As Bushindo advised, we'll start with exact 2-person solutions, and build up.

Kx = probability of x killing in one shot (Ka = 4/20, Kb =7/20, etc.)

Exy (Expected survival of X shooting first against Y)

(X kills Y + (X misses and Y misses, and we wind up in the same state as before))

= Kx + (1-Kx)*(1-Ky)*Exy

Since Exy appears on both sides, we can solve for it in terms of basic kill probabilities

Exy *(1-(1-Kx)*(1-Ky)) = Kx

(1) Exy = Kx / (1-(1-Kx)*(1-Ky))

Exyz (expected survival of X shooting first against YZ)=

X kills Y And Z misses: we get to Exz +

X misses Y and Y kills Z: we get to Exy +

X misses, Y misses, Z misses: we get to same state

Exyz = Kx*(1-Ky)*Exz +

(1-Kx)*Ky*Exy +

(1-Kx)*(1-Ky)*(1-Kz)*Exyz

Exyz * (1-(1-Kx)*(1-Ky)*(1-Kz)) = Kx*(1-Ky)*Exz + (1-Kx)*Ky*Exy

(2) Exyz = (Kx*(1-Ky)*Exz + (1-Kx)*Ky*Exy) / (1-(1-Kx)*(1-Ky)*(1-Kz))

Ewxyz (expected survival of W shooting first agains XYZ)

W kills X and Y kills Z: we get to Ewy +

W kills X and Y misses and Z misses: we get to Ewyz +

W misses X and X kills Y and Z misses: we get to Ewxz +

W misses X and X misses and Y kills Z: we get to Ewxy +

W misses X and X misses and Y misses and Z misses: we get to Ewxyz

Ewxyz = Kw*Ky*Ewy +

Kw*(1-Ky)*(1-Kz)*Ewyz +

(1-Kw)*Kx*(1-Kz)*Ewxz +

(1-Kw)*(1-Kx)*Ky*Ewxy +

(1-Kw)*(1-Kx)*(1-Ky)*(1-Kz)*Ewxyz

(3) Ewxyz =

(Kw*Ky*Ewy +

Kw*(1-Ky)*(1-Kz)*Ewyz +

(1-Kw)*Kx*(1-Kz)*Ewxz +

(1-Kw)*(1-Kx)*Ky*Ewxy ) /

(1-(1-Kw)*(1-Kx)*(1-Ky))

Evwxyz (expected survival of v shooting first against wxyz)

V kills w, x kills y, z misses, get to Evxz +

v kills w, x misses, y kills z, get to Evxy +

v kills w, x misses, y misses, z misses, get to Evxyz +

v misses, w kills x, y kills z, get to Evwy +

v misses, w kills x, y misses, z misses, get to Evwyz +

v misses, w misses, x kills y, z misses, get to Evwxz +

v misses, w misses, x misses, y kills z, get to Evwxy +

all miss, get to Evwxyz

(4) Evwxyz = (

Kv*Kx*(1-Kz)*Evxz +

Kv*(1-Kx)*Ky*Evxy +

Kv*(1-Kx)*(1-Ky)*(1-Kz)*Evxyz +

(1-Kv)*Kw*Ky*Evwy +

(1-Kv)*Kw*(1-Ky)*(1-Kz)*Evwyz +

(1-Kv)*(1-Kw)*Kx*(1-Kz)*Evwxz +

(1-Kv)*(1-Kw)*(1-Kx)*(1-Ky)*Evwxy ) /

(1 - (1-Kv)*(1-Kw)*(1-Kx)*(1-Ky)*(1-Kz))

-------------------

Finally, to answer the question:

Survivals:

(a) Evwxyz(a,b,c,d,e)

0.076

(b) A misses, then B starts the game:

(1-Ka)*Evwxyz(b,c,d,e,a)

0.194

© A kills B, then C starts with 4 players PLUS

A and B miss, then C starts the game:

Ka*Ewxyz(c,d,e,a) + (1-Ka)*(1-Kb)*Evwxyz(c,d,e,a,b)

0.181

(d) A Kills B, C misses, and D starts with 4 players, PLUS

A misses, B kills, D starts with 4 players PLUS

A, B, and C all miss, D starts the game

Ka*(1-Kc)*Ewxyz(d,e,a,c) +

(1-Ka)*Kb*Ewxyz(d,e,a,b)+

(1-Ka)*(1-Kb)*(1-Kc)*Evwxyz(d,e,a,b,c)

0.229

(e) A and C kill, E starts with 3 players PLUS

A kills, C and D miss, E starts with 4 players PLUS

A misses, B kills C, D misses, E starts with 4 players PLUS

A and B miss, C kills D, E starts with 4 players PLUS

A,B,C,D miss, E starts with 5 players

Ka*Kc*Exyz(e,a,c) +

Ka*(1-Kc)*(1-Kd)*Ewxyz(e,a,c,d)+

(1-Ka)*Kb*(1-Kd)*Ewxyz(e,a,b,d)+

(1-Ka)*(1-Kb)*Kc*Ewxyz(e,a,b,c)+

(1-Ka)*(1-Kb)*(1-Kc)*(1-Kd)*Evwxyz(e,a,b,c,d)

0.288