0)=1$.

My calculations give

$$\frac1{16}\left(\frac1{2^5}+\frac4{4^5}+\frac6{6^5}+\frac4{8^5}+\frac1{10^5}\right)$$ or about one in 443.7