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# Russian roulette

## Question

Peter and John challenged each other to a Russian roulette.Each one of them got exactely the same" Six-shooter revolver",with only one bullet in the uper most firing chamber.Peter was right-handed,were as John was left-handed.Each one spins the cylinder exactely(17 turns).They directed the muzzle against the head of each other and began to shot,both together!

Which one will survive? why?

Edited by wolfgang

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Assuming, since one is left handed and the other right, that they spun the cylinder in opposite directions the one who spun the cylinder in the same direction as it normally rotates fires his bullet on the second shot, the one who spun it counter to how it rotates would fire on the sixth shot if he wasn't already dead.

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kbrdsk is correct, but we need some extra information before determining who of Peter and John survive, if either. If we assume that both players hold the gun in their main hand, and both spin the cylinder on the inside of the gun with their offhand with a downward motion, then John would survive.

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if neither spun the cylinder the bullets would both fire on the sixth shot.

Now, to me, spinning the cylinder a turn means a full turn to me putting it back in the same spot. So we have the same situation as the initial condition meaning that both die on the sixth shot.

Now lets assume that spinning the cylinder a turn means the next chamber is now on top. If they are both given the exact same revolver the cylinder would come out on the same side for both guys meaning that they could only spin it one way, pushing from the top down.

If the cylinder comes out on the right side of the gun then they spin it clockwise making the chamber one clockwise from the bullet on top meaning that both guns will fire on the first shot since the next cylinder is the 'active' one.

If the cylinder comes out on the left side of the gun then they spin it counterclockwise making the champer on counterclockwise from the bullet on top meaning that both guns will fire on the fifth shot.

So in all cases it is a lose/lose situation.

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if neither spun the cylinder the bullets would both fire on the sixth shot.

Now, to me, spinning the cylinder a turn means a full turn to me putting it back in the same spot. So we have the same situation as the initial condition meaning that both die on the sixth shot.

Now lets assume that spinning the cylinder a turn means the next chamber is now on top. If they are both given the exact same revolver the cylinder would come out on the same side for both guys meaning that they could only spin it one way, pushing from the top down.

If the cylinder comes out on the right side of the gun then they spin it clockwise making the chamber one clockwise from the bullet on top meaning that both guns will fire on the first shot since the next cylinder is the 'active' one.

If the cylinder comes out on the left side of the gun then they spin it counterclockwise making the champer on counterclockwise from the bullet on top meaning that both guns will fire on the fifth shot.

So in all cases it is a lose/lose situation.

Not if it's one of these: http://www.trackofthewolf.com/imgPart/aal-344_2.jpg

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