In the game of Russian roulette, one inserts a single cartridge into the drum of a revolver.?

Question

Thus, the other 5 chambers of the drum are left empty. One then spins the drum, aim's at one head, pulls the trigger. a) What is the probability of being stil alive after playing the game N times? b) What is the probability of surviving (N-1) turns in this game and then being shot the nth time one pulls the trigger? c) What is the mean number of times a player gets the opportunity of pulling the trigger in this game?

Answer 1

Hmmm... somehow I would feel more comfortable if this problem were phrased with a 6-sided die instead of a revolver, but anyway:

a) (5/6)^N
b) (1/6)*(5/6)^(N-1)
c) 6

To find the answer for (c), note that for any event that happens with a probability p on each try, it takes 1/p tries on average for it to happen. For example, 2 is the average number of coin flips it takes to get the first head, and 6 is the average number of rolls of a 6-sided die that it takes get the first 1.

Answer 2

Technically, the answer to (a) should be 5/6, because the question as stated assumes you play the game the Nth time, and to do so you must have survived the first N-1 plays. In other words, the probability of surviving the first N-1 plays is 1, because we know you've already successfully done it.

BUT...

That's not really what they meant to ask. They meant to ask "What is the probability of playing the game N times and still being alive"?

With five out of six chambers empty, you have a 5/6 chance of surviving one game. In the 5/6 of cases in which you survive one game, you'll have a 5/6 chance of surviving a second. Your probability is thus 5/6 of 5/6, or (5/6) * (5/6). Each game you add, appends another factor of 5/6 to the probability of your survival. So in general, you have a (5/6)^N probability of playing and surviving N games.

For (b), it's (5/6)*(5/6)*(5/6)* ... *(5/6)*(1/6), where there are N-1 factors of (5/6) representing the N-1 games that are survived, the 1/6 factor representing the probability of the being-shot out come occurring in the Nth game.

For (c), you must take a weighted average: take the number of plays, times the probability of the person getting shot after that number of plays. Add these numbers up for all possible numbers of plays. So the first term in the sum is (1 play)*(1/6 probability).
The second is (2)*((5/6)*(1/6)). Keep going - you get an infinite sum. You can probably leave your answer like that, as the sum without evaluating it, but you should attempt to evaluate it to a single number. It is doable. Hint: call this sum A, then ask what would (5/6)*A look like?

Answer 3

1

Answer 4

When the player "aim's at one head" does it make a difference to whom that one head belongs?

Did you mean "aims at one's head"?

Probability of still living after N times: (5/6)^N

Probability of being shot on exactly the Nth turn:
(5/6)^(N-1)(1/6) = [5^(N-1)] / 6^N

Gimb's sum is wrong for c). It should be:

Sum 1 * (1/6) + 2 * (1/6)*(5/6) + 3 * (1/6)*(5/6)^(2) + ...

Answer 5

Playing Russian roulette is a very dangerous game and there really is no odds.... you can get the bullet the first time just as easily as you can the ninth time ......its not like a game of cards where there are odds s....this is a game of suicide intent my advice is don't start playing such a potentially dangerous game....cause no matter how you look at it the odds are against you

Answer 6

a) chanc eof living after one turn is 5/6
after 2 turns 5/6*5/6
etc so chances after N turns (5/6)^N = chance of still living

b) (1/6)*(5/6)^(N-1)

c) Sum 1* (1/6) + 2 * (1/6)*(5/6)^(2) + 3*(1/6)*(5/6)^(3) + ...

Answer 7

I really do not recommend playing this "game" as I personally know someone who did it and killed himself that way.

Get a Playstation 3 or something else to be occupied with.