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Russian Roulette - probability of winning
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I thought it was easy to calculate the probability for each player of winning a game of Russian roulette but my answer is not the as that given by a couple of books I have.
I would be be interested to see the reasoning of others on this; will they agree with me or my books; maybe both are wrong!
I would be be interested to see the reasoning of others on this; will they agree with me or my books; maybe both are wrong!
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Does each player re-spin?
The odds of the bullet being in the chamber for the first shot are not 1 in 6. The weight of the bullet means it is more likely to end up at the bottom ... not at the top. Only a slight edge, but still an advantage.
If the bullet is most likely to end up at the bottom then, unless they re-spin every time, the player most likely to get the bullet is player number 4.
The odds of the bullet being in the chamber for the first shot are not 1 in 6. The weight of the bullet means it is more likely to end up at the bottom ... not at the top. Only a slight edge, but still an advantage.
If the bullet is most likely to end up at the bottom then, unless they re-spin every time, the player most likely to get the bullet is player number 4.
Marty - I wondered whether I had overlooked some aspect the rules. But I think all were woking on the basis of players taking alternate turns until one shot himself.
Jayne - You are right that practical considerations may complicate things. But I am looking for a solution in the "ideal world", keeping things simple.
Jayne - You are right that practical considerations may complicate things. But I am looking for a solution in the "ideal world", keeping things simple.
I think if you do not spin the probabilities are 50/50 for each player.
Let A be the player who goes first. There is a 1/6 chance he dies, 5/6 chance he lives. if he lives, B goes next - there is a 1/5 chance he dies. So before the game starts his probability of dying in the 2nd round is (5/6) x (1/5) = 1/6.
Similar logic for rounds 3, 4, 5 6 says the ex ante odds are 1/6 for each round (e.g. if you get to the last round, B dies for sure but there is only a 1/6 chance to get to the last round.)
Let A be the player who goes first. There is a 1/6 chance he dies, 5/6 chance he lives. if he lives, B goes next - there is a 1/5 chance he dies. So before the game starts his probability of dying in the 2nd round is (5/6) x (1/5) = 1/6.
Similar logic for rounds 3, 4, 5 6 says the ex ante odds are 1/6 for each round (e.g. if you get to the last round, B dies for sure but there is only a 1/6 chance to get to the last round.)
Mmm. I'm not sure JJ.
For example the odds for the first round for player two also need to take account of the probability (5/6) that the first player hasn't died (because if the first payer had died then player two wouldn't need to play)
I saw the odds for this once and the results surprised me. I may have alook later
For example the odds for the first round for player two also need to take account of the probability (5/6) that the first player hasn't died (because if the first payer had died then player two wouldn't need to play)
I saw the odds for this once and the results surprised me. I may have alook later
Marty - thanks, yes, it is a matter of rules.
(We are considering games with just two players.) With no spinning of the cylinder between rounds, I reckon each player has an equal chance of dying (or surviving). After the cylinder has been spun at the start, the chances of the bullet being being in an "odd" chamber are the same as it being in an "even", therefore the first player has the same chance of shooting himself as the second. (Same conclusion as Dr b above, by a different route.)
I now realise that this does not agree with the book answers because they assume spinning the cylinder between firings, in which case the chances of the two players are not equal. I'm not sure that I follow the reasoning fully, I'll have to study it before posting anything (not helped with death being considered "victory"). I think I've been through this before, but forgot whaat I learnt! Thanks for all contributions so far.
(We are considering games with just two players.) With no spinning of the cylinder between rounds, I reckon each player has an equal chance of dying (or surviving). After the cylinder has been spun at the start, the chances of the bullet being being in an "odd" chamber are the same as it being in an "even", therefore the first player has the same chance of shooting himself as the second. (Same conclusion as Dr b above, by a different route.)
I now realise that this does not agree with the book answers because they assume spinning the cylinder between firings, in which case the chances of the two players are not equal. I'm not sure that I follow the reasoning fully, I'll have to study it before posting anything (not helped with death being considered "victory"). I think I've been through this before, but forgot whaat I learnt! Thanks for all contributions so far.
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