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Prompted by this discussion on the math exchange

My thought was that the added mass of a bullet in an otherwise empty revolver would bias the chamber spin such that the bullet would remain in one of the lower chambers.

Obviously not something I'm going to try in practice (+:

Anyway I find myself wondering whether my thought carries any weight. (pun not intended)

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The much larger mass of the rotating cylinder than that of the bullet, together with the friction between the cylinder and chamber position mechanism, would make the downward bias negligible.

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I am not familiar with playing Russian roulette, neither with revolvers.

In Russian roulette, the chamber is revolved rapidly, but they do not wait for friction to stop the chamber from rotating, this is done manually. So at this stage how large is the friction force compared to the gravitational force due to assymetric distribution of bullets? In this case, I think this additional force is negligible, as the moment of inertia of the revolving part will be way larger than the bullets.

Furthermore, where on one side the bullet accelerates the chamber, it will decelerate at the other side, this will more or less average out, so there will be no preference due to this issue.

Suppose you don't interfere with the revolvers rotation, then when it is almost not rotating anymore, at some point the gravitational force of this bullet will not be negligible any longer. However, at this point there will be mechanical restrictions on the revolver that force the chamber in one of six positions, which will probably prevent this from happening.

Maybe these roulette players point the barrel at this stage in the air or to the ground?

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+1: Mechanical restrictions didn't occur to me –  Everyone Jan 6 '12 at 7:07
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