# In Russian roulette, does the “initial advantage” of player #2 count? [closed]

Regarding the recent question about which player it is better to be in a game of Russian Roulette (http://math.stackexchange.com/questions/96331/in-russian-roulette-is-it-best-to-go-first), where it was answered that neither of the players had the advantage as the overall odds where the same.

A game that can stop on the first move (and has the same overall odds for each player by the end) surely must provide an "initial advantage" for one of the players...

The game begins with player #1 pulling the trigger (1/6 chance dying) and player #2 waiting (0/6 chance dying).

Clearly, player #2 has the initial advantage (yet same overall odds) in this game.

Isn't it best to be player #2, since player #2 is guaranteed to live on the first move (which is player #1's move) and this game can stop right here on this move.

Rules of the game: 2 players, more than 2 chambers (win/lose or continue), one bullet, no shuffle between moves, player can't shoot other player, player can only lose when his turn produces the chamber with the bullet on trigger pull.

Initial Advantage: beginning advantage in game (somewhat independant of overall / last-move odds).

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This question was already answered in the question you reference. – Graphth Jan 8 '12 at 15:01
Think of it this way: (1) Spin the barrel, (2) If the bullet is in an odd location, Player 1 dies, if it is in an even location, Player 2 dies. Player 2 has no advantage. – cardinal Jan 8 '12 at 15:19
Imagine there are only two chambers (not 6), can you see the game is fair now? The fact that player 2 cannot die on the very beginning does not give him any real "advantage". – sdcvvc Jan 8 '12 at 15:41
Player 2 has a slightly larger life expectancy, but that's it. The chances to survive the game are equal. – fedja Jan 8 '12 at 16:29
@fedja: How do you compute the life expectancy? Would playing Russian roulette against a terminally ill patient give an advantage? – Raskolnikov Jan 8 '12 at 17:07

## closed as off topic by joriki, Raskolnikov, Fabian, Asaf Karagila, t.b.Jan 9 '12 at 0:18

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The answer to the question is No.

There is no advantage of being player #2.

In this game, I can guarantee you that as player #2, you will live passed move #1. But after move #1, I cannot guarantee either player anything... Except that the last move will kill player #2 if all previous moves are empty chambers.

Hence the advantages between player #1 and #2 are the same when the initial "to-live" advantage tactic is negated by the final "will-die" dis-advantage.

The "initial advantage" of player #2 does not count since it (that initial advantage) has nothing to do with the probabilistic odds of winning this game, but is rather just a way for player #2 to always live on move #1...

Being player #2 is a tactic for a goal that is something other than "to win" ... to simply live pass the first move. This tactic and the goal of this game are independent of each other and have no relationship.

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