4 ms·

I'll assume that there are two players, and that you spin the cylinder at the beginning, but not after each turn. If this is the case, then at the start, the di

by wch 15y ago

I'll assume that there are two players, and that you spin the cylinder at the beginning, but not after each turn. If this is the case, then at the start, the die is cast; the bullet is in one of the chambers, each with equal probability. It therefore has an equal probability of being in an even or odd chamber. Depending on if you go first or second, you either take odd-numbered or even-numbered turns. Either way, you have the same probability of losing. On the other hand, if you spin the cylinder after each turn, it's best not to go first.

FredBrach 15y ago
Very very interresting, you make me doubt of my answer... But I finally continue to think that a player don't have to go first (if the barrel is spined once). First: for n=1 chamber, it is obvious - the clue. If we extend the n=1 case to n>1, we can feel what will happen: there is intrinsically the time in the game rule: you have to be the last. Then acting on this rule shall make a difference. Let's take n=2 chambers. The payer don't know where the bullet is. Then, from her point of view, she still have to let her opponent try the first one XD from the player point of view, the die is not cast, the fact that the barrel has been spined is virtual: the fact that the variable's drawing has been pre-made or not for each probability system is not relevant So, in short, when the player press the trigger, the die is casted once like in each probability system - but becareful, here, the die is changing after each try since the rule is that the barrel is spinned once=) Great great problem, I like it=) And so on for n>2.
- FredBrach 15y ago
  Ok no, I think now I'm wrong=) this is even, the die is cast once when the players choose to start or not and after, the game is made.
- muyuu 15y ago
  I'm pretty sure your first answer was wrong, but I don't get why would anyone downvote instead of correcting or giving a different answer.
  - FredBrach 15y ago
    It seems that I'm victim of some kind of ennemy's downvotes

by wch 15y ago

FredBrach 15y ago

Very very interresting, you make me doubt of my answer... But I finally continue to think that a player don't have to go first (if the barrel is spined once). First: for n=1 chamber, it is obvious - the clue. If we extend the n=1 case to n>1, we can feel what will happen: there is intrinsically the time in the game rule: you have to be the last. Then acting on this rule shall make a difference. Let's take n=2 chambers. The payer don't know where the bullet is. Then, from her point of view, she still have to let her opponent try the first one XD from the player point of view, the die is not cast, the fact that the barrel has been spined is virtual: the fact that the variable's drawing has been pre-made or not for each probability system is not relevant So, in short, when the player press the trigger, the die is casted once like in each probability system - but becareful, here, the die is changing after each try since the rule is that the barrel is spinned once=) Great great problem, I like it=) And so on for n>2.

FredBrach 15y ago
Ok no, I think now I'm wrong=) this is even, the die is cast once when the players choose to start or not and after, the game is made.
muyuu 15y ago
I'm pretty sure your first answer was wrong, but I don't get why would anyone downvote instead of correcting or giving a different answer.
- FredBrach 15y ago
  It seems that I'm victim of some kind of ennemy's downvotes