please help with THIS brain teaser?

Question

Two men are playing Russian roulette using a pistol with six chambers. Assuming that a single bullet is used and that the cylinder is spun after each turn, what is the probability that the 1'st man will lose the game?

Answer 1

The first man will lose if he fires the bullet on an odd numbered attempt.
First attempt (1/6)
Third attempt (5/6*5/6*1/6) - 1st and 2nd miss, then 1st hits
Fifth attempt (5/6*5/6*5/6*5/6*1/6) - 1 miss, 2 miss, 1 miss, 2 miss, 1 hit.
etc.

This sum looks like:
1/6 + 5/6*5/6*1/6 + 5/6*5/6*5/6*5/6*1/6, etc.

So this is:
1/6( 25/36^0 + 25/36^1 + 25/36^2 + ... )

The sum is of a geometric series is given by the formula below:
Sn = a(1 - r^n) / (1 - r)

Here a = 1/6 and r = 25/36
Sn = 1/6(1 - 25/36^n) / (1 - 25/36)

To figure the total probability, take the limit as n --> infinity:
S(infinity) = 1/6(1 / 1 - 25/36)
S(infinity) = 1/6(36/11)
S(infinity) = 6/11

So the probability that the first man will lose is 6/11. (You can easily figure the probability the 2nd man will lose by subtracting from 1).

P(1st man loses) = 6/11
P(2nd man loses) = 5/11

It's not an even 50-50 probability because there's slightly more risk to going first.

Answer 2

It is NOT 1 in 6. The chance of the bullet firing each time is one in 6. But as it is spun after each shot, it depends who loaded the gun as to the chances of the first player killing himself. If the first player loaded the gun, the chance of losing on his first shot can be nil if he so chooses.
Assuming the person who loaded the gun, had no knowledge as of where the bullet was. Assuming they play until one person shoots himself. There is a 50/50 chance that the first person will be the loser, the one to shoot himself.
If on the other hand they only have one shot each, and assuming no one knew where the bullet was before the first shot, then I am wrong and the possibility is 1 chance in 6.

Answer 3

He has a 1 in 6 chance on any given shot that he'll shoot himself, however that doesn't mean that he has a 1 in 6 shot to be the first to shoot himself. The other guy has a 1 in 6 shot every time he gets a chance. I'm not sure what the actual probability is, but it isn't 1 in 6.

"(1/6) + (5/6)^2(1/6) + (5/6)^4(1/6) + ... + (5/6)^n(1/6).
This equates to 6/11 or a 54.545% chance of the first player"

Answer 4

1 in 6

look at it this way - you do not consider anything but one factor - each time you pull the trigger there is a 1 in 6 chance of losing pulling it multiple times does not change the odds

the odds for the second man are a bit strange
before the first man pulls the trigger he has a 1 in 2 chance but as soon as he enters he has a 1 in 6 chance

note that the assumption that you have made and is making you think that the odds are 50-50 ( bet you do ) is that the gun will fire and someone will lose - you cannot make that assumption and still talk about odds

Answer 5

Speaking that you spin the barrel after each turn, it's 1 in 6.