3 ms·
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).
by 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 agoOk 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 agoI'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 agoIt seems that I'm victim of some kind of ennemy's downvotes