Bloody Roulette game

[jasen]

In a bloody Roulette game, usually a gun with even bullet's holes is used,
then the gun is fired alternately to the head of 2 player.
This is a fair game, because if the bullet is in odd hole so the first shooter will win,
but if the bullet is in even hole so the second shooter will win.

Now, How if we use 2 guns ?
first gun has 1 more hole then second gun.
(such as 1^st gun 6 hole and 2^nd gun 5 holes)
first player hold first gun and shoot first. 
both player shoot alternately.

Is this a fair game or not?
If not who has bigger chance to win?

Edited September 21, 2015 by jasen

[bubbled]

I think:

 

It's completely fair.

[jasen]

I think:

 

Hidden Content

Right answer, but I'm sure some of us still wonder why.

Let's cek the combinations
note (x,y) means : bullet at  x^th hole in gun1 and bullet at y^th hole in gun2
since gun1 is shot 1st so if x<= y then means 1st player win.
so (1,1) or (2,2) means 1^st player win.

for 2 holes and 1 hole guns
(1,1) 1st win
(2,1) 2nd win

for 3 holes and 2 holes guns
(1,1) 1st win (1,2) 1st win
(2,1) 2nd win (2,2) 1st win
(3,1) 2nd win (3,2) 2nd win

for 4 holes and 3 holes guns
(1,1) 1st win (1,2) 1st win (1,3) 1st win
(2,1) 2nd win (2,2) 1st win (2,3) 1st win
(3,1) 2nd win (3,2) 2nd win (3,3) 1st win
(4,1) 2nd win (4,2) 2nd win (4,3) 2nd win
 
for 5 holes and 4 holes guns
(1,1) 1st win (1,2) 1st win (1,3) 1st win (1,4) 1st win
(2,1) 2nd win (2,2) 1st win (2,3) 1st win (2,4) 1st win
(3,1) 2nd win (3,2) 2nd win (3,3) 1st win (3,4) 1st win
(4,1) 2nd win (4,2) 2nd win (4,3) 2nd win (4,4) 1st win
(5,1) 2nd win (5,2) 2nd win (5,3) 2nd win (5,4) 2nd win

 and so on....

if we pay attention at the combinations above, they create pattern of 2 triangle 

1st player win at upper triangle and 2nd player win in lower triangle
both triangle are same size, so both player have same chance to win.